IMFM
Institute of Mathematics, Physics and Mechanics Jadranska 19, 1000 Ljubljana, Slovenia
Preprint series Vol. 51 (2013), 1193
ISSN 2232-2094
CHARACTERIZING 2-CROSSING-CRITICAL GRAPHS
Drago Bokal Bogdan Oporowski R. Bruce Richter Gelasio Salazar
Ljubljana, December 13, 2013
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Characterizing 2-crossing-critical graphs
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Drago Bokal
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Bogdan Oporowski
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R. Bruce Richter
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Gelasio Salazar
Author address:
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University of Maribor, Maribor SLOVENIA E-mail address: drago.bokal@uni-mb.si
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Louisiana State University, Baton Rouge U.S.A. E-mail address: bogdan@math.lsu.edu
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University of Waterloo, Waterloo CANADA E-mail address: brichter@uwaterloo.ca
UA de San Luis Potosí, San Luis Potosí MEXICO E-mail address: gelasio.salazar@gmail.com
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	Contents	
List of Figures		v
Chapter	1. Introduction	1
Chapter	2. Description of 2-crossing-critical graphs with V10	5
Chapter	3. Moving into the projective plane	10
Chapter	4. Bridges	14
Chapter	5. Quads have BOD	17
Chapter	6. Green cycles	30
Chapter	7. Exposed spoke with	
	additional attachment not in Q0	42
Chapter	8. G embeds with all spokes in M	50
Chapter	9. Parallel edges	62
Chapter	10. Tidiness and global H-bridges	63
Chapter	11. Every rim edge has a colour	74
Chapter	12. Existence of a red edge and its structure	84
Chapter	13. The next red edge and the tile structure	99
Chapter	14. Graphs that are not 3-connected	120
14.1.	2-critical graphs that are not 2-connected	120
14.2.	2-connected 2-critical graphs that are not 3-connected	120
Chapter	15. On 3-connected graphs that are not peripherally-4-connected	127
15.1.	A 3-cut with two non-planar sides	127
15.2.	3-reducing to peripherally-4-connected graphs	129
15.3.	Planar 3-reductions	133
15.4.	Reducing to a basic 2-crossing-critical example	140
15.5.	Growing back from a given peripherally-4-connected graph	144
15.6.	Further reducing to internally-4-connected graphs	145
15.7.	The case of V8-free 2-crossing-critical graphs	150
Chapter	16. Finiteness of 3-connected 2-crossing-critical graphs with no V2n	156
16.1.	V2n-bridges are small	156
00 16.2. The number of bridges is bounded	161 i—l
Chapter 17. Summary	168
Bibliography	170
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Abstract
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It is very well-known that there are precisely two minimal non-planar graphs: K5 and K33 (degree 2 vertices being irrelevant in this context). In the language of crossing numbers, these are the only 1-crossing-critical graphs: they each have crossing number at least one, and every proper subgraph has crossing number less than one. In 1987, Kochol exhibited an infinite family of 3-connected, simple 2-crossing-critical graphs. In this work, we: (i) determine all the 3-connected 2-crossing-critical graphs that contain a subdivision of the Mobius Ladder Vi0; (ii) show how to obtain all the not 3-connected 2-crossing-critical graphs from the 3-connected ones; (iii) show that there are only finitely many 3-connected 2-crossing-critical graphs not containing a subdivision of Vi0; and (iv) determine all the 3-connected 2-crossing-critical graphs that do not contain a subdivision of Vg.
2010 Mathematics Subject Classification. Primary 05C10.
Key words and phrases. crossing number, crossing-critical graphs.
Bokal acknowledges the support of NSERC and U. Waterloo for 2006-2007, Slovenian Research Agency basic research projects L7-5459, J6-3600, J1-2043, L1-9338, J1-6150, research programme P1-0297, and an international research grant GReGAS. Richter acknowledges the support of NSERC. Salazar acknowledges the support of CONACYT Grant 106432.
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List of Figures
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2.1 2.2
2.3
2.4
3.1
3.2
3.3
5.1
5.2
5.3
6.1
7.1
7.2
7.3
7.4
8.1 8.2
The two frames. The thirteen pictures.
Each picture produces either two or four tiles. The different kinds of edges in the pictures.
The 3-connected, 2-crossing-critical graphs that do not embed in RP2. The 2-crossing-critical 3-representative embeddings in RP2. Standard labellings of the representativity 2 embeddings of Vi0.
The two possibilities for Di when j = i + 2. The two possibilities for D2. The two possibilities for D3.
The case e G ri+4 ri+5 for Qi being a (Qi U Mq.)-prebox. Only two of the three spokes are shown.
The subgraph K of G in RP2. The 1-drawing of K.
The 1-drawings D2[(K - («2)) U Po] and Ds[(K - («3)) U Po]. The 1-drawings D2[(K - («2)) U Po] and D3KK - («3)) U Po].
The two possibilities for D2. The two possibilities for D3.
11.1	The locations of e, f, we, wf, He, N, and Kf.
11.2	The spine and its constituent paths.
12.1	One of several examples of a A.
13.1	']
13.2	Definition ??.
14.1	The 2-crossing-critical graphs that are not 2-connected.
14.2	2-connected, not 3-connected, 2-crossing-critical graphs, 2 non-planar cleavage units
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00	14.3 2-connected, not 3-connected, 2-crossing-critical graphs, 3 cleavage units,
2 of which are non-planar.	123
15.1	The possible (T, U)-configurations.	146
15.2	The thick edge is a bear hug. The dotted edges tw and vz might be subdivided, and the dashed edge uw need not be present. If uw is not present, then {ux, uy} is a simultaneously deletable pair of bear hugs. 147
15.3	When s = b, Gi_i is a subgraph of the illustrated planar graph.	149
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CHAPTER 1
Introduction
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For a positive integer k, a graph G is k-crossing-critical if the crossing number cr(G) is at least k, but every proper subgraph H of G has cr(H) < k. In general, it is not true that a k-crossing-critical graph has crossing number exactly k. For example, any edge-transitive non-planar graph G satisfies cr(G — e) < cr(G), for any edge e of G, so every such graph is k-crossing-critical for any k satisfying cr(G — e) < k < cr(G). If G is the complete graph Kn, then cr(Kn) — cr(Kn — e) is of order n2, so Kn is k-crossing-critical for many different values of k.
Insertion and suppression of vertices of degree 2 do not affect the crossing number of a graph, and a k-crossing-critical graph has no vertices of degree 1 and no component that is a cycle. Thus, if G is a k-crossing-critical graph, the graph G' whose vertex set consists of the nodes of G (i.e., the vertices of degree different from 2) and whose edges are the branches of G (i.e., the maximal paths all of whose internal vertices have degree 2 in G) is also k-crossing-critical. Our interest is, therefore, in k-crossing-critical graphs with minimum degree at least 3.
By Kuratowski's Theorem, the only 1-crossing-critical graphs are K33 and K5. The classification of 2-crossing-critical graphs is currently not known. The earliest published remarks on this classification of which we are aware is by Bloom, Kennedy, and Quintas [7], where they exhibit 21 such graphs. Kochol [20] gives an infinite family of 3-connected, simple 2-crossing-critical graphs, answering a question of Siran [33] who gave, for each n > 3, an infinite family of 3-connected n-crossing-critical graphs. Richter [29] shows there are just eight cubic 2-crossing-critical graphs.
About 15 years ago, Oporowski gave several conference talks about showing that every large peripherally-4-connected, 2-crossing-critical graph has a very particular structure which was later denoted as 'being composed of tiles'. The method suggested was to show that if a peripherally-4-connected, 2-crossing-critical graph has a subdivision of a particular V2k (that is, k is fixed), then it has the desired structure and that only finitely many peripherally-4-connected, 2-crossing-critical graphs do not have a subdivision of V2k. (The graph V2n is obtained from a 2n-cycle by adding the n diagonals. Note that V4 is K4 and V6 is K3 3.)
Approximately 10 years ago, it was proved by Ding, Oporowski, Thomas, and Vertigan [13] that, for any k, a large (as a function of k) 3-connected, 2-crossing-critical graph necessarily has a subdivision of V2k. It remains to show that having the V2k-subdivision implies having the desired global structure. Their proof involves first showing a statement about non-planar graphs that is of significant independent interest: for every k, any large (as a function of k) "almost 4-connected" non-planar graph contains a subdivision of one of four non-planar graphs whose sizes grow with k. One of the four graphs is V2k. This theorem is then used for the crossing-critical application mentioned above.
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00	Tiles have come to be a very fruitful tool in the study of crossing-critical graphs.
Their fundamentals were laid out by Pinontoan and Richter [27], and later they turned out to be a key in Bokal's solution of Salazar's question regarding average degrees in crossing-critical graphs [8, 28, 31]. These results all rely on the ease of establishing the crossing number of a sufficiently large tiled graph, and they generated considerable interest in the reverse question: what is the true structure of crossing-critical graphs? How far from a tiled graph can a large crossing-critical graph be? Hlineny's result about bounded path-width of k-crossing-critical graphs [18] establishes a rough structure, but is it possible that, for small values of k, tiles would describe the structure completely? It turns out that, for k = 2, the answer is positive. A more detailed discussion of these and other matters relating to crossing numbers can be found in the survey by Richter and Salazar [30].
Our goal in this work, not quite achieved, is to classify all 2-crossing-critical graphs. The bulk of our effort is devoted to showing that if G is a 3-connected 2-crossing-critical graph that contains a subdivision of V1q, then G is one of a completely described infinite family of 3-connected 2-crossing-critical graphs. These graphs are all composed from 42 tiles. This takes up Chapters 3 - 13. This combines with [13] to prove that a "large" 3-connected 2-crossing-critical graph is a member of this infinite family.
The remainder of the classification would involve determining all 2-crossing-critical graphs that either are not 3-connected or are 3-connected and do not have a subdivision of V1o. In Chapter 14, we deal with the 2-crossing-critical graphs that are not 3-connected: they are either one of a small number of known particular examples, or they are 2-connected and easily obtained from 3-connected examples.
There remains the problem of determining the 3-connected 2-crossing-critical graphs that do not contain a subdivision of V10. In the first five sections of Chapter 00	15, we explain how to completely determine all the 3-connected 2-crossing-critical
graphs from peripherally-4-connected graphs that either have crossing number 1 or are themselves 2-crossing-critical. In the sixth and final subsection, we determine which peripherally-4-connected graphs do not contain a subdivision of V8 and either have crossing number 1 or are themselves 2-crossing-critical. Combining the two parts yields a definite (and practical) procedure for finding all the 3-connected 2-
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crossing-critical graphs that do not contain a subdivision of V8. This leaves open the problem of classifying those that contain a subdivision of V8 but do not have a subdivision of V10. In Sections 16.1 and 16.2, we show that there are only finitely many. (Although this follows from [11], the approach is different and it keeps our work self-contained.)
There is hope for a complete description. In her master's essay, Urrutia-Schroeder [36] begins the determination of precisely these graphs and finds 326 of them. Oporowski (personal communication) had previously determined 531 3-connected 2-crossing-critical graphs, of which 201 contain a subdivision of V8 but not of V10. Austin [3] improves on Urrutia-Schroeder's work, correcting a minor error (only 214 of Urrutia-Schroeder's graphs are actually 2-crossing-critical) and finding several others, for a total of 312 examples. Only 8 of Oporowski's examples are not among the 312. A few have been determined by us as stepping stones in our classification of those that have a subdivision of V10. We have hopes of completing the classification.
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00	The principal facts that we prove in this work are summarized in the following
statement.
Theorem 1.1 (Classification of 2-crossing-critical graphs). Let G be a 2-cros-sing-critical graph with minimum degree at least 3. Then either:
•	if G is 3-connected, then either G has a subdivision of V1o and a very particular tile structure or has at most 3 million vertices; or
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•	G is 2- but not 3-connected and is obtained from a 3-connected example by replacing digons by digonal paths.
We remark again that vertices of degree 2 are uninteresting in the context of crossing-criticality, so we assume all graphs have minimum degree at least 3.
Chapters 2-13 of this work contain the proof of the following, which is the main contribution of this work. (The formal definitions required for the statement given below are presented in Chapter 2.)
Theorem 1.2 (2-crossing-critical graphs with V10). Let G be a 3-connected, 2-crossing-critical graph containing a subdivision of V10. Then G is a twisted circular sequence (T1, T2,. .. ,Tn) of tiles, with each T coming from a set of 42 possibilities.
This is part of the first item in the statement of Theorem 1.1.
Chapter 14 is devoted to 2-crossing-critical graphs that are not 3-connected. (We remind the reader of Tutte's theory of cleavage units and introduce digonal paths in Chapter 14.) The results there are summarized in the following.
Theorem 1.3 (2-crossing-critical graphs with small cutsets). Let G be a 2-crossing-critical graph with minimum degree at least 3 that is not 3-connected.
(1)	If G is not 2-connected, then G is one of 13 graphs. (See Figure 14.1.)
(2)	If G is 2-connected and has two nonplanar cleavage units, then G is one of 36 graphs. (See Figures 14.2 and 14.3.)
(3)	If G is 2-connected with at most one nonplanar cleavage unit, then G has precisely one nonplanar cleavage unit and is obtained from a 3-connected, 2-crossing-critical graph by replacing pairs of parallel edges by digonal
S	paths.
Chapter 15 shows how to reduce the determination of 3-connected 2-crossing-critical graphs to "peripherally-4-connected" 2-crossing-critical graphs. A graph G is peripherally-4-connected if G is 3-connected and, for every 3-cut X in G, any partition of the components into nonnull subgraphs H and J has one of H and J being a single vertex. The main result here is the following.
Theorem 1.4. Every 3-connected, 2-crossing-critical graph is obtained from a peripherally-4-connected, 2-crossing-critical graph by replacing each degree 3 vertex
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with one of at most 20 different graphs, each having at most 6 vertices.
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We combine this with Robertson's characterization of V8-free graphs to explain how to determine all the 3-connected 2-crossing-critical graphs that do not have a subdivision of V8. This requires a further reduction to "internally 4-connected" graphs.
Chapter 16 shows that a 3-connected, 2-crossing-critical graph with a subdivision of V8 but no subdivision of V10 has at most three million vertices. The general result we prove there is the following.
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00	THEOREM 1.5. Suppose G is a 3-connected, 2-crossing-critical graph. Let n > 3
be such that G has a subdivision of V2n but not of V2(n+i). Then \V(G)| = O(n3).
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CHAPTER 2
Description of 2-crossing-critical graphs with V10
In this section, we describe the structure of the 2-crossing-critical graphs that contain Vi0. As mentioned in the introduction, they are composed of tiles. This concept was first formalized by Pinontoan and Richter [27, 28] who studied large sequences of equal tiles. Bokal [8] extended their results to sequences of arbitrary tiles, which are required in this section. In those results, "perfect" tiles were introduced to establish the crossing number of the constructed graphs. However, this property required a lower bound on the number of the tiles that is just slightly too restrictive to include all our graphs. As we are able to establish the lower bound on the crossing number of all these graphs in a different way (Theorem 5.5), we summarize the concepts of [8] without reference to "perfect" tiles. Where the reader feels we are imprecise, please refer to [8] for details.
Definition 2.1. (1) A tile is a triple T = (G, A, p), consisting of a graph G and two sequences A and p of distinct vertices of G, with no vertex of
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G appearing in both A and p.
(2)	A tile drawing is a drawing D of G in the unit square [0,1] x [0,1] for which the intersection of the boundary of the square with D[G] contains precisely the images of the vertices of the left wall A and the right wall p, and these are drawn in {0} x [0,1] and {1} x [0,1], respectively, such that the y-coordinates of the vertices are increasing with respect to their orders in the sequences A and p.
(3)	The tile crossing number tcr(T) of a tile T is the smallest number of crossings in a tile drawing of T.
(4)	The tile T is planar if tcr(T) = 0.
(5)	A k-drawing of a graph or a k-tile-drawing of a tile is a drawing or tile-drawing, respectively, with at most k crossings.
It is a central point for us that tiles may be "glued together" to form larger tiles. We formalize this as follows.
Definition 2.2. (1) The tiles T = (G, A, p) and T' = (G',A',p') are compatible if |p| = |A'|.
(2)	A sequence (T0, Ti,..., Tm) of tiles is compatible if, for each i = 1, 2,..., m, Tj_i is compatible with Tj.
(3)	The join of compatible tiles (G, A, p) and (G', A', p') is the tile (G, A, p) ( (G',A',p') whose graph is obtained from G and G' by identifying the sequence p term by term with the sequence A'; left wall is A; and right wall is p'.
(4)	As ( is associative, the join (T of a compatible sequence T = (T0, Ti,..., Tm) of tiles is well-defined as T0 ( Ti ( • • • ( Tm.
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or vertices of degree two. If we are interested in 3-connected graphs, we may suppress vertices of degree two, but we keep the multiple edges.
We have the following simple observation.
OBSERVATION 2.3. Let (To,Ti, .. . ,Tm) be a compatible sequence T of tiles. Then
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tcr((T) < ^ tcr(Tj).	■
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An important operation on tiles that we need converts a tile into a graph.
Definition 2.4. (1) A tile T is cyclically compatible if T is compatible with itself.
(2) For a cyclically-compatible tile T, the cyclization of T is the graph oT obtained by identifying the respective vertices of the left wall with the right wall. A cyclization of a cyclically-compatible sequence of tiles is defined as oT = o((T).
The following useful observation is easy to prove. Typically, we will apply this to the tile (T obtained from a compatible sequence T of tiles.
Lemma 2.5 ([8, 28]). Let T be a cyclically compatible tile. Then cr(oT) < tcr(T).	■
We now describe various operations that turn one tile into another.
Definition 2.6. (1) For a sequence w, w denotes the reversed sequence.
(2)	• The nght-mverted tile of a tile T = (G, X, p) is the tile T$ = (G, X, p); • the left-inverted tile is $T = (G, A, p);
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(3)	A tile T is k-degenerate if T is planar and, for every edge e of T, tcr(T$ - e) < k.
Note that our k-degenerate tiles are not necessarily perfect, as opposed to the definition in [8]. However, the following analogue of [8, Cor. 8] is still true.
Lemma 2.7. Let T = (To,.. ., Tm), m > 0, be a cyclically-compatible sequence of k-degenerate tiles. Then ((T) is a k-degenerate tile.
Proof. By Lemma 2.5, (T is planar. Let e be any edge of (T. Let Ti be
0,...,Ti_i,Ti$ - e, $Ti+1$
the tile of T containing e. Let T' = (T0,..., Ti-1,Ti$ - e, $Ti+1$,..., $Tm$), so
(T' = (T$ — e; in particular, they have the same tile crossing number. As Ti$ is k-degenerate, tcr(Ti$ — e) < k. Since all other tiles of T' are planar, Lemma 2.5 implies tcr((T$ - e) < tcr(Ti$ - e) < k.	□
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The following is an obvious corollary.
Corollary 2.8. Let T be a k-degenerate tile so that cr(o(T$)) > k. Then
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(T$)
is a k-crossing-critical graph.	■
Definition 2.9. (1) T is a compatible sequence (T0, T1,..., Tm), then: • the reversed sequence T" is the sequence (T", T"_ 1,..., T");
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•	the i-flip Tj is the sequence (T0,..., T^,
•	the i-shift T is the sequence (Tj,..., Tm, T0,
,Tm); and
-1, Ti+2 ...,Ti+i).
(2) Two sequences of tiles are equivalent if one can be obtained from the other by a series of shifts, flips, and reversals.
Note that the cyclizations of two equivalent sequences of tiles are the same graph.
Definition 2.10. The set S of tiles consists of those tiles obtained as combinations of two frames, illustrated in Figure 2.1, and 13 pictures, shown in Figure 2.2, in such a way, that a picture is inserted into a frame by identifying the two squares. A given picture may be inserted into a frame either with the given orientation or with a 180° rotation (some examples are given in Figure 2.3).
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Figure 2.1. The two frames.
Figure 2.2. The thirteen pictures.
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We remark that each picture produces either two or four tiles in S; see Figure
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Lemma 2.11. Let T be a tile in the set S. Then both T and ^T,^ are 2-degenerate.
Proof. Figure 2.4 shows that all the tiles are planar. The claim for T implies the result for ^Tj^, so it is enough to prove the result for an arbitrary T G S. Let e be an arbitrary edge of T. We consider cases, depending on whether e is either dotted, thin solid, thick solid, thin dashed, or thick dashed in Figure 2.4. Using this classification, we argue that tcr(T — e) < 2.
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Figure 2.3. Each picture produces either two or four tiles.
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Figure 2.4. The different kinds of edges in the pictures.
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If e is a dotted edge, then T — e has a wall with a single vertex and tcr(T^ — e) =
0.
If e is a thin solid edge, then there is a 1-tile-drawing of T^ with two dotted edges of T crossing each other.
If e is a thick solid edge, then there is a unique thin dashed edge f adjacent to e, and there exists a 1-tile-drawing of T^ — e with f crossing the dotted edge not on the same horizontal side of T as f.
If e is a thin dashed edge, then there is a unique thick dashed edge e' such that e and e' are in the same face of the exhibited planar drawing of T, as well as a unique dotted edge f, that is not in the same horizontal side of T as e. For such e and e', there exists a 1-tile-drawing of T^ — e with e' crossing f, as well as a
00	1-tile-drawing of T2 - e' with e crossing f. As each thick dashed edge corresponds
to at least one thin dashed edge, this concludes the proof.	□
We now define the set of graphs that is central to this work.
Definition 2.12. The set T(S) consists of all graphs of the form o(((T)2), where T is a sequence (T0, 2TT2, ..., 2T2m-1, T2m) so that m > 1 and, for each i = 0,1, 2,..., 2m, Ti G S.
The rim of an element of T(S) is the cycle R that consists of the top and bottom horizontal path in each frame (including the part that sticks out to either side) and, if there is a parallel pair in the frame, one of the two edges of the parallel pair.
00	The following is an immediate consequence of Lemmas 2.7 and 2.11.

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Corollary 2.13. Let G G T(S). For every edge e of G, cr(G - e) < 2.
In Theorem 5.5, we complete the proof that each graph G in T(S) is 2-crossing-critical by proving there that cr(G) > 2.
We are now able to state the central result of this work.
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Theorem 2.14. If G is a 3-connected 2-crossing-critical graph containing a subdivision of Vio, then G G T(S).
This theorem is proved in the course of Chapters 3 - 13. We remark that not every graph in T(S) contains a subdivision of Vlo.
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CHAPTER 3
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Moving into the projective plane
It turns out that considering the relation of a 2-crossing-critical graph to its embeddability in the projective plane is useful. This perspective was employed by Richter to determine all eight cubic 2-crossing-critical graphs [29]. It is a triviality that, if G has a 1-drawing, then G embeds in the projective plane (put the crosscap on the crossing). Therefore, any graph G that does not embed in the projective plane has crossing number at least 2. Moreover, Archdeacon [1, 2] proved that it contains one of the 103 graphs that do not embed in the projective plane but every proper subgraph does. Each obstruction for projective planar embedding has crossing number at least 2. Of these, only the ones in Figure 3.1 are 3-connected and 2-crossing-critical. (The non-projective planar graphs that are not 3-connected are found by different means in Section 14.) These are the ones labelled — left to right, top to bottom — D17, E20, E22, E23, E26, F4, F5, F10, F12, F13, and G1 in Glover, Huneke, and Wang [15].
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Figure 3.1. The 3-connected, 2-crossing-critical graphs that do not embed in RP2.
Definition 3.1. Let G be a graph embedded in a (compact, connected) surface £. Then:
(1)	the representativity rep(G) of G is the largest integer n so that every non-contractible, simple, closed curve in £ intersects G in at least n points (this parameter is undefined when £ is the sphere);
(2)	G is n-representative if n > r(G);
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(3) G is embedded with representativity n if rep(G) = n.
Representativity is also known as face-width and gained notoriety in the Graph Minors project of Robertson and Seymour. We only require very elementary aspects of this parameter; the reader is invited to consult [12] or [26] for further information on representativity and Graph Minors.
Barnette [4] and Vitray [37] independently proved that every 3-representative embedding in the projective plane topologically contains one of the 15 graphs ([37, Figure 2.2]). Vitray pointed out in a conference talk [38] that each of these 15 graphs has crossing number at least 2. Therefore, any graph that has a 3-representative embedding in the projective plane has crossing number at least 2. One immediate conclusion is that there are only finitely many 2-crossing-critical graphs that embed in RP2 and do not have a representativity at most 2 embedding in RP2, and, not only are there only finitely many of these, but they are all known and are shown in Figure 3.2. Vitray went on to show that the only 2-crossing-critical graph whose crossing number is not equal to 2 is C3nC3, whose crossing number is 3.
Figure 3.2. The 2-crossing-critical 3-representative embeddings in RP2.
Since every graph that has an embedding in the projective plane with represen-tativity at most 1 is planar, it remains to explore those 2-crossing-critical graphs that have an embedding in RP2 with representativity precisely 2. To cement some terminology and notation, we have the following.
Definition 3.2. Let n > 3 be an integer. The graph V2n is the Mobius ladder consisting of:
•	the rim R of V2n, which is a 2n-cycle (v0, v1, v2,..., v2n-1, v0); and,
•	for i = 0,1, 2, .. ., n — 1, the spoke VjVn+i.
Suppose V2n = H C G. (The notation L = H means that H is a subdivision of L. Thus, V2n = H C G means H is a subgraph of G and is also a subdivision of V2n.)
•	The H-nodes are the vertices of H corresponding to v0, v1, ..., v2n-1 in V2n; the H-nodes are also labelled v0, v1, ..., v2n-1.
•	For i = 0,1, 2,..., 2n — 1, the H-rim branch r is the path in H corresponding to the edge VjVi+1 of V2n.
•	For i = 0,1, 2,..., n — 1, the H-spoke is the path Sj in H corresponding to the edge VjV„+i in V2„.
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V2n.
Whenever we discuss elements of a subdivision H of the Mobius ladder V2n, we presume the indices are read appropriately. For the H-nodes vk and the H-rim branches rk, the index k is to be read modulo 2n. For the H-spokes s^, the index i is to be read modulo n. Thus, for example, s5+n = s5 and vs+2n = vs, while rg+n = rg.
Let G be a 2-crossing-critical graph embedded in RP2 with representativity 2. Let y be a simple closed curve in RP2 meeting G in precisely the two points a and b. We further assume V2n = H C G, with n > 3. Because G — a and G — b have 1-representative embeddings in the projective plane, they are both planar. We note that, for n > 3, V2n is not planar; therefore, a, b G H.
Remark 3.3. Throughout this work, we abuse notation slightly. If K is any graph and x is either a vertex or an edge of K, then we write x G K, rather than the technically correct x G V(K) or x G E(K). We have taken care so that, in any instance, the reader will never be in doubt about whether x is a vertex or an edge.
If n > 4, the deletion of a spoke of V2n leaves a non-planar subgraph; thus, when n > 4, we conclude a, b G R. If y does not cross R at a, say, then deleting the H-spoke incident with a (if there is one), and shifting y away from a leaves a subdivision of K3 3 in RP2 that meets the adjusted y only at b. But then this K3 3 has a 1-representative embedding in RP2, showing K3,3 is planar, a contradiction. Therefore, y must cross R at a and b. As any two non-contractible curves cross an
cm
odd number of times, R is contractible and so bounds a closed disc D and a closed Mobius strip M.
Let P and Q be the two ab-subpaths of R, let a = y H D and fl = y H M. (We alert the reader that the notations D, M, a, and y will be reserved for these objects.) Since each spoke is internally disjoint from y, the spoke is either contained in D or contained in M. Since the spokes interlace on R, at most one can be embedded in D.
Moreover, observe that a divides D into two regions, one bounded by PU a and the other bounded by Q U a. Thus, if a spoke — label it s0 — is embedded in D, then s0 has both attachments in just one of P and Q, say P. In this case, P contains either all the H-nodes v0, vi,..., vn or all the H-nodes vn, vn+i,..., v2n-i, v0. It follows that, for n > 4, there are only two (up to relabelling) representativity 2 embeddings of V2n in the projective plane. See Figure 3.3. We remark that it is possible that one or both of a and b might be an H-node.
We introduce a notation that will be used extensively in this work.
M3
Definition 3.4. The set of 3-connected, 2-crossing-critical graphs is denoted
m
It is a tedious (and unimportant) exercise to check the observation that none of the graphs in Mf found among the obstructions to having a representativity 2 embedding in RP2 has a subdivision of Vi0. We record it in the following assertion.
Theorem 3.5. Let G G M3 and Vi0 = H C G. Then G has a representativity 2 embedding in RP2.	I
We will also need information about 1-drawings of V2n, for n > 4. These are similarly straightforward facts that can be proved by considering K3 3's in V2n.
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LEMMA 3.6. Let n > 4 and let D be a 1-drawing of V^n- Then there is an i so that Ti crosses one of Ti+n_i, Ti+n, and Ti+n+i.	I
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CHAPTER 4
Bridges
The notion of a bridge of a subgraph of a graph is a valuable tool that allows us to organize many aspects of this work. This section is devoted to their definition and an elucidation of their properties that are relevant to us. Bridges are discussed at length in [35] and, under the name J-components, in [34].
Definition 4.1. Let G be a graph and let H be a subgraph of G.
(1)	For a set W of vertices of G, ||W|| consists of the subgraph of G with vertex set W and no edges.
(2)	An H-bridge in G is a subgraph B of G such that either B is an edge not in H, together with its ends, both of which are in H, or B is obtained from a component K of G - V(H) by adding to K all the edges from vertices in K to vertices in H, along with their ends in H.
(3)	For an H-bridge B in G, a vertex u of B is an attachment of B if u G V(H); att(B) denotes the set of attachments of B.
(4)	If B is an H-bridge, then the nucleus Nuc(B) of B is B - att(B).
(5)	For u, v G V(G), a uv-path P in G is H-avoiding if P n H C ||{u, v}||.
CM	(6) Let A and B be either subsets of V(G) or subgraphs of G. An AB-path is
00	a path with an end in each of A and B but otherwise disjoint from A U B.
CM	If, for example, A is the single vertex u, we write uB-path for {u}B-path.
We will be especially interested in the bridges of a cycle.
Definition 4.2. Let C be a cycle in a graph G and let B and B' be distinct C-bridges.
CO	(1) The residual arcs of B in C are the B-bridges in C U B; if B has at least
two attachments, then these are the maximal B-avoiding subpaths of C.
(2)	The C-bridges B and B' do not overlap if all the attachments of B are in the same residual arc of B'; otherwise, they overlap.
(3)	The overlap diagram OD(C) of C has as its vertices the C-bridges; two C-bridges are adjacent in OD(C) precisely when they overlap.
(4)	The cycle C has bipartite overlap diagram, denoted BOD, if OD(C) is bipartite; otherwise, C has non-bipartite overlap diagram, denoted NBOD.
CO	The following is easy to see and well-known.
Lemma 4.3. Let C be a cycle in a graph G. The distinct C-bridges B and B' overlap if and only if either:
(1)	there are attachments u, v of B and u', v' of B' so that the vertices u, u', v, v' are distinct and occur in this order in C (in which case B and B' are skew C-bridges); or
(2)	att(B) = att(B') and |att(B)| = 3 (in which case B and B' are 3-equivalent).	■
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work.
Definition 4.4. Let C be a cycle in a graph G and let B be a C-bridge. Then
B is a planar C-bridge if C U B is planar. Otherwise, B is a non-planar C-bridge.
Note that there is a difference between C U B being planar and, in some embedding of G in RP2, C U B being plane, that is, embedded in some closed disc in RP2. If C U B is plane, then B is planar, but the converse need not hold.
We now present the major embedding and drawing results that we shall use. The theorem is due to Tutte, while the corollary is the form that we shall frequently use.
Theorem 4.5. [35, Theorems XI.48 and XI.49] Let G be a graph.
(1)	G is planar if and only if either G is a forest or there is a cycle C of G having BOD and all C-bridges planar.
(2)	G is planar if and only if, for every cycle C of G, C has BOD.	M
For the corollary, we need the following important notion.
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Definition 4.6. Let H be a subgraph of a graph G and let D be a drawing of G in the plane. Then H is clean in D if no edge of H is crossed in D.
Corollary 4.7. Let G be a graph and let C be a cycle with BOD. If there is a C-bridge B so that every other C-bridge is planar and there is a 1-drawing of C U B in which C is clean, then cr(G) < 1.
I
Proof. Let x denote the crossing in a 1-drawing D of C U B in which C is clean. As C is not crossed in D, x is a crossing of two edges of B. Let Gx denote the graph obtained from G by deleting those two edges and adding a new vertex adjacent to the four ends of the deleted edges. Then C has BOD in Gx and every C-bridge in Gx is planar. By Theorem 4.5 (2), Gx is planar. Any planar embedding of Gx easily converts to a 1-drawing of G.	■
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We will also need the following result.
Lemma 4.8 (Ordering Lemma). Let G be a graph, C a cycle in G, B a set of non-overlapping C-bridges. Let P and Q be disjoint paths in C, with V(C) = V(P U Q). Suppose that each B G B has at least one attachment in each of P and Q. Let Pb and Qb be the minimal subpaths of P and Q, respectively, containing P fl B and Q f B, respectively. Then:
(1) the {Pb } and {Qb } are pairwise internally disjoint and there is an ordering
(Bi,...,Bfc)
ofB so that
P = Pb, ...Pb2 ■■■Pb, ■■■Pbfc and Q = Qb, ...Qb2 ■■■Qb,...Qbk;
and
(2) if, for each B, B' G B, att(B) = att(B'), the order is unique up to inversion.
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Then B and B' have attachments X1, X2, X1. x'2 in both components of P — e and attachments x,x' in Q. If |{x1, x\, x2, x', x, x'}| = 3, then they have 3 common attachments and so overlap, a contradiction. Otherwise, some y e {x1,x',x'} is not in {x1, x2, x}. Then y is in one residual arc A of x1, x2, x in C and not both of the other two of {x1,x',x'} are in A. So again B, B' overlap, a contradiction from which we conclude PB and PB> are internally disjoint.
Let C = P-1R1QR2. Suppose B,B' e B are such that P = ...PB ...PB, ... and Q = ... QB> ... QB .... We claim that either PB = PB> or QB = QB>. If not, then there is an attachment uP of one of B and B' in P that is not an attachment of the other and likewise an attachment uq of one of B and B' in Q that is not an attachment of the other. Note that uP and uq are not attachments of the same one of B and B', as otherwise the orderings in P and Q imply B and B' overlap.
For the sake of definiteness, we assume uP e att(B), so that uq e att(B'). Let wP e att(B') n P and let wq e att(B) n Q. The ordering of B and B' in P and Q imply that, in C, these vertices appear in the cyclic order wP,uP,uq,wq. Since uP,uq,wp,wq are all different, we conclude that B and B' overlap on C, a contradiction.
It follows that, by symmetry, we may assume PB = PB>. As PB and PB> are internally disjoint, they are just a vertex. So if P = ... PB ... PB> ... and Q = ... Qbi ... Qb ..., we may exchange PB and P'B, to see that P = ... PB> ... PB ... and Q = ... QBr ... QB ... . We conclude there is an ordering of B as claimed. Let (B1,...,Bk) and (Bn(1),..., Bn(k)) be distinct orderings so that P =
PB! ,...,PBk , P = PBn(1)	PBn(k) , Q = QB! ...QBk and Q = Q B n (1) , . . . , Q B „ (fc) .
There exist i < j so that n(i) > n(j). We may choose the labelling (P versus Q) so that the preceding argument implies that PBi = PBj = u. If QBi = QBj, then QB = QB = w and att(Bj) = att(Bj), which is (2). Therefore, we may assume there is an attachment y of one of Bi and Bj that is not an attachment of the other. Let z be an attachment of the other. Since Q is either (Q1, y, Q2, z, Q3) or
(Q-1, z, Q-1,y, Q-1), the only possibility is that n is the inversion (k,k — 1,..., 1). ■
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CHAPTER 5
Quads have BOD
There are two main results in this section. One is to show that each graph in the set T(S) is 2-crossing-critical and the other, rather more challenging and central to the characterization of 3-connected 2-crossing-critical graphs with a subdivision of V10, is to show that all H-quads and some H-hyperquads have BOD. We start with the definition of quads and hyperquads.
Definition 5.1. Let G be a graph and V10 = H C G.
(1)	For a path P and distinct vertices u and v in P, [uPv] denotes the uv-subpath of P, while [uPv) denotes [uPv] — v, (uPv] is [uPv] — u, and (uPv) is (uPv] — v.
(2)	When concatenating a uv-path P with a vw-path Q, we may write either PQ or [uPvQw]. If u = w and P and Q are internally disjoint, then both PQ and [uPvQu] are cycles. The reader may have to choose the appropriate direction of traversal of either P or Q in order to make the concatenation meaningful.
(3)	If L is a subgraph of G and P is a path in G, then L — (P) is obtained from L by deleting all the edges and interior vertices of P. (In particular, this includes the case P has length 1, in which case L — (P) is just L less
cm
one edge.)
(4)	For i = 0,1, 2, 3,4, the H-quad Qj is the cycle rj sj+1 rj+5 sj.
(5)	For i = 0,1, 2, 3,4, the H-hyperquad Qj is the cycle (Qj-1 U Qj) — (sj).
(6)	The Mobius bridge of Qj is the Qj-bridge Mqi in G such that H C Qj U

(7) The Möbius bridge of Qi is the Qrbridge Mq in G for which (H — (sj)) C
Qiu Mq .
The following notions will help our analysis.
Definition 5.2. Let G be a graph, V2n = H C G, n > 3, and let K be a subgraph of G. Then:
(1)	a claw is a subdivision of K13 with centre the vertex of degree 3 and talons the vertices of degree 1;
(2)	an {x, y, z}-claw is a claw with talons x, y, and z;
(3)	an open H-claw is the subgraph of H obtained from a claw in H consisting of the three H-branches incident with an H-node, which is the centre of
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the open H-claw, but with the three talons deleted;
(4)	K is H-close if K n H is contained either in a closed H-branch or in a open H-claw.
(5)	A cycle C in K is a K-prebox if, for each edge e of C, K — e is not planar.
The following is elementary but not trivial.
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00	Lemma 5.3. Let C bean H-close cycle, for some H = Vq. Then C is a (C U H )-
prebox.
Proof. For e G E(C), if e G H, then evidently (C U H) - e contains H, which is a V6; therefore (C U H) - e is not planar. So suppose e G H. Since C is H-close, C n H is contained in either a closed H-branch b or an open H-claw Y. There is an H-avoiding path P in C - e having ends in both components of either b - e or Y - e. In the former case, (H - e) U P, and hence (C U H) - e, contains a V6. In o	the latter case, (Y - e) U P contains a different claw that has the same talons as
Y, so again (H - e) U P, and (C U H) - e, contains a V6.	■
Lemma 5.4. Let K be a graph and C a cycle of K. If C is a K-prebox, then, CO	in any 1-drawing of K, C is clean.
Proof. Let D be a 1-drawing of K and let e be any edge of C. Since K - e is not planar, D(K - e) has a crossing. It must be the only crossing of D(K) and, therefore, e is not crossed in D(K).	■
We can now show that any of the tiled graphs described in Section 13 in fact have crossing number 2, thereby completing the proof that they are all 2-crossing-critical.
Theorem 5.5. If G g T (S), then G g Mf.
Proof. By Lemmas 2.7 and 2.11 and Corollary 2.8, we know that if K is a proper subgraph of G, then cr(K) < 1. Thus, it suffices to prove that cr(G) > 2.
There are two edges in a tile that are not in the corresponding picture and are not part of a parallel pair. An edge of G is a A-base if it is one of these edges. A A-cycle is a face-bounding cycle in the natural projective planar embedding of
cm
G containing precisely one A-base. Recall that the rim R of G is described in Definition 2.12.
There are at least three A-cycles contained in G and any two are totally disjoint. From each A-cycle we choose either of its RR-paths (by definition, these are R-avoiding) as a "spoke", and, with R as the rim, we find 8 different subdivisions of V6. There are two of these that are edge-disjoint on the spokes, so if D is a 1-drawing of G, the crossing must involve two edges of R.
Claim 1. If e is a rim edge in one of the 13 pictures, then e is in an H'-close cycle Ce, for some H' = V6 in G.
The point of this is that Lemmas 5.3 and 5.4 imply that Ce is clean in D. This is also obviously true for the other edges of the rim that are in digons. The conclusion is that we know the two crossing edges must be from among the A-bases. We shall show below that no two of these can cross in a 1-drawing of G, the desired contradiction.
Proof of Claim 1. Let e be in edge in the rim R of G that is in the picture T, let r be the component of T n R containing e, and let r' be the other component of T n R. There is a unique cycle in T - r' containing e; this is the cycle Ce. Let e' be the one of the two A-bases incident with T that has an end in r. Choose the RR-subpath of the e'-containing A-cycle that is disjoint from r. For any other two of the A-cycles, choose arbitrarily one of the RR-subpaths. These three "spokes",
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The proof is completed by showing that no two A-bases can cross in a 1-drawing of G. If there are at least five tiles, then it is easy to find a subdivision of V8 so that the two A-bases are on disjoint H-quads and therefore cannot be crossed in a 1-drawing of G. Thus, we may assume there are precisely three tiles and the crossing A-bases e1 and e2 are, therefore, in consecutive A-cycles.
Let T be the picture incident with both e1 and e2. Choose a subdivision H' of V6 containing R but so that T n H' = T n R. There is a unique 1-drawing D of H' with e1 and e2 being the crossing pair. For i = 1, 2, let the H'-branch containing ej be 6j. The end u of e4 that is in T is in the interior of b.
The vertices u1 and u2 are two of the four attachments of T in G. Let w1 and w2 be the other two, labelled so that w1 is in the same component of T n R as u2. It follows that w2 is in the same component of T n R as u1. In T, there is a unique pair of totally disjoint R-avoiding u1w1- and u2w2-paths P1 and P2, respectively. The crossing in D is of e1 with e2, so [u161w2] and [u262w1] are both not crossed in D. Therefore, D[P1] and D[P2] are both in the same face F of D.
Since the two paths P1 and P2 are totally disjoint(text deleted), D[P1] and D[P2] are disjoint arcs in F; the contradiction arises from the fact that their ends alternate in the boundary of F, showing there must be a second crossing.	■
One important by-product of cleanliness is that it frequently shows a cycle has BOD.
Lemma 5.6. Let C be a cycle in a graph G. Let D be a 1-drawing of G in which C is clean. If there is a non-planar C-bridge, then C has BOD and exactly one non-planar bridge.
Proof. Let B be a non-planar C-bridge. Then D[CU B] has a crossing, and, since C is clean in D, the crossing does not involve an edge of C. Therefore, it involves two edges of B. This is the only crossing of D, so inserting a vertex at this crossing turns D into a planar embedding of a graph Gx. As C is still a cycle of Gx, C
CO
has BOD in Gx and all C-bridges in Gx are planar. But ODGx (C) is the same as ODg(C) and all C-bridges other than B are the same in G and Gx.	■
We shall routinely make use of the following notions.
Definition 5.7. Let G be a connected graph and let H be a subgraph of G. Then:
(1)	H# is the subgraph of G induced by E(G) \ E(H); and
(2)	if G is embedded in
RP2, then
an H-face is a face of the induced embedding of H in RP2.
We will often use this when B is a C-bridge, for some cycle C in a graph G, in which case B# is the union of C and all C-bridges other than B. The following two lemmas are useful examples.
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Lemma 5.8. Let G be a graph embedded in RP2 with representativity 2 and let Y be a non-contractible curve in RP2 so that Gn7 = {a, b}. Let C be a contractible cycle in G and let B be a C-bridge so that Nuc(B)n{a, b} = 0. Then B# is planar.
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Proof. This is straightforward: B# = G — Nuc(B) C G — ({a, b} n Nuc B) and the latter has a representativity at most 1 embedding in RP2. Therefore it is planar.
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The following result, when combined with the (not yet proved) fact that H-
S	quads and some H-hyperquads have BOD, yields the fact, often used in the sections
to follow, that deleting some edge results in a 1-drawing in which a particular H-quad or H-hyperquad must be crossed.
LEMMA 5.9. Let G be a graph with cr(G) > 2 and let C be a cycle in G. If C has BOD in G, then, for any planar C-bridge B, C is crossed in any 1-drawing of B#.
Proof. Suppose there is a 1-drawing D of B# with C clean. Since C has BOD and G is not planar, there is a non-planar C-bridge B'. Because C is clean, any crossing in D[C U B'] involves two edges of B'. The only crossing in D involves two edges of B', so every other C-bridge in is planar. Since B is planar, it follows from Corollary 4.7 that cr(G) < 1, a contradiction.	■
We remark that Mq is a non-planar Q-bridge whenever Q is an H-quad or H -hyperquad.
_
Proof. Let B be a planar Qj-bridge. Because G is 2-crossing-critical, there is a 1-drawing D of B#. By Lemma 5.9, Qj is crossed in D. Note that H — (sj} C
Corollary 5.10. Let G G M3 and V10 = H C G. If the H-quad Q, and H-hyperquad Qj are disjoint, Qj has BOD, and there is a planar Qj-bridge B, then Q, has BOD and there is precisely one non-planar Q,-bridge.
B#. In any 1-drawing of H — (sj} in which Qj is crossed, the crossing is between rj-2 Urj_1 Urj Urj+1 and r„+j_2 Ur„+j_1 Ur„+j Urn+j+1. Since Q, is edge-disjoint from these crossing rim segments, Q, is clean in D.
The two graphs ODG(Qj) and ODB#(Q,) are isomorphic: the Q,-bridges in both G and B# are the same, except MQi in G becomes MQi — Nuc(B) in B# and they have the same attachments. Since Q, is clean in D, ODB# (Q,) is bipartite. Furthermore, the crossing in D is between two edges of Qj, so D shows that every Qj-bridge other than MQi is planar.	■
We next introduce boxes, which are cycles that, it turns out, cannot exist in a 2-crossing-critical graph G. On several occasions in the subsequent sections, we prove a result by showing that otherwise G has a box.
Definition 5.11. Let C be a cycle in a graph G. Then C is a box in G if C has BOD in G and there is a planar C-bridge B so that C is a B#-prebox.
Lemma 5.12. Let G G Mq- Then G has no box.
e	Proof. Suppose C is a box in G. Then C has BOD and there is a planar C-bridge
B so that C is a B#-prebox. As B# is a proper subgraph of G, there is a 1-drawing D of B#. By Lemma 5.4, D[C] is clean. This contradicts Lemma 5.9.	■
We can now determine the complete structure of a 2-connected H-close subgraph.
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00	Lemma 5.13. Let G G M3 and V2n = H C G with n > 4. If K is a 2-connected
H-close subgraph of G, then K is a cycle.
Proof. If KnH consists of at least two vertices, then we include in K the minimal connected subgraph of the H-branch or open H-claw containing K n H. Since K is H-close, there is a K-bridge MK in G so that H C K U MK. Let e be an edge of any H-spoke totally disjoint from K. Note that MK - e is a K-bridge in G - e and that MK has the same attachments in G as MK - e has in G - e.
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Since K is 2-connected, every edge of K is in an H-close cycle contained in K. Thus, for any 1-drawing D of G - e, Lemmas 5.3 and 5.4 imply that D[K] is clean. There is a face F of D[K] containing D[MK - e]. As D[K] is clean and K is 2-connected, F is bounded by a cycle C of K.
Lemma 5.3 implies the cycle C is a (C U H)-prebox. If K is not just C, then there is a C-bridge B contained on the side of D[C] disjoint from MK. Evidently B is a planar C-bridge.
Lemma 5.6 implies C has BOD. Since C is a (CUH)-prebox, C is a B#-prebox. We conclude that C is a box, contradicting Lemma 5.12. This shows that K = C.
o	■
The second of the following two corollaries is used several times later in this work. We recall from Definition 4.1 that, for a set W of vertices, ||W|| is the subgraph with vertex set W and no edges.
Corollary 5.14. Let G g M2, let V2n = H C G 'with n > 4, let B be an H-bridge.
(1)	If x, y G att(B) are such that ||{x,y}|| is H-close, then there is a unique H-avoiding xy-path in G.
(2)	There do not exist vertices x, y, z G att(B) so that ||{x,y, z}|| is H-close.
Proof. Suppose P1 and P2 are distinct H-avoiding xy-paths. There is either a closed H-branch or an open H-claw containing an xy-path; this subgraph of H contains a unique xy-path P. Then P U P1 U P2 is a 2-connected H-close subgraph of G and so, by Lemma 5.13, is a cycle. But it contains three distinct xy-paths, a contradiction.
For the second point, suppose by way of contradiction that such x,y, z exist. Let Y be an {x, y, z}-claw in B. There is a minimal connected subgraph Z of H contained either in a closed H-branch or in an open H-claw and containing x, y, and z. We note that Z is either a path or an {x, y, z}-claw. Thus, Y U Z is 2-connected and is H-close. It is a cycle by Lemma 5.13, but the centre of Y has degree 3 in Y U Z, a contradiction.	■
Corollary 5.15. Let G G M2, let Vlo = H C G, and let B be a Q-local H-bridge, for some H-quad Q. If s is an H-spoke and r is an H-rim branch, both contained in Q, then |att(B) n s| < 2 and |att(B) n (Q - [r])| < 2.
Proof. The first claim follows immediately from Corollary 5.14. For the second, suppose there are three such attachments x, y, and z. Corollary 5.14 implies they are not all in the other H-rim branch r' of Q, so at least one of x, y, and z is in the interior of some H-spoke of Q.
Suppose first that some H-spoke s in Q is such that (s) n {x, y, z} = 0. Then let H' = H - (s), let B' be the H'-bridge containing B, and let r' and s' be the
o
CM
00
m
u
a
CD U
00	two H-branches in Q other than r and s. Then x, y, and z are all attachments of
B' and they are all in the same open H'-claw containing (r' U s') — r, contradicting Corollary 5.14.
Otherwise, we may suppose both H-spokes s and s' in Q have one of x, y, and z in their interiors. We may suppose s has no other one of x, y and z. Choose the labelling so that x G (s). Let r' be the H-rim branch in Q other than r and again let H' = H — (s) and B' be the H'-bridge containing B. Then y and z are attachments of B', as is the H-node in s fl r'. But now these three attachments of
CD O
B' contradict Corollary 5.14.
We want to find cycles having BOD in our G G M3 that is embedded with representativity 2 in the projective plane. The following will be helpful.
Lemma 5.16. Let G be a graph embedded in RP2 and let C be a contractible cycle in G. Suppose B is a C-bridge so that C U B has no non-contractible cycles and let F be the C-face containing B. If B' is another C-bridge embedded in F, then B and B' do not overlap on C.
Proof. Let x and y be any distinct attachments of B and let P be a C-avoiding xy-path in B. Then C U P has three cycles, all contractible by hypothesis. We claim that one bounds a closed disc A so that C U P C A. If P is contained in the disc A bounded by C, then we are done. In the remaining case, let C' be one of these cycles containing P. If the closed disc A' bounded by C' contains C, then we are done. Otherwise, A H A' is a path in C and then A U A' is the desired closed disc.
As no other C-bridge in F can have attachments in the interiors of both the two xy-subpaths of C and, therefore, there is no C-bridge embedded in F that is skew (see Lemma 4.3 (1)) to B.
Likewise, if x, y, z are three distinct attachments of B, then there is a disc A' containing the union of C with a C-avoiding {x,y, z}-claw in B. This disc shows that no other C-bridge embedded in F can have all of x, y, z as attachments and, therefore, no C-bridge embedded in F is 3-equivalent (see Lemma 4.3 (2)) to B. ■
CO
The following is an immediate consequence of Lemma 5.16 and the fact that C has only two faces.
COROLLARY 5.17. Let G be a graph embedded in RP2 and let C be a cycle of G bounding a closed disc in RP2. If at most one C-bridge B is such that C U B contains a non-contractible cycle, then C has BOD and, for every other C-bridge B', C U B' is planar.	H
The following result is surprisingly useful in later sections.
Lemma 5.18. Let G G M3 and suppose G is embedded with representativity 2 in the projective plane. Let y be a non-contractible curve in the projective plane so that |yHG| = 2 and let C be a cycle of G so that yHC = 0. If there is a non-planar
CD
C-bridge B, then y H G C B, C has BOD, and, for every other C-bridge B', CU B' is planar.
Proof. Let a and b be the two points in y n G. We note that G — a and G — b are planar, as they have representativity 1 embeddings in RP2. Thus, if, for example, a ^ B, then C U B Ç G — a and so C U B is planar, a contradiction.
o
00
o
csr 00
$H
a
CD Jh
00	If B' is any other C-bridge, then a, b G C U B' and, therefore, C U B' is disjoint
from 7. Since any non-contractible cycle must intersect 7, C U B' has no non-
contractible cycles. The result is now an immediate consequence of Corollary 5.17. ■
Here is a simple result that we occasionally use.
Lemma 5.19. Suppose G G M3 and V2n = H C G, with n > 4. Let B be an H-bridge.
(1)	Then |att(B)| > 2.
(2)	If |att(B)| = 2, then B is isomorphic to K2.
(3)	If |att(B)| = 3, then B is isomorphic to K13.
Proof. Note that att(B) = Bf B# and G = BUB#. If |att(B)| < 1, then G is not 2-connected. If |att(B)| = 2 and Nuc(B) has a vertex, then G is not 3-connected.
Now suppose |att(B)| = 3 and B is not isomorphic to K13. Let Y be an att(B)-claw contained in B. As B# U Y is a proper subgraph of G, it has a 1-drawing D1; Y is clean in D1, as H must be self-crossed. On the other hand, if s is an H-spoke disjoint from B, there is a 1-drawing D2 of G — (s). Again, the crossing in D2 involves two edges of H — (s), so B is clean. We can substitute D2[B] for D1[Y] to convert D1 into a 1-drawing of G, a contradiction.	■
^
The following lemma is the last substantial one we need before proving that every H-quad has BOD.
Lemma 5.20. Let G be a ara.nh that is embedded in RP2
Lemma 5.20. Let G be a graph that is embedded in RP2 and let C be a cycle of G. Let B be a C-bridge so that Nuc(B) contains a non-contractible cycle. Then
C is contractible, C has BOD, and every C-bridge other than B is planar.
Proof. Let N be a non-contractible cycle in Nuc(B) and let B' be a C-bridge different from B. Then C U B' is disjoint from N. Since any two non-contractible cycles in RP2 intersect, C U B' does not contain a non-contractible cycle. Clearly this implies C is contractible and the remaining items are an immediate consequence
£
of Corollary 5.17.
CO
We prove below that every H-quad has BOD and that at least two hyperquads have BOD. A standard labelling of the embedded V10 will help make the details of the statement comprehensible. We have seen that, up to relabelling, there are two representativity 2 embeddings of V10 in RP2. There is a simple non-contractible curve 7 in RP2 meeting G in two points a and b. These are both in the rim R of H and either none or one of the H-spokes is outside the Mobius band M bounded by R. Let a and fl be the two ab-subarcs of 7, labelled so that fl C M.
Definition 5.21. Let G be a graph and let V10 = H C G. If G is embedded in RP2 so that one H-spoke is not in M, then H has an exposed spoke and the exposed spoke is the H-spoke not in M.
In this case, the standard labelling is chosen so that the exposed spoke is s0 and so that v0, v1, v2,v3, v4, v5 are all incident with one of the two faces of H U 7 incident with s0.
The faces of H U 7 are bounded by the cycles:
(1) [a,r9,v0] S0 [v5,r5,b, a, a];
o
CM
o
CO CD
Jh
a
CD U
(2)	ro r1 r3 rs r4 so;
(3)	[a, rg,vo] ro S1 [v6, r5,b, a];
(4)	Q1, Q2, Qs;
(5)	r4 [v5,r5,6,^, a, rg,vg] S4; and
(6)	[b,r5,v6] r6 r7 rg [vg, rg, a, a, b].
This case is illustrated in the diagram to the left in Figure 3.3. In the case all the H-spokes are in M, the labelling of H may be chosen so that the faces of H U 7 are bounded by:
(1)	[a, rg, vo]ro n r2 rs [v4, r4, b, a, a];
(2)	[a, rg, vo, so, v5, r4, b, a];
(3)	Qo, Q1, Q2, Qs;
(4)	[v4, r4, b, a, rg, vg, S4, v4]; and
(5)	[b,r4,v5] r5 r6 r7 rg [vg, rg, a, a, b].
This case is illustrated in the diagram to the right in Figure 3.3. We need one more technical lemma before the main result of this section.
Lemma 5.22. Let G G M\, let V1o ^ H C G, and let i, j G {0,1, 2, 3,4} be such that Qj and Qj have precisely one H-spoke in common. If Qj has BOD and sj is in a planar Qj-bridge, then (Mq )# is planar.
Proof. Let e be any edge of sj and let D be a 1-drawing of G — e. By Lemma 5.9, Qj is crossed in D. Thus, the crossing of D involves an edge of Mq , showing that
(Mq )# is planar.	■
CM
The following is the main result of this section.
cm	g	3
Theorem 5.23. Let G G M2 and V1o = H C G. Let G be embedded with representativity 2 in the projective plane, with the standard labelling. Then:
(1)	each H-quad Q of G has BOD and exactly one non-planar bridge;
(2)	Q2 has BOD;
(3)	for each i G {0, 1, 3, 4}, (Mq )# is planar;
CO	(4) if there is an exposed spoke, then Q3 has BOD;
CO	(5) if there is no exposed spoke, then at least one of Q1 and Q3 has BOD.
(6) if there is no exposed spoke and Q1 does not have BOD, then there is a Q1 -bridge B different from Mq so that B C D and either:
(a)	a = vo and B has an attachment at a, an attachment in r5 r6, and att(B) C {a} U r5 r6; or
(b)	b = v5 and B has an attachment at b, an attachment in ro r1, and att(B) C {b} U ro r1. (The analogous statement holds for Q3 in place
of Q1.)
The following definitions will be useful throughout the remainder of this work.
Definition 5.24. Let G be a graph embedded in RP2 and let C be a cycle of G bounding a closed disc A in RP2. A C-bridge B is C-interior if B is contained in A and C-exterior otherwise.
Proof of Theorem 5.23. We distinguish two cases. Case 1: H has an exposed spoke.
o
00
o
csr
i
csr 00 csr csr
CD
$H
CD
00	We adopt the standard labelling, so so is the exposed spoke. We note that Q2
is disjoint from Gny and, therefore, Lemma 5.18 implies Q2 has BOD and precisely one non-planar bridge, which is part of (1).
The arguments for Q1,Q3,Q2,Q3 are all analogous and so we do Q2. Since s0 is exposed, the cycle [a, rg,v0] s0 r4 s4 [vg,rg,a] is not contractible and is disjoint from Q2. Lemma 5.20 shows Q2 has BOD and precisely one non-planar bridge, proving (2) and (4). We have also proved (3) for j = 3 and (1) for Q1 and Q3.
To complete the proof of (1) in Case 1, it remains to deal with Q0 and Q4. These two cases are symmetric and so it suffices to prove Q0 has BOD and only one non-planar bridge. We note that Q3 is completely disjoint from Q0 and we have shown that Q3 has BOD. Let B be the Q3-bridge containing s3. As Q3 is contractible and B is Q3-interior, we conclude that B is planar. Therefore, Corollary 5.10 implies Q0 has BOD, and each Q0-bridge except Mq0 is planar, as required for (1).
For (3), it remains to prove that, for j e {0,1,4}, (M— )# is planar. We apply Lemma 5.22: for j = 0 or 4, we take i = 2; for j = 1, we take i = 3. In all cases, the result follows.
Case 2: H has no exposed spoke.
Lemma 5.18 shows Q1, Q2, and Q2 all have BOD and just one non-planar bridge. This proves (2) and part of (1). We use this in Corollary 5.10 to see that Q4 has BOD and just one non-planar bridge, another part of (1). Also, taking i = 2 and j e {0, 4} in Lemma 5.22, we see that (M— )# is planar, part of (3).
If Q3 has BOD, then Corollary 5.10 implies Q0 has BOD, so in order to show Q0 has BOD, we may assume Q3 has NBOD. There is an analogous situation for Q3 and Q1. We first prove (6) for Q3; we will use this to prove both Q0 has BOD and (5).
If v4 = b and v9 = a, then Lemma 5.18 shows that Q3 has BOD and exactly one non-planar bridge. So suppose either (or both) v4 = b or v9 = a. If every Q3-bridge other than Mq has only contractible cycles, then Q3 has BOD by Corollary 5.17. Thus, some Q3-bridge B other than M— is such that Q3 U B contains a CO	non-contractible cycle. Evidently, B is Q3-exterior. If B C M, then again Q3 U B
CO	has only contractible cycles. Thus, B C D.
Any Q3-exterior bridge B contained in the face of H U 7 bounded by
[a, rg, v0]r1 r2 r3 [v4, r4, b, a, a]
has all its attachments in {a} U r2 r3. Note that B is planar; moreover, if a is not an attachment, then Q3 U B has no non-contractible cycle and, therefore, does not overlap any other Q3-exterior bridge. We have the analogous conclusions if B is contained in the face of H U 7 bounded by [b, r5, v6]r6 r7 r8[vg, rg, a, a, b].
We conclude that either B has a as an attachment and also has an attachment in r2 r3 or, symmetrically, B has b as an attachment and also has an attachment in r7 r8. This proves (6).
We now prove (5). If {v0,v5} n {a, b} = 0, then Q1 has BOD and just one non-planar bridge; likewise if {v4, vg} n {a, b} = 0, then Q3 has BOD and just one non-planar bridge. Up to symmetry, the only other possibility is that v0 = a and v4 = b.	_
Now suppose that Q1 also has NBOD. Then (6) implies that there must be, up to symmetry, a Q1-bridge B1 different from M—^ having attachments at a and in
00 1-H
o
CM 00
r5 r6. Likewise, there is an H-bridge B3 different from Mq having attachments at b and in r7 r8. As B1 cannot have an attachment at b, B1 = B3. Considering the embedding of G in RP2, we see that both B1 and B3 must be embedded in the face of HUy incident with [b, r4, v5]r5 r6 r7 r8[vg, r9, a, a, b]. If B1, say, has an attachment other than a and v7, then the H-avoiding path in B3 from b to any attachment in r7 r8 crosses B1, a contradiction. So att(B1) = {a, v7}, att(B3) = {b, v7}, and, by Lemma 5.19, both B1 and B2 are just edges.
ONow recall that Q2 has BOD and, letting B2 be the Q2-bridge containing s2, - #
Lemma 5.9 implies Q2 is crossed in a 1-drawing D of B#. The crossing must be between the paths r0 r1 r2 r3 and r5 r6 r7 r8.
There are two maximal uncrossed subpaths of R in D and we know that v0 and v9 are on one uncrossed segment, say S1, of R, while v4 and v5 are on S2. Suppose first that v7 is on S1. Then the cycle [v0,B1,v7] r6 r5 r4 s4 r0 separates v8 from v3 in D, yielding the contradiction that s3 is crossed in D. On the other hand, if v7 is on S2 , then the same cycle separates v6 from v1, yielding the contradiction that s1 is crossed in D.
We conclude that not both Q1 and Q3 can have NBOD which is (5). By symmetry, we may assume Q1 has BOD. Then Lemma 5.22 shows (Mq )# is planar. Furthermore, Corollary 5.10 implies Q3 has BOD and precisely one non-planar bridge.
What remains is to prove that Q0 has BOD and precisely one non-planar bridge and that there is precisely one non-planar Q1-bridge. Recall that symmetry implies this will show the same things for Q3 and Q3, completing the proofs of (1) and (3).
CM
From (6), we may assume that vg = a and that there is a Q3-bridge B3 attaching at a and in r2 r3. Let w be any attachment of B3 in r2 r3, let P be an H-avoiding vgw-path in B3, and let Q be the subpath of r2 r3 joining w to v4. Then the cycle [vg,P,w,Q,v4,s4, vg] is non-contractible in RP2 and is disjoint from Q0. By Lemma 5.20, Q0 has BOD and has just one non-planar bridge.
As for Q1, we consider two cases. If Q3 has BOD, then Lemma 5.22 implies (Mq )# is planar. If Q3 has NBOD, then (6) implies either vg = a or v4 = b. In both cases, nuc(Mq ) n {a, b} = 0, so Lemma 5.8 implies (Mq )# is planar, as required.	■
HH
The following technical corollary of Theorem 5.23 and Lemmas 5.6 and 5.9 will be used in a few different places later.
Corollary 5.25. Let G G M2 and V10 = H C G. With indices read modulo 5, suppose, i G {0,1, 2, 3, 4} is such that Qi has BOD and, where {j, k} = {i + 2, i + 3}, suppose further that Qj has NBOD. Then si is in a planar Q¿-bridge Bi and Qk has BOD. Moreover, if ei is any edge of Bi and Di is a 1-drawing of G - ei, then either ri-1 ri crosses whichever of ri+3 and ri+6 is in Qj or ri+4 ri+5 crosses whichever of ri-2 and ri+1 is in Qj.
The two possibilities for Di in the case j = i + 2 are illustrated in Figure 5.1.
W
Proof of Corollary 5.25. By way of contradiction, suppose si is not in a planar Qi-bridge. We observe that s0 must be exposed, as otherwise we have the contradiction that, for every i G {0,1, 2, 3, 4}, s£ is in a planar Q^-bridge. It follows that, for i G {2, 3}, s£ is in a planar Q^-bridge. Thus, i G {2, 3}.
00
o
I
cm 00 cm cm
o
00
u
CD
CD O CD
00
Vi+3
Vi+8
Vi+2
Vi+4 Vi+7
Vi+9
Figure 5.1. The two possibilities for Dj when j = i + 2.
0 o
1
00
£ CO CO
Let i € {2, 3} be such that i and i are not consecutive in the cyclic order (0,1, 2, 3, 4). Let e^ be the edge of s^ incident with v^ and let D^ be a 1-drawing of G — e^. By Lemma 5.9, Q£ is crossed in D^.
If Q^ is self-crossed in D^, then D^ shows that the Qrbridge containing sj is planar. Thus, we have that Q£ is not self-crossed in D^. One of S£_1 and S£+1 is exposed in D^. If this exposed spoke is not also in Qj, then again sj is in a planar Qj-bridge; therefore, we must have that the exposed spoke is in Qj. For the sake of definiteness, we assume that S£_1 is exposed, which implies that i = i + 2.
As the only non-planar Qj-bridge is Mq , we must have an H-avoiding path P from the interior of sj to the interior of one of r^ and r^+5 r^+6. The drawing D^ restricts the possibility to the interior of one of r^ and r^+4 r^+5. But now the embedding in RP2 implies i = 0. This implies j € {2, 3}; however, neither Q2 nor Q3 has NBOD. Therefore, sj is in a planar Qj-bridge.
Because Mq — ej and Mq have the same attachments, ODG_ei (Qj) and
ODG(Qj) are isomorphic. As the latter is not bipartite, neither is the former. By Lemma 5.6, Qj is not clean in Dj. Thus, either rj-1 rj or rj+4 rj+5 is crossed in D2. These are edge-disjoint from Qj.
Lemma 5.9 implies that Qj is also crossed in Dj. Since Qj is crossed and, from the preceding paragraph, something outside of Qj is crossed, either
rj_1 rj crosses
rj+3 U rj+6
rj+4 rj+5 crosses rj_2 U rj+1
m
CD
$H
CD
m
u
a
CD U
as required.
Since Q2 always has BOD, Corollary 5.25 implies at least one of Q0 and Q4 has BOD. Together with the fact that, in all cases, at least one of Q1 and Q3 has BOD, we conclude that at least three of the H-hyperquads have BOD.
The last result in this section will be useful early in the next section.
COROLLARY 5.26. Let G € M3 and let V10 = H C G and suppose G has a representativity 2 embedding in the projective plane, with the standard labelling. Suppose, for some i, B is an H-bridge having an attachment in both (rj_1 Sj_1) and (r„+j Sj+1).
or
o
00
u
CD
CD O CD
00
(1)	if i = 0, then B Ç D.
(2)	// i = 0, then either Q3 has NBOD or B consists only of the edge v^vg.
Proof. For (1), we may assume B Ç M. The two representativity 2 embeddings of V10 in RP2 show that B can only be embedded in a face bounded by either [a, rg, v0]ri si [v6, b, a] or [b, a, rg, vg] s4 r4 [v5, r5, b] and that s0 is necessarily exposed in RP2. Notice that i = 0 in both cases, proving (1).
Now assume i = 0 and suppose Q3 has BOD. From Theorem 5.23, we know that Q2 also has BOD. For j G {2, 3}, let ej be the edge of sj incident with vj and let Dj be a 1-drawing of G — ej. Because sj is in a Qj-interior bridge, from Lemma 5.9, we know that Qj is crossed in Dj.
If Q0 is clean in Dj, then no face of Dj is incident with vertices in both (rg s4) and (s 1 r5). Therefore, Dj[B] cannot be crossing-free in Dj, a contradiction. Thus, Q0 is crossed in Dj. The two possibilities for D2 are shown in Figure 5.2, while the two possibilities for D3 are shown in Figure 5.3.
0 o
1
00 £
CO CO
v4
vg
Figure 5.2. The two possibilities for D2.
Let P be an H-avoiding path in G joining a vertex in each of (rg s4) and (si r5). The left-hand version of D2 has no face incident with both these paths, and so we must have the right-hand version of D2. Thus, D2 implies P has one end in (v0,rg, vg] and one end in (v1, s1, v6]. The right-hand version of D3 has no face incident with these paths, so it must be the left-hand version of D3. The only possibility there for the ends of P are and vg, as claimed.	■
CO CD
$H
CD CO
u
a
CD U
CO
1-H
o
cm
CO
u
cd
i
cm CO cm
m
CHAPTER 6
Green cycles
In this section, we begin our study of the rim edges of H. Ultimately, we will partition them into three types: "green", "yellow", and "red", and it will be the red ones that we focus on to find the desired tile structure. In this section, however, we begin with the study of green edges. We shall show that the cycles C we label green and yellow cannot be crossed in any 1-drawing of H U C.
Definition 6.1. An edge e of a non-planar graph G is red in G if G — e is planar.
o
We will eventually prove that every edge of R is either in a green cycle, or in a yellow cycle, or red. The main result in this section, one of the three main steps of the entire proof, is that no edge of R is in two green cycles.
Definition 6.2. Suppose G is a graph and V10 = H C G. Suppose further that G is embedded in RP2 with representativity 2 and that M is the Mobius band bounded by the H-rim R.
(1) A cycle C in G is H-green if C is the composition P1P2P3P4 of four paths, such that:
(a)	Pi C R and Pi has length at least 1;
(b)	P2P3P4 is R-avoiding;
(c) P2 U P4 Ç H
£ CO CO
(d)	P3 is H-avoiding (and, therefore, is either trivial or contained in an H-bridge); and
(e)	either
(i)	Pi contains at most 3 H-nodes or
(ii)	Pi is exceptional, that is, for some i G {0,1, 2,..., 9} and indices read modulo 10,
Pi = r ri+i ri+2 .
(2)	An edge of R is H-green if it is in an H-green cycle.
(3)	A vertex v of R is H-green if both edges of R incident with v are in the same H-green cycle.
m
There is a natural symmetry between P2 and P4: if C is an H-green cycle, consisting of the composition P1P2P3P4 as in Definition 6.2, then Pr1P4-1P3-1P2-1
_1
is another H-green cycle. Thus P4 and P2 can both be considered to be P2. As
cd
the orientations of the individual Pj will not be of any importance (except in as much as they are required to make C a cycle), we may say P2 and P4 are symmetric. Note that the exceptional case 1(e)ii is the only one in which P1 has 4 H-nodes.
LEMMA 6.3. Suppose G is a graph and V10 = H Ç G. Let C be any H-green cycle expressed as the composition P1P2P3P4 as in Definition 6.2.
u
a
30
$H
o
CM
(1)	If i G {2, 4}, then Pi has an end in R and is either trivial or contained in an H-spoke.
(2)	The path P3 is not trivial.
(3)	If P2 and P4 are both non-trivial, then they are contained in different H-spokes.
Proof. (1) For sake of definiteness, we assume i = 2. If P2 is not trivial, then there is an edge e in P2. From the definition, e is in H but not in R. Therefore, there is a spoke s containing e. If P2 has a vertex u not in s, then P2 is a path contained in H and containing e and u. This implies that one end of s, a vertex of R, is internal to P2, contradicting the fact that P2P3P4 is R-avoiding. So P2 C s, as required. Since P1 C R and P2 has an end in common with P1, P2 has an end in R.
(2)	Suppose P3 is trivial. Then P2P4 is an R-avoiding path joining the ends of P1. Each of P2 and P4 is either trivial or in a spoke and, since P2P4 is R-avoiding, either both are trivial or P2P4 is contained in a single spoke. If both are trivial, then P1 is the cycle P^1P2P3P4, which is impossible, since P1 is properly contained in the cycle R. Each of P2 and P4 has an end in R (or is trivial) and P2P4 has both ends in common with P1, so P2P4 is the entire spoke. But then P1 contains six H-nodes, a contradiction.
(3)	For j = 2, 4, Pj is non-trivial by hypothesis. Therefore, (1) shows it is contained in an H-spoke s. As it has a vertex in common with P1, Pj has a vertex in R. This vertex is an H-node incident with s. If P2 and P4 are contained in the same spoke s, then, as in the proof of (2), they contain different H-nodes. But then P1 contains six H-nodes, contradicting Definition 6.2.	■
There is a small technical point that must be dealt with before we can successfully analyze the relation of an H-green cycle to the embedding of G in RP2.
CM
Definition 6.4. Let n be a representativity 2 embedding of a graph G in RP2 and let V1o = H C G. Then n is H-friendly if, for each H-green cycle C of G and any non-contractible simple closed curve 7 in RP2 meeting n(G) in precisely two points, n[C] is contained in the closure of some face of n[H] U 7.
Lemma 6.5. Suppose G G M3 and V1o = H C G. Let n be any representativity 2 embedding of G in RP2, let 7 be a non-contractible simple closed curve in RP2 meeting n(G) in precisely two points, and let C be an H-green cycle in G. Give H the standard labelling relative to 7.
(1)	Either n[C] is contained in the closure of some face of n[H] U 7 or vgvg is an edge of G embedded in M and C = r6 r7 rs [vg, v6vg, v6]. In particular, if n[H] C M, then n is H-friendly.
(2)	If n is not H-friendly, then there is an H-friendly embedding of G in RP2 obtained from n by reembedding only v^vg.
(3)	In particular, there is an H-friendly embedding of G in RP2.
Proof. Suppose n[C] is not contained in the closure of any face of n[H] U7 and let P1P2P3P4 be the decomposition of C as in Definition 6.2. As P3 is (HU7)-avoiding and non-trivial by Lemma 6.3 (2), there is an (H U 7)-face F3 containing P3. Note that, if P2 is not trivial, then Lemma 6.3 (1) asserts it is contained in an H-spoke s and it contains an end of P3, so P2 is contained in the boundary of F3. Likewise for P4. We assume by way of contradiction that P1 C cl(F3).
00 1—1
o
Claim 1. Then:
i—l
(1)	P1 = r6 rr rg;
(2)	s0 is exposed;
CD
CO
O
u
a
CD U
(3)	either a = vg or b = v6; and
(4)	if F3 C D, then both v6 = b and vg = a.
Proof. We first consider the case F3 C D. Both ends of P1 are contained in one of the ab-subpaths of R. If P1 is not contained in the boundary of F3, then it must contain the other complete ab-subpath of R. As each of these has at least 4 H-nodes, the only possibility is that it is precisely 4 H-nodes. In this case, P1 must be exceptional and s0 must be exposed. In particular, P1 = r6 rr r8 and P3 has ends v6 and vg. The paths P2 and P4 are both trivial. Moreover, as P1 is not incident with F3, we must have v6 = b and vg = a.
In the other case, F3 C M. If F3 is contained in the interior of an H-quad, then P1 joins two vertices in the same quad and is not contained in the quad. In this case, P1 must have at least 5 H-nodes, which is impossible. Therefore, F3 is not contained in the interior of an H-quad, and so is bounded by one of [a,rg,v0]r0 S1 [v6, r5, b, fl, a] and [a, rg,vg] S4 r4[v5, r5,b, fl, a]. (Recall fl = 7 fl M.) Notice that s0 is exposed.
These cases are symmetric; for sake of definiteness, we presume F3 is bounded by [a, rg, v0]r0 s1 [v6, r5, b, fl, a]. The path P1 has at most 4 H-nodes and joins two vertices on Q0. If P1 C Q0, then n[C] is contained in the closure of one of the two (n[H] U 7)-faces whose boundary is contained in n[Q0] U 7; thus, p C Q0. Therefore, P1 has at least 4 H-nodes; by definition it has at most 4, so P1 has precisely 4 H-nodes. In particular, P1 can only be r6 rr r8 and vg = a.	□
CO	- -
Because s0 is exposed, Theorem 5.23 implies that both Q2 and Q3 have BOD. Let e be any edge in s2 and let D2 be a 1-drawing of G — e. Since Q2 has BOD, Lemma 5.9 shows Q2 is crossed in D2, so r0 r1 r2 r3 crosses r5 r6 rr rg. This implies that neither s0 nor s4 is exposed in D2 and, therefore, P3 cannot be in the same (H — (s0))-bridge as S0.
Let B0 and B be the (H — (s0))-bridges containing s0 and P3, respectively. These evidently overlap on Q0 and they both overlap Mq — e (in G — e). Therefore, Q0 has NBOD. Since Mq —e is a non-planar Q0-bridge in G — e, Lemma 5.6 implies that Q0 is not clean in D2.
As Q0 and Q2 have only s1 in common and both are crossed in D2, s1 must be exposed in D2. It follows that D2[P3] is in the face of D2[H — (s2)] bounded by S1 r6 rr rg rg r0.
The same arguments apply with Q3 in place of Q2, showing that D3[P3] is in the face of D3[H — (s3)] bounded by s4 r4 r5 r6 rr rg. These two drawings imply that att(B) C r6 rr r8.
If F3 C D, then F3 is bounded by rg s0 r5 [v6, a, vg] (recall a = 7 fl D). Thus, att(B) = {v6,vg} and Lemma 5.19 implies that P3 is just the edge v6vg. In this case, Claim 1 implies P3 can obviously be embedded in the other face of H U 7 contained in D and incident with both v6 and vg.
If F3 C M, then F3 is bounded by either
[a, rg,v0]r0 S1 [v6, r5, b, fl, a] or [a,rg,vg] S4 r4 [v5, r5, b, fl, a]
o
CM
CD O
00
s
CO CO
C/3 CD
00	Again, this implies that att(B) C {v6, vg}, so P3 is just the edge v6vg. In this case,
Claim 1 implies only that either v6 = b or vg = a. Again these cases are symmetric, so we assume vg = a.
We remark that if v G A n B, then (v) is an AB-path and this is the only path containing v that is an AB-path. We now return to the proof.
We wish to reembed v6vg in the (H U 7)-face incident with v6, v7, vg, and vg. We need only verify that there is no H-avoiding [b,r5,v6) (v6, r6, v7, r7, vg, rg, vg]-path. But such a path would have to appear in D3, where it can only also be in the face of D3[H — (s3)] bounded by s4 r4 r5 r6 r7 rg. But then it crosses v6vg in D3, a contradiction completing the proof.	■
We are now prepared for our analysis of H-green cycles.
LEMMA 6.6. Let G G M2, V10 = H ç G, and let n be an H-friendly embedding of G in RP2. Let C be an H-green cycle expressed as the composition P1P2P3P4 as in Definition 6.2. Then:
(1)	Pi is contained in one of the two ab-subpaths of R;
(2)	if C Ç M and s is any H-spoke contained in M that is totally disjoint from C, then C is a (C U (H — {s)))-prebox;
(3)	if C is not contained in M and s is any H-spoke contained in M having one end in the interior of Pi, then C is a (C U (H — {s)))-prebox;
(4)	there is a C-bridge MC so that H Ç C U MC;
(5)	C is contractible, C has BOD, and all C-bridges other than Mc are planar;
(6)	C is a (C U H)-prebox;
o	_______________
(7)	Mc is the unique C-bridge (that is, there are no planar C-bridges);
(8)	C bounds a face of n;
CM
Proof. Because n is H-friendly, there is a face F of (HU7) whose closure contains C.
(9) there are at most two H-nodes in the interior of Pi; and (10) in any 1-drawing of H U C, C is clean.
(1)	This is an immediate consequence of Definition 6.4, as the boundary d of any face of H U 7 has each component of d n R contained in one of the ab-subpaths of R.
(2)	and (3) Note that H — (s) contains a subdivision of Vg. In particular, if e is an edge of C not in R, then H — (s) is a non-planar subgraph of (CU(H — (s))) — e, as required. If e G C is in R, then we claim the cycle R' = (R — (P)) U P2P3P4 is the rim of a V6. We see this in the two cases.
Case 1: (2) In this case, there are three H-spokes t1, t2, t3 other than s contained in M. Each tj has an end vj in R — (P1) and a maximal R'-avoiding subpath tj containing vj. It is straightforward to verify that R' U t1 U t2 U t3 is a subdivision of V6 , as required.
Case 2: (3) In the exceptional case P1 = rj rj+1 rj+2, s is different from all of sj, sj+3, and sj+4, so R' U sj U sj+3 U sj+4 is the required V6. (Note that one of sj and sj+3 can be the exposed spoke and part of that spoke might be in either P2 or P4, but whatever part is not in P2 U P4 makes the third spoke.)
In the remaining case, there are two H-spokes sj and sj+1 that are completely disjoint from C. Any other H-spoke s', different from s, sj, and sj+1, and contained in M, will connect to R' to make a third spoke, either because both its ends are in
o
CD O
00
s
CO
00	R' or because one end is in R' and the other end is in P1 and one of the paths in
P1 — e joins the other end of s to a vertex in R'.
(4)	Let MC be the C-bridge containing the ab-subpath Q of R that is P1-avoiding. We claim H C C U MC. Observe that the maximal P^avoiding subpath Q' of R containing Q is contained in MC and, therefore, R C C U MC. Note that every H-spoke has at least one end in Q' that is not in P1 and, therefore, that end is in Nuc(MC). Thus, if P3 is not contained in M, it is obvious that H C C U MC. So suppose P3 is contained in M. The H-spokes other than those that contain P2 and P4 are obviously in MC, and the ones containing P2 and P4 are in the union of MC and C.
(5)	If either P1 has at most 3 H-nodes, or s0 is not exposed, or P1 is neither r1 r2 r3 nor r6 r7 r8, then there is an H-spoke s contained in M and totally disjoint from C. The spoke s combines with the one of the two subpaths of R joining the ends of s that is disjoint from P1 to give a non-contractible cycle disjoint from C. The claim now follows immediately from Lemma 5.20.
We now treat the case s0 is exposed and P1 is either r1 r2 r3 or r6 r7 r8. In this case, F is a face of H U 7 contained in D. Let B' be a C-bridge other than MC. If B' C cl(F), then CUB' C cl(F) and cl(F) is a closed disc in RP2. Therefore, CUB' has no non-contractible cycles in RP2. Otherwise, B' is contained in the closure of one of the H-faces bounded by Q1 or Q2 or Q3. For each i e {1, 2, 3}, let Fj, be the H-face bounded by Qi. Then cl(Fi) n cl(F) is a path and, therefore, cl(Fi) U cl(F) is a closed disc containing C U B' and again C U B' has no non-contractible cycles. The result now follows from Corollary 5.17.
(6)	In the case P3 C M, at most the H-spokes containing P2 and P4 meet
C.	There are at least two others contained in M that are disjoint from C; let s be one of these. By (2), for any edge e of C, (C U (H — (s))) — e is not planar, so
00	(C U H) — e is not planar.
Now suppose P3 C D. If some H-spoke s contained in M has an end in the interior of P1, then (3) implies that, for any edge e of C, (C U (H — (s))) — e is not planar, so (C U H) — e is not planar.
In the alternative, no H-spoke contained in M has an end in the interior of P1. If e is not in P1, then H n M, which is a V8 or V10, is contained in (C U H) — e, so
CO
we may assume e e P1. But then (R — (P1)) U P2P3P4 and the H-spokes contained in M make a V8 or V10, showing (C U H) — e is not planar.
(7)	Observe that (5) shows any other C-bridge is planar and that C has BOD. If B is any other C-bridge, then C is a B#-prebox by (6) and, therefore, is, by definition, a box, contradicting Lemma 5.12.
(8)	This is an immediate consequence of the facts that C is contractible (5) and there is only one C-bridge (7).
(9)	Suppose by way of contradiction that vi_1,vi,vi+1 are internal to P1. Notice that P1 is not exceptional. We claim that Qi is a box, contradicting Lemma 5.12.
For s e {si_1, s^ si+1}, s is contained in one of the two faces of R (i.e., the Mobius band M and the disc D). By (8), C is the boundary of some face F of G. Clearly F and s are in different R-faces, so one is in M and the other is in
D.	Therefore, all of si_ 1, si, and si+1 are contained in the same one of M and D. Since D contains at most one H-spoke, it must be that all three are contained in M. Clearly, this implies F C D and, therefore, P2P3P4 C D.
iH
o
00
u
CD
CD O CD
00
0 Ö
o
1
00
£ CO CO
CO CD
$H
CD CO
Jh
a
CD U
There is another H-spoke s contained in M that is totally disjoint from Q^ As P2P3P4 C D, R U P2P3P4 U s contains a non-contractible cycle including both P2P3P4 and s that is totally disjoint from Q^ Thus, Lemma 5.20 implies Qj has BOD and all Qrbridges except Mq are planar.
We claim Qj is a (Qj U Mq )-prebox. Note that Qj U Mq contains H — (sj} and so the deletion of any edge in sj_i U sj+1 leaves a V6. By (3), C is a CU (H — (s^)-prebox, so the deletion of any edge e in rj_i U r leaves a non-planar subgraph in (C — e) U (H — (sj}), which is contained in (Qj — e) U Mq . That is, if e G rj_1 U rj, then (Qj — e) U Mq is not planar.
We must also consider an edge in rj+4 U rj+s (these indices are read modulo 10). Let R' be the cycle made up of the following four parts: the two paths in R — (Pi} — (rj+4 rj+5}, P2P3P4, and sj_1 rj_1 rj sj+1. To get the V6, add to R' both H-spokes totally disjoint from P1 and either of the two R'-avoiding subpaths of P1 whose ends are in R'. Thus, if e G rj+4 rj+5, then (Qj — e) U Mq is not planar, completing the proof that Qj is a (Qj U Mq )-prebox. (See Figure 6.1.)
ri+4 ri+5 ri+6
Figure 6.1. The case e g rj+4 rj+5 for (Qj being a (Qj U M,^.)-prebox. Only two of the three spokes are shown.
Since the Qj-bridge B containing sj is contained in the closed disc in RP2 bounded by Qj, B is planar and, therefore, Qj is a box, the desired contradiction.
(10) Let D be a 1-drawing of H U C. Let P1P2P3P4 be the decomposition of C into paths as in Definition 6.2, so P1 C R and P3 is H-avoiding. If C is crossed in D, then it is P1 that is crossed, while P2P3P4, being R-avoiding, is not crossed in D. We claim that there is an H-spoke vjvj+5 disjoint from C that is not exposed in D. The existence of s and the fact that C is crossed in D shows that no face of R U s is incident with both ends of P1 and, therefore, P2P3P4 must cross R U s in D, the desired contradiction.
To prove the claim, we consider two cases. If P1 has at most 3 H-nodes, then this is obvious, since only one H-spoke can be exposed. In the alternative, P1 is exceptional, say P1 = rj rj+1 rj+2. As the spoke exposed in D is incident with an end of the H-rim branch that is crossed, we see that sj+4 is not the exposed spoke and is disjoint from P1, as required.	■
The next result is the main result of this section and the first of three main steps along the way to obtaining the classification of 3-connected, 2-crossing-critical graphs having a subdivision of V1q. The other two major steps are, for G G M2
o
00
u
a
CD U
00	containing a subdivision H of V10: (i) G has a representativity 2 embedding in
RP2 so that H C M; and (ii) G contains a subdivision of V10 with additional properties (that we call "tidiness"). It is this tidy V10 for which the partition of the edges of the rim into the red, yellow, and green edges that allows us to find the decomposition into tiles.
Theorem 6.7. If G G M3 and V10 = H C G, then no two H-green cycles have an edge of R in common.
Proof. Suppose e0 G R is in distinct H-green cycles. By Lemma 6.5 (3), there is an H-friendly embedding n of G in RP2. By Lemma 6.6 (8), any H-green cycle bounds a face of n[G]. As e0 is in R and R is the boundary of both the (closed) Mobius band M and the (closed) disc D, one of these faces, call it Fm, is contained in M, while the other, call it Ff, is contained in D. For n G {M, D}, let Cn be the green cycle bounding Fn and let PfP^Pj.1 Pf be the path decomposition of Cn as in Definition 6.2; in particular, Pf C R and P3 is H-avoiding.
Note Pf P3dP4d is disjoint from M (except for its ends) and P2MP3MP4M is contained in M. Thus, Cd n Cm = PD n PM. Lemma 6.6 (9) implies that, for n G {M, D}, Pf has at most 4 H-nodes. We conclude that Pf U PM is not all of R, and so Cf n Cm is a path. Therefore, there is a unique cycle C in Cf U Cm not containing e0 and, furthermore, C bounds a closed disc in RP2 having e0 in its interior.
On the other hand, Lemma 6.6 (1) shows there is an a6-subpath A1 of R that contains Pf. Since e0 G Pf H PM, it is also the case that PM C A1. Let A be the other a6-subpath of R, so that A is (Cd U CM)-avoiding. In particular, there is a C-bridge MC containing A. By Lemma 6.6 (7), for n G {M, D}, A is in the unique Cn-bridge MC . Since MC (and therefore A) is not contained in the face of 00	G bounded by Cn, we conclude that A is not in the disc bounded by C. Therefore,
Mc is different from the C-bridge Be containing e0.
Claim 1. For each H-spoke s, some H-node incident with s is not in Cm U Cf.
Proof. By Lemma 6.6 (9), there exists an i so that PD C r rj+1 rj+2. In
CO
particular, e0 is in r U rj+1 U rj+2. Thus, PM has an edge in at least one of rj, ri+1, and ri+2.
Lemma 6.6 (8) implies that Cm bounds a face of G. Therefore, Cm is contained in the closure cl(F) of a face F of n[H] and F C M. Thus, PM is contained in one of the two components of cl(F) n R. Since such a component is contained in consecutive H-rim branches, if PM contains an edge in rj, then PM is contained in either rj_1 rj or rj rj+1. From the preceding paragraph, PM is contained in one of rj_1rj, rjri+1, ri+1ri+2, and ri+2ri+3.
We conclude that PD U PM is contained in either ri-1 rj rj+1 rj+2 or rj ri+1 rj+2 ri+3 showing that no H-spoke has both ends in PD U PM.	□
CLAIM 2. (1) H C C U MC U BC.
(2) If s is an H-spoke contained in M disjoint from Cm, then (C U MC) — (s)
CD
is not planar.
Proof. For (1), we note that it is clear that R C C U MC U BC. Now let s be an H-spoke. Suppose first that s C M. By Claim 1, there is an H-node v incident with s and not in Cm U Cf .If s n Cm is at most an end of s, then it is evident that s C MC. If s H Cm is more than just an end of s, then s consists of a CM-avoiding
o
CM
00
o Ö
00 cm cm
00	subpath s' joining v to a vertex in Cm, together with the path Cm n s (which is by
Lemma 6.3 (1)) either PM or PM). But then it is again evident that s C C U MC.
Otherwise, s is exposed, in which case we have the same argument, but replacing Cm with Cd, completing the proof of (1).
For (2), a V6 is found whose rim is (R - (PM)) U P2mP3mP4m. The spokes are
S	contained in the three other spokes in M, namely they are the parts that are not
in P2m U P4m.	□
Claim 3. C has BOD.
CD
Proof. Let S be the set of H-spokes contained in M and disjoint from Cm. As Cm meets at most two H-spokes in M, |S| > 2. If some s G S is also disjoint from Cd , then R U s contains a non-contractible cycle disjoint from C, in which case Lemma 5.20 shows C has BOD, as claimed.
So we may assume that no element of S is also disjoint from Cd . Let s be any element of S; then s n Cd is a vertex v of PD. Let e be the edge of s incident with v. In order to show that C has BOD, we will show that: (i) the overlap diagrams ODG-e(C) and ODq(C) are the same; and (ii) ODG-e(C) is bipartite. For (i), note that Cd bounds a face in RP2 and that (s) is in the boundary of two (H U y)-faces. Thus, there can be no C-bridge that overlaps MC in G because of its attachment at v. That is, ODG-e(C) and ODG(C) are the same.
For (ii), Lemma 6.6 (2) applied to Cm and (3) applied to Cd, combined with Lemma 5.4, shows Cd and Cm are both clean in De. Therefore, C is clean in De. By Claim 2 (2), (C U MC) - e is not planar, so Lemma 5.6 shows C has BOD in G - e. Therefore, C has BOD in G.	□
Claim 4. C is a C U H-prebox.
Proof. Note that CD U CM C C U H. If e G C, then let i G {M, D} be such that e G Ci. Lemma 6.6 (6) says that Ci is a (Ci U H)-prebox and, therefore, (Ci U H) - e is not planar. Since (Ci U H) - e C (C U H) - e, we conclude that C is a (C U H)-prebox.	□
CLAIM 5. G = C U MC U BC.
CO
Proof. By way of contradiction, suppose there is another C-bridge B'. Let F be the (H U Y)-face containing B'. Then C U B' is contained in the closed disc that is the union of the closure of F and the disc bounded by C, showing B' is planar. By Claim 4 and the fact that CU H C B'#, Lemma 5.4 says that C is clean in a 1-drawing of B'#, of which there is at least one, since G is 2-crossing-critical. This yields a 1-drawing of C U MC with C clean. By Claim 3, C has BOD, BC is planar because it is contained in the closed disc bounded by C, and above we showed that every other C-bridge is planar; Corollary 4.7 implies cr(G) < 1, a contradiction.	□
We are now on the look-out for a box in G; it is not true that C is necessarily one. Our next claim gives a sufficient condition under which we can find some box and the following two claims show that, in all other cases, C is a box.
Claim 6. Suppose all of the following:
(1)	there is an i so that PM is in a Qi-local H-bridge;
(2)	PM contains vi and is a non-trivial subpath of si; and
o
CM
CD O CD
Qï
o
cm
I
cm cm
(3) vi+2 is in the interior of PD. Then G has a box.
Proof. We note that (2) implies s, C M.
Subclaim 1. Both si+1 and si+2 are contained in M.
Proof. Suppose first that si+2 is exposed. Then (3) implies PD and P® are both trivial. That is, C® = PDPD. But PD is H-avoiding and overlaps si+2 on R (because PD has at most four H-nodes, only two of which can be in the interior of PD). Thus, P® and si+2 cross in RP2, a contradiction. Therefore, si+2 C M.
Next, suppose si+1 is exposed. Then, by symmetry, we may assume i = 4 or i = 9. In either case, PM and PD are in different a6-subpaths of R and so do not
00
have an edge in common, a contradiction. Hence si+1 is also contained in M. □
Let u be the common end of PM and PM and let w be the common end of PM and PM. By (2), u G (s,) and, by (1) and (2), w G r,. Observe that the edge e0 common to Cm and C® is in [v,, r,, w].
Let C' be the cycle [vi+5, s,, u, P^Pp, w, r,, ri+i si+2 ri+6 ri+5. We note that there are two obvious C'-bridges: the C'-interior bridge Bc containing the edge of sj+i incident with Vj+6; and the C'-exterior bridge Mc for which H — (sj+i) C C' U Mc. To show C' is a box, it suffices to show that C' has BOD and C' is a (C' U Mc )-prebox.
Notice that v,+2 is in the interior of PD by hypothesis and v,+ i is in the interior of PD because e0 G r,. Lemma 6.6 (9) implies that the only H-nodes in the interior of P® are v,+i and v,+2. In particular, v, and v.j+3 are in
R — (P?> ,
as are all the ends of s,+3 and s,+4.
To see that C' has BOD, we produce a non-contractible cycle in Nuc(MC/). Lemma 5.20 then implies C' has BOD and precisely one non-planar bridge. We start with the two paths PDP3DPP and s,+4, and easily complete the required cycle using two paths in R, one containing r,+3 and the other containing ^+9.
It remains to show that C' is a (C' U Mc)-prebox. Since V8 = H — (s^i) C C' U Mc, it is obvious that, if e G C' and e G R, then (C' U Mc) — e contains a V6 and so is not planar. So suppose e G C' and e G R. There are two cases.
If e G r, r^, then take (R — (P®)) U P^fPfPf as the rim. We choose as spokes s,, s,+3, and s,+4.
If e G r,+5 r,+6, then the rim consists of the two paths PDP3DPP and C' — (r,+5 r,+6), together with the two subpaths of R joining them, one containing v,+3, v,+4, and v,+5, and the other containing «,+7, v,+9, and v,. In this case, the spokes are s,+3, s,+4, and PM.	□
In the remaining case, we show that C is a box. The following simple observations get us started, the first being the essential ingredient.
CD
Claim 7. Either:
(1)	there is an i so that
•	PM is in a Q,-local H-bridge;
•	s, contains an edge of Cp; and
•	v,+2 is in the interior of PD; or (symmetrically)
(2)	there is an i so that
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•	PM is in a Qj-local H-bridge;
•	Sj+1 contains an edge of Cm; and
•	vi-1 is in the interior of PD;
or
(3) there are three H-spokes not having an edge in Cm and not having an incident vertex in the interior of PD.
CD	Proof. Lemma 6.6 (9) implies there are at most two H-nodes in the interior
of PD. Therefore, if no H-spoke contains an edge of Cm, then (3) holds. So we may suppose Cm has an edge in some H-spoke.
Suppose first that s0 is exposed, Cm has an edge in s1 and e0 is in either [a, rg, v0, r0, v1] or [b, r5,v6]. Therefore, PD has one end in either [a, rg, v0, r0, v1) or [b,r5,v6). Lemma 6.6 (9) implies at most two H-nodes can be in the interior of PD, so no end of s3 can be in the interior of PD. We conclude that s0, s3 and s4 are the required three spokes yielding (3).
Symmetry treats the same case on the other side.
In the remaining case, PM is contained in a Qj-local H-bridge and both Sj and Sj+1 are contained in M. The edge e0 is in either rj or rj+5. If the only H-nodes in the interior of PD are incident with either sj or sj+1, then the other three H-spokes suffice for (3).
Thus, by symmetry we may assume an end of sj+2 is in the interior of PD. This implies that an end of sj+1 is also in the interior of PD. Lemma 6.6 (9) shows these are the only H-nodes in the interior of PD. If sj does not contain an edge of Cm, then the three spokes other than sj+1 and sj+2 suffice for (3), while if sj does contain an edge of Cm, then we have (1).	□
Claims 6 and 7 show we need only consider the third possibility in Claim 7 to find a box.
Claim 8. If there are three Hf-spokes not having any edge in CM and not having an incident H-node in (PD), then C is a box.
Proof. By Claim 3 and the fact that BC is a planar C-bridge, it suffices to show C is a (C U MC)-prebox. For each e G C, we show that (C U MC) — e contains a V6.
We note that 3-connection and the fact that Cm and C® both bound faces implies Cm f C® is just e0 and its ends. That is, BC consists of just e0 and its ends. Thus, Claim 5 implies that G — e0 = C U MC. In particular, every spoke is in C U MC.
Let w be any H-node that is not in C. There are two wC-paths in R — e0; let them be Rx with end x G C and Ry with end y G C. Thus, R consists of the C-avoiding path Rx U Ry, a subpath of C, the edge e0, and another subpath of C. The cycle C consists of two xy-paths; let them be ND containing P^P^P® and NM containing P2MP3MP4M. We note that ND C D and NM C M.
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Subclaim 2. Let s be an H-spoke with no edge in Cm and not having an incident H-node in (PD).
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(1)	If s C M, then s f C is either empty, x, or y.
(2)	If s C D (that is, s = s0 is exposed), then s f C contains at most one of v0 and v5.
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hypothesis, s has no edge in Cm and, therefore, s has no edge in C. Being in N p the vertex u is either in R or in PpPpPp.
CD	Suppose that u is in
u is an end of P3p, and so is in P2p U P4p. Thus, if u is in P2pP3pP4p, then u is in P2p U P4p. Since both Pp and Pp are contained in H, are R-avoiding, and neither has an edge of s, the one containing u is trivial and u is in R.
Thus, in every case u is in R and so is an H-node. It follows that one of [x, Np,u] and [u, Np,y] contains PpPpPp and the other is contained in R. We choose the labelling so that [x, Np, u] C R.
As we follow R - eo from w to x and continue to u along Np, we see there is an edge of C incident with x and not in R. That it is in ND implies it is in P^P^P®. All the vertices in x, N p , u are incident with two rim edges in what we have just traversed. In particular, eo is not incident with any of these vertices and, therefore, x, N p , u] is contained in CD. More precisely, [x, N p , u] is contained in P®. As we continue along R past u, we either find eo is incident with u or the other edge of C incident with u is in R. In either case, u is in (PD), a contradiction.
For (2), suppose vo and v5 are both in C. Then Pp UPD contains both vo and v5. By Definition 6.2 (1e), vo and v5 are not both in the same one of Pp and PD, so one is in Pp and the other is in PD. By symmetry, we may assume vo is in Pp. Because n is H-friendly, Pp is contained in either [a, rg, vo, ro, v1] or, if a = vo, rg (these being the only two faces of n[H] U 7 in M that can be incident with vo).
Recall that eo is in both Pp and PD. If Pp C [a, ro, vo, ro, v1 ], then eo is in either rg or ro and PD is, by Definition 6.2 (1e), contained in either [a, rg, vo]ro r1 [v2, r2, v3) or ro r1 r2 [v3, r4, v4), and v5 is not in C. If Pp C rg, then eo is in rg, so PD is contained in rg r8 r7 [v7, r6, v6), and again v5 is not in C.	□
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The case e G ND is easy: the rim of the V6 is (R - (Pp)) U PpPpPp and cm	we choose as spokes any three of the H-spokes that are contained in M. (If one
intersects Cp, then only the part of the spoke that is Cp-avoiding will be the actual spoke of the V6.)
If e G Np, then the rim R' of the V6 is (R - (P1D)) U P2DP3DP4D and the spokes are the three H-spokes from the hypothesis. If all three hypothesized H-spokes are contained in M, then it is evident from Subclaim 2 (1) that we have indeed described a V6 in (C U MC) - e.
So suppose that one of the H-spokes in the hypothesis is the exposed spoke so. From Subclaim 2 (2), either so is disjoint from C or precisely one H-node incident with so is in C. We may choose the labelling so that vo is not in C.
If v5 is not in C, then so is disjoint from C. Subclaim 2 (1) shows the other two hypothesized H-spokes meet C in at most x or y; it is now obvious that the three hypothesized H-spokes combine with R' to make a V6.
Finally, suppose v5 is in C. Because C® is H-green, PD C r2 r3 r4 [v5, r5, b]. In particular, s1 is disjoint from Cp. If s2 has no edge in Cp, then R' U s1 U s2, together with the portion of so from vo to C® is a V6 avoiding Np. If s2 has an edge in Cp, then Cp is in the n[H]-face bounded by Q2. In this case, we may replace s2 with s4 r4 to obtained the desired V6.	□
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Evidently, Claims 6, 7, and 8 show that G has a box, contradicting Lemma 5.12.	■
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CHAPTER 7
Exposed spoke with additional attachment not in Q0
The main result of this section is the proof of the following technical theorem, which limits possibilities for the V10-bridges. This will be used in the next section when we get our second major step by showing that there is a representativity 2 embedding of G in RP2 for which all the H-spokes are contained in the Mobius band.
Theorem 7.1. Suppose G G and V10 = H C G. Let n be an H-friendly embedding of G in RP2, with the standard labelling. Then there is no H-bridge having attachments in both (so) and (r1 r2 r3).
At one point in the proof of this theorem, we need the following lemma. Most of it is used again several times.
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Lemma 7.2. Let G be a graph and let Vg = H C G. Let P be an H-avoiding path in G joining distinct vertices x and y of R and let P' be one of the two xy-subpaths of R. Let D be a 1-drawing of H U P.
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(1)	If P' has at most two H-nodes or, for some i, P' = rj rj+1, then P' is not
CM	crossed in D.
(2)	If there are only the two H-nodes vj, vj+1 in the interior of P' and P' has at most one other H-node, then rj+4 is not crossed in D.
(3)	Suppose rj rj+1 C P', P' C rj rj+1, but P' C rj rj+1 [vj+2, rj+2, vj+3).
(a)	Then rj rj+1 is not crossed in D.
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(b)	If P' is crossed in D, then sj+3 is exposed in D and P' n rj+2 crosses rj-1.
Proof. Let x and y be the ends of P and let R' = (R — (P')) UP. For (1) and (2), we find three spokes to add to R' to find a subdivision of V6 disjoint from P' — or at least some part of P'. The part of P' disjoint from the V6 cannot be crossed in any 1-drawing of H.
For (1), if P' contains at most one H-node, then this is easy: any three H-spokes not having an end in P' will suffice. If P' = rj rj+1, then the three H-spokes sj, sj+2, and sj+3 suffice.
In the remaining case, P' has precisely two H-nodes. We may express P' in
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the form
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P' = [x,rj-1,vj] rj [vj+1,rj+1,y],
where either of [x, rj-1,vj] and [vj+1, rj+1, y] might be a single vertex. In this case, the spokes are sj+2, sj+3 and sj+1 [vj+1, rj+1, y], showing that [x, rj-1,vj]rj is not crossed in D, while replacing sj+1 [vj+1, rj+1, y] with [x, rj-1,vj] sj shows [vj+1, rj+1, y] is not crossed in D. This completes the proof of (1).
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For (2), replace R' with (R' — (rj+4)) U (sj r sj+1). We now need three spokes. If there is a third H-node in P', then symmetry allows us to assume it is vj-1. In either case, we choose sj-1, [vj+1, rj+1, y], and sj+2 as the three spokes for the V6. This V6 avoids rj+4, showing it is not crossed in D.
For (3), x = vj and the hypotheses imply that y G (rj+2). For (3a), we may use the spokes sj, sj+2 [vj+2, rj+2, y], and sj+3 to see that rj rj+i is not crossed in D, as required.
For (3b), suppose P' is crossed in D. Part (3a) shows that it must be P' n rj+2 that is crossed and (2) shows that rj+5 = rj-3 is not crossed in D. We need only show that rj-2 is also not crossed in D. If it were, then [vj+2, rj+2, y] crosses rj-2. But then the cycle rj+3 rj+4 rj_3 rj-2 sj-1 separates vj = x from y in D, showing that P is also crossed in D, a contradiction.	■
Proof of Theorem 7.1.
suppose so is exposed.
This is obvious if no spoke is exposed in n, so we may
Claim 1. There is no H-avoiding (so) (v1,r1,v2]- or (s0) [v3,r3,v4)-path.
Proof. By symmetry, it suffices to prove only one. By way of contradiction, we suppose that there is an H-avoiding path P from x G (s0) to y G (v1, r1, v2].
Let e G s3 and consider a 1-drawing D of G — e. By Lemma 5.9 and Theorem 5.23 (4), we know that Q3 is crossed in D. This implies that r1 r2 r3 r4 crosses r6 r7 r8 rg. This already implies neither s0 nor s1 is exposed in D. Furthermore, the crossing is of two edges in R and, since P is H-avoiding, we conclude that D[P] is not crossed in D. Therefore, the end of P in (v1, r1, v2] must occur in the interval of r1 r2 r3 r4 between the crossing and that is, the crossing must involve an edge of r1. In particular, r2 r3 r4 r5 is not crossed in D.
Since Q3 is crossed in D and r1 is crossed in D, the other crossing edge is in r7 r8. Thus it is in r6 r7 r8. It follows that s2 is exposed in D. Thus, the cycle
r4 r5 s1 r0 rg s4 separates x from y in D, showing P is crossed in D, a contradiction.
□
It follows from Claim 1 that, if there is an H-avoiding path P0 joining x G (s0) to y G (r1 r2 r3), then y G (r2). Let K = H U P0. See Figure 7.1.
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Figure 7.1. The subgraph K of G in RP2
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Let J1 and J2 be the two cycles r0 ri [v2, r2, y, P0, x, s0, v0] and r4 r3 [v3, r2,y, P0, x, s0, v5], respectively.
Claim 2. The cycles J1 and J2 both bound faces of G in RP2.
Proof. These cycles are both H-green, so this is just Lemma 6.6 (8). □
The following claim completes the determination of the (H n M)-bridge con-
CD	taining so.
Claim 3. The (H - (s0))-bridge containing s0 is s0 U P0.
Proof. Suppose not and let B be the (H - (s0))-bridge containing s0. Then Lemma 5.19 implies that B has an attachment z other than v0, y, and v5. By Claim 2, z G [a, rg, v0) U (v5,r5, b]; by symmetry we may assume the former. Let P be a K-avoiding z (s0)-path.
Suppose z = vg. Let e be the edge of s0 incident with v0. We show that cr((K U P) - e) > 2. As this is a proper subgraph of G, we contradict the fact that G is 2-crossing-critical. In P U (s0 - e) U P0, there is a claw Y with talons z = vg, y and v5. We show cr((H - (s0)) U Y) > 2.
By way of contradiction, we suppose D is a 1-drawing of (H - (s0)) U Y. As H -(s0) = V8, Lemma 7.2 (1) implies that (using the labelling from H) [y, r2,v3] r3 r4 is not crossed in D, while (2) of the same lemma implies neither r6 nor r8 is crossed in D. Part (3a) implies rg r0 r1 is not crossed, while (3b) implies (since r9 is not crossed) that [v2,r2,y] is not crossed. The only remaining possibilities for crossed (H - (s0))-rim branches are r5 and r7. But no 1-drawing of H - (s0) has these two rim-branches crossed, the desired contradiction.
So z = vg. But then we may replace s0 with the zv5-path s0 in P U s0 and replace P0 with the ys0-path in P0 U s0 to get a new subdivision H' of V10. We notice that Lemma 6.5 (1) implies that n is H'-friendly. However, the analogue Ji of J1 does not bound a face, contradicting Claim 2.	□
Claim 4. There is a unique 1-drawing of K. In this 1-drawing, s0 is exposed.
The 1-drawing of K is illustrated in Figure 7.2.
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Proof. If D is a 1-drawing of K, then Claim 2 and Lemma 6.6 (10) imply neither J1 nor J2 is crossed in D. It follows that none of r0, r1, r2, r3, and r4 is crossed in D. Lemma 3.6 implies r7 cannot be crossed in D, so Q2 is clean in D. Therefore, s0 must be in a face of D[R U Q2] incident with r2. This is only possible if s0 is exposed, which determines D.	□
For j G {2, 3}, let Dj be a 1-drawing of G - (sj).
Claim 5. The crossing in D2[(H - s2) U P0] is of r5 with [y, r2, v3]. Likewise, the crossing in D3[(H - s3) U P0] is of rg with [v2, r2, y].
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The 1-drawings of Claim 5 are illustrated in Figure 7.3.
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Proof. We treat the case j = 2; the case j = 3 is very similar. By Theorem 5.23 (2), Q2 has BOD, so Lemma 5.9 implies Q2 is crossed in D2. This implies that s0 is not exposed in D2. The H-avoiding path P0 joins x G (s0) to y G (r2), so y must be on a face incident with s0. It follows that Q0 must be crossed in D2. This
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Figure 7.2. The 1-drawing of K.
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Figure 7.3. The 1-drawings D2[(K - (s2)) U P0] and D3[(K -(S3» U Po].
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implies that si is exposed. We deduce that either r5 crosses ri U r2 or r0 crosses Ur7. In the latter case, D2[P0] must cross D2[H — s2], a contradiction, so it must be the former.
As D2[Po] is not crossed, y occurs between v1 and the crossing in r1 U r2, as required.	□
The following claims help us obtain the structure of (Mq )#; we will use this to find a 1-drawing of G, which is the final contradiction.
Claim 6. Suppose B is a Q0-bridge having an attachment in each of r9 and
r5. Then B is one of M^ , v«vg, v0v6, and v5v9. Q0
Proof. We note that s0 U P0 Ç MQ . Either B = MQ , or, in the drawing
Q0 Q0
D2, B is in a face of D2 [(H — S2 ) U Po] incident with both rg and r5. There are only
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00	two such faces, namely F, bounded by Q4, and F', the other face incident with
rg. Whichever face B is in, its attachments are in the intersection of Q0 with the boundary of the containing face. Thus, if B is in F, then att(B) C r4 s4 rg. In this case, the only possibility for an attachment in r5 is v5, so v5 G att(B). If, on the other hand, B is in F', then att(B) C rg r0 s1. In this case, v6 G att(B). Similarly, D3 shows either B = Mq , or att(B) C r0 s1 r5 and v0 G att(B), or att(B) C s4 r4 r5 and vg G att(B). Comparing these possibilities, we conclude that one of the following four cases holds for att(B): att(B) = {v0,v5}; att(B) = {v6,vg}; v5, vg G att(B) and att(B) C r4 U s4; and v0, v6 G att(B) and att(B) C r0 U s1.
We claim v0v5 is not an H-bridge. For if it were, let D be a 1-drawing of G — v0v5. Then s0 U P0 is not crossed in D and Claim 3 says the (H — (s0))-bridge containing s0 is s0 U P0. In particular, s0 consists of the two edges v0x and xv5, and x has degree 3 in G. Thus, we can draw v0v5 alongside s0, yielding a 1-drawing of G, a contradiction.
We must show that, if v0,v6 G att(B) and att(B) C r0 U s1, then B = v0v6. Likewise, if v5, vg G att(B) and att(B) C r4 U s4, then B = v5vg. We consider the former case, the latter being completely analogous. Corollary 5.15 shows that B can have at most one other attachment. Lemma 5.19 shows that either B = v0v6 or B is a claw with talons v0, v6, and z G (v0, r0, v1, s1, v6). Since we are trying to show B = v0v6, we assume the latter. Let e be the edge of B incident with z and let D be a 1-drawing of G — e. Since K C G — e, D extends the 1-drawing illustrated in Figure 7.2. We modify D to obtain a 1-drawing of G, which is impossible.
Observe that B — z is an H-avoiding v0v6-path P (having length 2); there is only one place D[P] can occur in Figure 7.2. Notice that B is a Q0-local H-bridge and, furthermore, P overlaps Mq0 .
Theorem 5.23 shows Q0 has BOD in G; let (B, M) be the bipartition of OD(Q0), with B G B. Then Mq0 G M. Every Q0-bridge is drawn in D, with the exception that we have B — e in place of B.
csr	Because we cannot add e back into D to get a 1-drawing of G, there must be
an H-avoiding path P' in G — e joining the two components of [v0, r0, v1, s1, v6] — z so that D[P'] is on the same side — henceforth, the inside — of D[Q0] as P. Let B' be the Q0-bridge containing P'. If B' has just v0 and v6 as attachments, then let D be a 1-drawing of G — v0v6. As we did above for v0v5, we can add v0v6 alongside P to recover a 1-drawing of G. Therefore, B' does not have just v0 and v6 as attachments.
It follows that B' overlaps B, so it is in M. Therefore, it does not overlap Mq0 ; in particular, it cannot have an attachment in both [v6, s1,v1) and [v0,r0, v1). We conclude that, for some q G {r0,s1} ; and (ii) att(B') C q. Let q' be such that
{qs q'} = {r0,s1}.
Let B1,B2,.. .,Bk be a path in OD(Q0) — {Mq0 , B} so that B' = B1.
Subclaim 1. For i = 1,2,... ,k, att(Bj) C q.
Proof. Above, we chose q to contain att(B'), which is the case i = 1. Notice that B1, B3, ... are all on the same side of D[Q0] as B' and P, while B2, B4, ... are all on the other side of D[Q0]. The former are all in M, while the latter are in B. Let i be least so that Bj has an attachment outside q. Then it also has an attachment in (q) (in order to overlap Bj-1).
If Bj is inside D[Q0], then Bj does not overlap Mq0 , so it has no attachment in q' — q. As Bj cannot cross P in D, att(Bj) C q, a contradiction.
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If Bi is outside D[Q0], then either att(Bj) Ç s1; so q = si and we are done, or att(Bj) Ç r0 U [v0, s0,x], so, in particular, q = r0. Furthermore, Bi does not overlap B. Therefore, Bi has no attachment in (v0, s0,x], so att(Bi) Ç r0.	□
Let L be the component of OD(Q0) — {Mq0, B} containing B'. We can flip the Q0-bridges in L so that they exchange sides of D[Q0], yielding a new 1-drawing of G — e with fewer Q0-bridges in M on the same side of D[Q0] as P. Inductively, this shows there is a 1-drawing D' of G — e in which all Q0-bridges in the face of D'[K U P] bounded by r0 s1 P are in B. As none of these overlaps B, we may add e into D' to obtain a 1-drawing of G, a contradiction.	□
Let e5 be the edge in r5 that is crossed in D2 and let e9 be the edge in r9 that is crossed in D3. For i = 5, 9, let ui be the end of ei nearer to vi in ri and let wi be the other end of ei. See Figure 7.4. We highlight some relevant "cut" properties of these edges in the next three claims.
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Figure 7.4. The 1-drawings D2[(K - (s2)) U P0] and D3[(K -(ss>) U Po].
Claim 7. Any rg-avoiding (s4 r4] (r0 si]-path in (Mq )# contains e5. In particular, there are not two edge-disjoint rg-avoiding (s4 r4] (r0 s1]-paths in (Mq )#.
Proof. Suppose P is a rg-avoiding (s4 r4] (r0 s1]-path. Let e be any edge of s2 and let D be any 1-drawing of G — e. By Claim 5, D2 [(H — (s2)) U P0] is illustrated in Figure 7.3. But here we see that the cycle C = [v0,s0,x]P0 [y, r2,v3]s3 r8 rg separates (s4 r4] and (r0 s1]. Note that C consists of rg and a Q0-avoiding v0vg-path in Mq . Therefore, P is disjoint from C, and so it must cross C in D2. As this can only happen at the crossing in D2, it must be that the edge of r5 crossed in D2 is in P.	□
Analogously, deleting e G s3 provides a proof of the following claim.
Claim 8. Any r5-avoiding [s4 r4) [r0 s1)-path in (Mq )# contains eg. In particular, there are not two edge-disjoint r5-avoiding [s4 r4) [r0 s1)-paths in (Mq )#. □
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The final claim is a central point about Mq .
Claim 9. Let Pi and P2 be the two paths of Q0 — {e5, e9}. Then there is no PiP2-path in
(Mq)# — {e5, e9, V6V9} .
CD	Proof. Assume that there is a PiP2-path P in (Mq )# — {e5, e9}. For i = 1, 2,
let z be the end of P in P,.
Q	Suppose first that is in (s4 r4). If z2 is in [v6,r5,w5], then P[z2,r5,v6] is an
r9-avoiding (s4 r4] (r0 si]-path in (Mq )# that also avoids e5, contradicting Claim 7. If z2 is not in [v6, r5, w5], then there is an r5-avoiding [s4 r4) [r0 si)-path in (Mq )# that also avoids e9, contradicting Claim 8. Therefore, is in Pi — (s4 r4); that is is in [v9,r9,u9] U [v5,r5,u5]. Symmetrically, z2 is in [w9,r9,v0] U [w5,r5, v6].
If is in [v5,r5,u5], then Claim 7 implies z2 is not in [w5,r5,v5]. Therefore, z2 is in [w9,r9,v0]. By Claim 6, P is one of v6v9, v0v6, and v5v9. Clearly, neither nor z2 is v6 and neither is v9, so none of these outcomes is possible.
Therefore, is in [v9,r9,u9]. Claim 8 implies z2 is not in [w9,r9,v0]. By Claim 6, the only possibility is that = v9 and z2 = v6 and P is just the edge v6v9, as required.	□
We will show that there is an embedding n' of G in
RP2 and a non-contractible
simple closed curve 7' in RP2 so that 7' n G consists of one point in each of the
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interiors of n'[e5] and n'[e9]. Standard surgery then implies that cr(G) < 1 (see, for example, [29]).
Consider the two faces of n[K] incident with both e5 and e9. Let Fq be the one bounded by Q0. Let F' be the other; it is bounded by the cycle s0 r5 r6 r7 r8 r9, which we call C'. Both Q0 and C' contain both e5 and e9. What we would like to prove is that, for each such face F with boundary C, there is no K-avoiding path contained in F and having an end in each of the two components of C — {e5, e9}. Although not necessarily true for n, it is true for an embedding obtained from n by possibly re-embedding the edges v0v6 and v5v9.
Let us begin with the possible re-embeddings. We deal with v0v6; the argument for v5v9 is completely analogous. If v0v6 is not embedded in F', then do nothing with it. Otherwise, it is embedded in F' and we claim we can re-embed it in Fq .
The embedding n shows that v0v6 is contained in one of the two faces of K U 7 into which F' is split. Therefore, v0 and v6 must be on the same ab-subpath of R. This implies that either v0 = a or v6 = b, or both. In order not to be able to embed v0v6 in Fq, there must be a Q0-avoiding path P contained in Fq joining (r0 si) to (r5 r4 s4 r9).
We first consider where D2[P] can be. There are only two possibilities: it is either in the face of D2[K — (s2)] bounded by [v2,r2, x,r5,v6]si ri; or in the face incident with both r0 and si. The latter cannot occur, as v0v6 is also in that face and they overlap on the boundary of this face. So it must be the former.
However, in this case, both v0v6 and P are in the face of D3[K — (s3)] bounded by Q0, and they overlap on Q0, the final contradiction that shows that P does not exist, so we can re-embed v0v6 in Fq . Let n' be the embedding of G obtained by any such re-embeddings of v0v6 and v5v9.
o
CM
CD
O CD
CO CD
CD CO
a
CD U
00	The faces FQ and F' of n[K] are also faces of n'[K] with the same boundaries;
we will continue to use these names for them, while Q0 and C' are still their boundaries.
We now show that there is no K-avoiding path in Fqjoining the two paths Pi and P2 of Q0 — {e5, eg}. Such a path is necessarily in (Mq )#. By Claim 9, such a path is necessarily vgvg. But n is H-friendly, so vgvg is not embedded in M and so, in particular, is not embedded in Fq . Thus, V6«g is also not in this face of n', whence there is no PiP2-path in Fq , as required.
Now consider the possibility of a K-avoiding path in F' having its ends in each of the two paths in C' — {e5, eg}. Such a path is in a C'-bridge B embedded in F'. By Claim 3, B has no attachment in (s0). Thus, B has an attachment either in 00	[vo,rg,wg] or in [vs,rs, us].
We claim it must also have an attachment in (r6 r7 r8). If not, then all its attachments are in
[vo,rg,Wg] U [V5 ,r5,U5] U [W5,r5 ,V6] U [vg,rg,ug] .
But then B is a Q0-bridge. If it has an attachment in both r5 and rg, then Claim 6 implies B is one of v0v6, v5vg, and V6«g. The first two are not embedded in the n'-face F' and the last does not have attachments in both components of C' — {e5, eg}. In the alternative, either att(B) C r5 or att(B) C rg, and then we contradict either Claim 7 or Claim 8.
So B has an attachment in (r6 r7 r8). If B has an attachment in [v0,rg,wg], then D3[B] must have a crossing, which is not possible. If B has an attachment in [v5,r5,u5], then D2[B] must have a crossing, which is not possible. Therefore, there is no such B, as claimed.
For each of the faces Fq and F' of n' and any points x and y in the interiors of n' [e5] and n'[eg], the preceding paragraphs show that there is a G-avoiding simple xy-arc in the face. The union of these two arcs is a simple closed curve 7' in G that meets n'[G] in just the two points x and y.
In a neighbourhood of x, there are points of e5 on both sides of 7'. If 7' were contractible in RP2, then {e5, eg} would be an edge-cut of size 2 in the 3-connected graph G, which is impossible. So 7' is non-contractible. But this is also impossible, as it meets G precisely in x and y, showing that G has a 1-drawing, the final contradiction.	■
00 1-H
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00
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cm 00 cm cm
CHAPTER 8
G embeds with all spokes in M
In this section, we prove that if G G M2 and V10 = H C G, then G has a representativity 2 embedding in RP2 with H C M. This is an important step as it provides the embedding structure we need to find the tiles.
It turns out that we need something stronger than H C M. We must also show that, in addition to H C M, the representativity 2 embedding of G is such that Mq4 is the only Q4-local H-bridge B for which Q4 U B contains a non-contractible cycle. (We remind the reader that Q4 is special. Each H-quad bounds a face of n[H]. In the standard labelling, the only one of these five faces that contains an arc of y is the one bounded by Q4.)
Theorem 8.1. Suppose G G M2 and, V10 = H C G. Then G has a representativity 2 embedding n in RP2 so that, with the standard labelling:
(1)	s0 is not exposed in n, that is, n[H] C M; and,
(2)	if B is a Q4-local H-bridge other than Mq4, then n[Q4 U B] has no non-contractible cycle.
In principle, these two arguments are consecutive: we first show we can arrange H C M, and then deal with the Q4-bridges. However, the arguments are essentially the same. Therefore, we shall have parallel statements and arguments, one for getting the five H-spokes in M and one for getting such an embedding with Q4 nicely behaved. (If we knew that G had an embedding with H not contained in M, then we could do both simultaneously.)
Definition 8.2. A friendly, standard quadruple, denoted ((G, H, n, y)), consists of G G M2, V10 = H C G, an H-friendly embedding n of G, and a non-contractible, simple closed curve y meeting n[G] in precisely two points, used as the reference for giving H the standard labelling relative to n. We abbreviate friendly, standard quadruple as fsq.
Observe that Theorem 3.5 implies G has a representativity 2 embedding in RP2. Lemma 6.5 (3) implies G has an H-friendly embedding n. Any non-contractible simple closed curve y in RP2 meeting G in precisely two points yields a standard labelling of H relative to n and y. Summarizing, we have the following observation.
Lemma 8.3. If G G M2 and V10 = H C G, then there is an fsq ((G, H, n, y)).
■
■
—
Let Q* be Q0 if s0 is exposed in n and let Q* be Q4 if s0 is not exposed in n, that is, if n[H] C M. Our first step is to show that OD(Q*) is (nearly) bipartite. Theorem 5.23 (1) implies OD(Q4) is bipartite. For Q* = Q0, this is more involved. In the following statement, v1v4 and v6vg are meant to be possible Q0-bridges consisting of a single edge joining the two indicated vertices. They need not exist in G.
o
u
CD
CD
CD ■ i J-H
CD W
U
a
CD U
CO	Lemma 8.4. Let ((G, H, n, y)) be an fsq. If s0 is exposed in n, then OD(Q0) —
{v1v4,v6vg} is bipartite.
The following observations will be needed throughout the proof of Theorem 8.1 and, in particular, the proof of Lemma 8.4.
Definition 8.5. Let ((G, H, n,Y)) be an fsq and let Q* be either Q0 (if s0 is exposed) or Q4 (otherwise). Then N — a function of ((G, H, n, y)) — denotes the set of Q*-bridges B other than Mq* for which n[Q* U B] has a non-contractible cycle. In the case Q* = Q0, any of v1v4 and V6«g that occurs in G is a Q0-bridge B for which n[Q0 U B] has a non-contractible cycle, and we do not include these in
N.
We remark that, if s0 is exposed in n, then Theorem 7.1 implies the (H n M)-
bridge B0 containing s0 is distinct from MQ . In this case, B0 G N. If s0 is not
Q0
exposed in n, then Q* = Q4. If N = 0, then n satisfies the conclusions of Theorem 8.1. Therefore, in this case, we may assume N = 0.
Before we can prove Lemma 8.4, we need some results common to both cases. An easy corollary of the following lemma will be used to deal with the main case in the proof of Lemma 8.4.
Lemma 8.6. Let D be a 1-drawing of V8 (with the usual labelling) in which Q1 is crossed.. Then:
(1)	Q3 bounds a face of D; and
(2)	if Q0 is crossed in D, then either r1 crosses r4 or r5 crosses r0.
Proof. As Q1 is crossed in D, either r1 crosses r4 r5 r6 in D or r5 crosses r0 r1 r2 in D. This already shows that Q3 bounds a face of D. CO	As Q0 is crossed in D, either r7 r0 or r3 r4 is crossed in D. Compare each of
these with the possible crossing of Q1. In the former case, r0 crosses r5, while in the latter case r4 crosses r1.	■
The following is the simple corollary that we will use.
Corollary 8.7. Let G G M2 and V10 = H C G. Let D2 be a 1-drawing of G — (s2). Then:
(1)	Q4 bounds a face of D2H — S2]; and
(2)	if Q0 is crossed in D2, then either r6 r7 crosses r1 or r1 r2 crosses r5 (see Figure 8.1 for the possibilities for D2H — (S2)]).
Likewise, if D3 is a 1-drawing of G — (S3) in which Q0 is crossed, then
the two possibilities for D3[H — (S3)] are illustrated in Figure 8.2.
_ _
Proof. Theorem 5.23 implies Q2 has BOD. Lemma 5.9 implies Q2 is crossed in D2. The results now follow immediately from Lemma 8.6.	■
Let r* denote rg U r0 in the case Q* = Q0 and rg in the case Q* = Q4. We also let r+5 denote the other component of Q* n R.
Lemma 8.8. Let ((G, H, n,Y)) be an fsq. If B G N, then n[B] C D, att(B) C r* U r+5, and B has an attachment in each of r* and r+5.
Proof. If n[B] C M, then n[Q* U B] is contained in a closed disc and, therefore, has only contractible cycles, a contradiction. Thus, n[B] C D. It now follows
o
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u
CD
CD O CD
00
Figure 8.1. The two possibilities for D2.
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1
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£ CO CO
CD
$H
CD C0
u
a
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Figure 8.2. The two possibilities for D3.
that att(B) is contained in the intersection of Q° with the boundary of D; that is, att(B) C r* U r+5.
Suppose by way of contradiction that att(B) C r*. Let f* be a minimal subpath of r* containing att(B). Then there is a non-contractible cycle C contained in BUf *.
Let F be the closed (n[H]UY)-face containing n[B]. Then F contains n[BUf*], so the non-contractible cycle n[C] is contained in the closed disc F, a contradiction. So att(B) is not contained in r* and, likewise, it is not contained in r+5.	■
Let ((G, H, n, 7)) be an fsq, with s° exposed in n. Suppose D2 is a 1-drawing of G — (s2) in which Q° is crossed. Corollary 8.7 implies that D2[H — (s2)] is one of the two drawings illustrated in Figure 8.1. The outside of D2[Q°] is the face of D2[Q°] containing D2[s3]. The inside is the other face of D2[Q°]. Likewise, if D3 is a 1-drawing of G — (s3) in which Q° is crossed, then the outside of D3[Q°] is the face of D3[Q°] containing D3[s2].
LEMMA 8.9. Let ((G, H, n, y)) be an fsq, with s° exposed in n. For i = 2, 3, let Di be a 1-drawing of G — (sj) in which Q° is crossed.. Suppose B is a Q0-bridge in
N.
(1)	If D2[B] is outside of D2[Q°],
(2)	If D3[b] is outside of D3[q°],
then B G {V1V5,v°v6}. then B G {v°V4,«5^9}.
o
00 Gï
O
CSI CSI
s
CO CD
Jh
CD
CO	Proof. We prove (1); (2) is completely analogous. We remark that B = B0 as
D2[s0] is inside D2[Q0]. Lemma 8.8 shows that either: (i) att(B) C [b, r5,v6] U [vg,rg,a] and B has attachments in both [b, r5,v6] and [vg,rg,a]; or (ii) att(B) C [a, r9,v0]r1 Ur4[v5,r5,b] and B has attachments in both [a, r9,v0]r1 and r4 [v5,r5,b].
Suppose first that D2 is the left-hand possibility illustrated in Figure 8.1. Considering D2, we see that v1 is one attachment of B and the others are in r4 r5.
Now consider the possibilities for D3 [B]. We see that D3[B] can be outside D3[Q0] in only one of the two possible D3's, namely the right-hand one, and then only if att(B) = {v1, v4}. But in this case B is just the edge v1v4, which is not in N. So D3[B] is inside D3[Q0]. It now follows from this and the previous paragraphs that att(B) C {v1} U r5.
Putting this information into n, we see that the only possibility for B, which is embedded in D and not in M, is that B = v1v5.
In the case D2 is the right-hand possibility in Figure 8.1, D2 shows that att(B) C {v6} U r9 r0. Since v6v9 e N, B = v6v9, so D3[B] is not outside D3[Q0]. Therefore, D3 shows att(B) C {v6} U r0.
Again we recall that B is embedded in D in RP2. If B is embedded in the face bounded by [a, r9, v0, s0, v5, r5, b, a, a], then b = v6 and the only other possible attachment for B is v0, as required. If B is embedded in the face bounded by [b, r5,v6]r6 r7 r8[v9, r9, a, a, b], then a = v0 and again this is the only possible attachment other than v6, as required.	■
Let N be the graph |J B .
BeN
LEMMA 8.10. Let ((G, H, n, 7)) be an fsq. Then there are not disjoint (N n r*)(N n r+5)-pa,ths in N. In particular, if Q* = Q0 and |N| > 2, then either every B e N has only v0 as an attachment in r9 r0 or every B e N has only v5 as an attachment in r4 r5.
Proof. Suppose by way of contradiction that Pi and P2 are disjoint r*r+5-paths in N, with, for j = 1, 2, Pj having the end pj in r* and the end qj in r+5. Choose CO	the labelling so that, in r*, p1 is closer to v9 than p2 is. There are three possibilities
CO	for how Pi and P2 are embedded by n: both in the (closed) disc contained in D
bounded by [a, r9,v0]r0 r1 r2 r3 r4 [v5,r5, b]a (recall that a = 7 n D); both in the disc in D bounded by [b, r5, ve]r6 r7 r8[v9, r9, a]a; or one in each of these discs. In all cases, we conclude that q1 is closer in r+5 to v6 than q2 is. Summarizing, we have the following.
Fact 1 Any two disjoint r*r+5-paths in N overlap on Q*.
For Q* = Q4 we are done: Corollary 8.7 implies D2[Q4] bounds a face of D2[H — (s2>]. Both P1 and P2 have ends in both r* and r+5, so both must be inside D2[Q4], yielding the contradiction that they cross in D2[Q4].
Now suppose Q* = Q0. For i = 2, 3, Dj[Q0] is not self-crossing; thus Fact 1 implies that Di[P1] and Dj[P2] are on different sides of Dj[Q0]. If Q0 is clean in Dj, then we have a contradiction, as no face of Dj[H — (sj>] is incident with both r* and r+5 except the ones bounded by Q4 and Q0.
Thus, Q0 is crossed in Dj. By Lemma 8.9, the one that is outside is one of v0v4, v0v6, v1v5, and v5v9. We treat in detail that this one is v0v4, as the other cases are completely analogous. It is in D3 that v0v4 is outside D3[Q0].
o
00

00
00	Because q1 is closer to v6 than q2 is, q1 cannot be v4; it follows that it is P2
that is v0v4. Lemma 8.9 also implies that P2, that is v0v4, is not outside D2[Q0] and, therefore, it is inside D2[Q0]. Thus, P1 is outside D2[Q0]. By Lemma 8.9, P1 is one of v0v6 and v1v5. By choice of the labelling, it cannot be that v1 is an end of P1, so P1 = v0v6, which is not disjoint from P2 = v0v4, a contradiction. We conclude that there are not such disjoint paths.
For the "in particular", there is a cut vertex u of N separating N H (rg r0) and N H (r4 r5) in N, as claimed. As s0 is a ([rg r0]) ([r4 r5])-path in N, we deduce u G s0. If B0 is not the only member of N, then any other element B of N shares the vertex u with B0, so u is an attachment of both. But u G s0 implies u G {v0, v5}.
As a final preparatory remark, we have the following.
Lemma 8.11. Let ((G, H, n, 7)) be an fsq. Let B and B' be distinct elements of N. Then:
(1)	B and B' do not overlap on Q*; and
(2)	either B overlaps Mq* on Q* or Q* = Q4 and B is either V4V9 or V0V5.
Proof. In the case Q* = Q4, Corollary 8.7 and Lemma 8.8 imply B and B' are both drawn inside the face of D2[H — (s2)] bounded by Q4 and, therefore, they do not overlap, yielding (1) for Q4.
For Q* = Q0, if both B and B' are in the same face of either D2 [Q0] or D3[Q0], then they obviously do not overlap on Q0. Thus, we may assume one is outside D2[Q0] and the other is inside D2[Q0] and that one is outside D3[Q0] and the other is inside D3[Q0].
By Lemma 8.9, the one outside D2[Q0] is either v1v5 or v0v6, while the one outside D3[Q0] is either v0v4 or v5vg. Thus, we may assume B G {v1 v5,v0v6} and B' G {v0v4,v5vg}. But none of the four possibilities is an overlapping pair, which is (1) for Q0.
As for overlapping Mq* , we suppose first that B has an attachment x in the interior of one of r* and r+5. (The "in particular" part of Lemma 8.10 implies this is always the case when Q* = Q0.) In this case, it is a simple exercise to see that x, together with any attachment of B in the other one of r* and r+5, are skew to at least one of the pairs of diagonally opposite corners of Q* (in the case of Q4 these pairs are {vg,v5} and {v4,v0}; for Q0, they are {vg,v6} and {v4,v1}). Thus, B overlaps Mq* .
In the remaining case, Q* = Q4 and att(B) C {vg, v0, v5, v4}. If both vg and v5 are attachments, then B is again skew to Mq* ; the same happens if both v0 and v4 are attachments. The only remaining cases are: att(B) = {v4,vg} and {v0,v5}, as claimed.	■
m
The next result contains the essence of the proof of Lemma 8.4.
Lemma 8.12. Let ((G, H, n,Y)) be an fsq. Suppose B1 G N, Bk = Mq* , and B1, B2, .. ., Bk is an induced cycle in OD(Q*). Then either
(1)	Q* = Q0, k = 3, and B2 G {v1v4,v6vg} or
(2)	k is even and Bfc_1 G NU {V1V4, V6Vg}.
• I
Proof. Case 1. k is odd.
o
CD O
csr csr
CO	Theorem 5.23 implies OD(Q4) is bipartite. Therefore, Q* = Q0 and so is
exposed in n.
For i = 2, 3, let ej be the edge of sj incident with vj and let Dj be a 1-drawing of G — ej. Theorem 5.23 implies Qj has BOD; Lemma 5.9 implies Qj is crossed in
Dj.	_	_
If, for some i G {2, 3}, Q0 is clean in Dj, then Lemma 5.6 implies Q0 has BOD, yielding the contradiction that k is even. Therefore, Q0 is crossed in both D2 and D3.
CD	Claim 1. If some Bj is either viv4 or v6v9, then i = 2 and k = 3.
Proof. Since both v1v4 and v6v9 overlap Mq , neither is in N, B1 is in N, and the cycle is induced, it must be that i = k — 1. For sake of definiteness, we CO	suppose Bk-i = V1V4; the alternative is treated completely analogously.
Because Bk-1 = v1v4, we deduce that D2 is the left-hand one of the two drawings in Figure 8.1, while D3 is the right-hand drawing in Figure 8.2; in both drawings, Bk-1 is outside Q0.
We note that B0 overlaps v1v4, so if B1 is B0, then k = 3, as claimed. Otherwise, B1 G N\ {B0}. By Lemma 8.10, either the only attachment of B1 in r9 r0 is v0 or the only attachment of B1 in r4 r5 is v5. For sake of definiteness, we assume the former; the latter is completely analogous. In order not to overlap v1v4, the only attachment for B1 in r4 r5 is v4. Therefore, either k = 3 and we are done, or B1 is just the edge v0v4. We show that B1 = v0v4 is not possible.
Suppose that B1 = v0v4. Because we know D2, we see that D2[B1] = D2[v0v4] is inside D2[Q0], while D2[Bk-1] = D2[v1v4] is outside. In D3, both are outside. But this is impossible, as B1, B2, B3,..., Bk-2, Bk-1 alternate sides of Q0 in both D2 and D3.
We conclude that B1 = v0v4 is impossible and therefore k = 3, as claimed. □
It remains to show that no other possibility can occur with k odd. So suppose no Bj is either v1v4 or v6v9. Suppose some Bj other than B1 is in N. As Bj overlaps Mq and the cycle B1, B2,..., Bk is induced, Lemma 8.11 implies i = k — 1. The same lemma implies k > 5. Therefore, Lemma 5.16 implies B1, B2,..., Bk-2, Bk-1 alternate sides of n[Q0]. Since k is odd, B1 and Bk-1 are on different sides of n[Q0], contradicting the fact that both are in N. Hence no other Bj is in N.
By Lemma 8.9, for at least one i G {2, 3}, Dj[B1] is inside Dj[Q0]. For the sake of definiteness, we consider the case i = 2 and D2 is the left-hand drawing of H — (s2) in Figure 8.1; the remaining cases are completely analogous. Thus, either B1 is B0 or B1 is either a Q0- or a Q1-bridge.
Since k is odd, Bk-1 is on the other side of D2[Q0] from B1. Therefore, Bk-1 is outside D2[Q0]. In order to understand how Bk-1 can overlap Mq in D2, we analyze D2 Mq ].
Let e be the edge of Mq that is crossed in D2. The end w of e outside D2[Q0] is in Nuc(Mq). If the other end u of e is not in Nuc(Mq), then u = v6 and [x,r6, v6] is the only part of Mq inside D2[Q0]. Otherwise, Nuc(Mq ) — {e2,e} is not connected. Since Nuc(Mq )^e2 is connected, Nuc(Mq ) — {e2, e} consists of the component inside D2[Q0] and the component O outside. In particular, Mq —{e2, e} consists of two Q0-bridges in G — {e2, e}. Let I be the one contained inside D2[Q0] and let O be the one outside. All attachments of M^ are attachments of either I
a
Q
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o o
cm 00
¡5
00	or O, and possibly both. In the case u = v6, we take I to be the portion of e from
x to v6.
We observe that D2 shows that, except for one end of e, all the attachments of
I are in Q0. On the other hand, Theorem 7.1 implies that MQ , and, therefore I,
Q0
has no attachment in (s0). The embedding n shows that I has no attachment in (r0): otherwise, I is not just [x,e6, v6] and u = v6. Thus, the simple closed curve si ri r2 r3 s4[v9, r9, a]a[b, r5, v6] bounds a closed disc in RP2 separating u from (r0) and is disjoint from Nuc(1) U (r0). Unless v0 = a, the same simple closed curve separates u from v0; thus, if v0 is an attachment of I, then a = v0.
Because Bk-i is outside D2[Q0] and att(Bk-i) C Q0, there are four candidates for the face of D2[H — (s2)] that contains Bk-i. The one bounded by Q3 is not possible: if Bk-i were in that face, it would not overlap Mq , as all the Mq
attachments there would be in s4 and, therefore, all in O and not in I; both Bk-i and O being outside D2[Q0] shows they do not overlap.
The face of D2[H — (s2)] incident with [x,r0,vi] is not a possibility for Bk-i for exactly the same reason: the only attachment of I there can be and is not
part of a pair of attachments of Mq that are skew to two attachments of i,
Q0
which are all contained in [x,r0,vi].
The face of D2 [H — (s2)] incident with r8 r9 is also not a possibility for Bk-i. To see this, v0 is the only possible attachment of I in the boundary of this face. Thus, v0 is an attachment of I and Bk-i must have attachments in each of [v9, r9, v0) and (v0, r0, x]. However, in n we must have a = v0 and then there is no way to embed Bfc_i.
cm	Therefore, Bk-i is in the face of D2[H — (s2)] incident with r5 si.
By way of contradiction, suppose Bk-i is outside D3[Q0]. Identical arguments as those just above show that Bk-i is in the face of D3[H — (s2)] incident with r9 s4. Because the previous paragraph shows att(Bk-i) C r4 r5 si, it cannot overlap Mq
using an attachment of the portion of Mq that is inside D3[Q0] and, therefore, it cannot overlap Mq at all, a contradiction. Therefore, Bk-i is inside D3[Q0]. This implies Bk-i is either a Q0- or Q4-bridge. CO	If Bfc_i is a Q4-bridge, then att(Bfc_i) C r4 (because of D2). Letting r denote
CO	the minimal subpath of r4 containing att(Bk-i), D2 shows that no attachment of I
is in (r) and, because O and Bk-i do not overlap (in D2), O also has no attachment in (r). Consequently, Bk-i does not overlap Mq , a contradiction. Therefore, Bk-i is a Q0-bridge.
Because Bk-2 is inside D2[Q0], has no attachments in s0, and overlaps Bk-i as Q0-bridges, we see that Bk-2 is also a Q0-bridge. Continuing back, we see that each of Bk-3, ..., B2 is a Q0-bridge and that Bi is outside D3[Q0]. By Lemma 8.9, Bi is either v0v4 or v5v9. But neither of these overlaps B2. This contradiction shows that, except for the case described in Claim 1, k is even.
Case 2. k is even.
For each i = 2, 3,..., k — 2, B, UQ* has no non-contractible cycle in RP2. Thus, Lemma 5.16 implies Bi and Bk-i are on the same side of Q* in RP2; since Bi is Q*-exterior, we have that Bk-i is Q*-exterior. If n[Q* UBk-i] has no non-contractible cycle, then Lemma 5.16 shows that it cannot overlap Mq* , a contradiction. In the case Q* = Q4, this implies that Bk-i is in N, while if Q* = Q0, then Bk-i is in
NU{viV4,V6 V9}.	I
o
CM
CD
CD
00
CO cm cm
CO	Proof of Lemma 8.4. We show that any odd cycle C in OD(Q0) contains either
v1 v4 or v6v9. Theorem 5.23 (3) implies that OD(Q0) — Mq is bipartite. Therefore, C contains Mq . Lemma 8.12 shows that any odd cycle in OD(Q0) containing Mq and an element of N has length 3 and contains one of v1v4 and v6v9, as required.
Thus, we may suppose C avoids NU{v1v4, v6v9}; let C = (B1, B2,..., B2k, Mq ) For each i = 1, 2,..., 2k, n[Bj U Q0] has no non-contractible cycles in RP2. Lemma CD	5.16 implies Bj and Bj+1 are on different sides of n[Q0]. From this, parity implies
that B1 and B2k are on opposite sides of n[Q0]. On the other hand, they are both on the side of n[Q0] not containing Mq , a contradiction.	■
We are now prepared for the proof of Theorem 8.1.
Proof of Theorem 8.1. By Theorem 3.5, G has a representativity 2 embedding n in RP2. For (1), if no spoke is exposed in n, then we are done; thus, with the standard labelling, we may suppose that s0 is exposed in n. From Theorem 7.1, we know that the Q0-bridge B0 containing s0 is different from Mq . From Lemma 8.4, we know that OD(Q0) — {v1v4,v6v9} is bipartite and from Theorem 5.23 (3), we know that (Mq )# is planar.
We need to modify n so that the set N (Definition 8.5) becomes empty. We start with terminology that will be useful for the next claims.
Definition 8.13. Let L be a graph. A path (v1, v2,..., vk) in L is chordless in L if there is no edge vjvj- of L that is not in P except possibly v1 vk.
I
The following is a simple consequence of Lemma 8.12.
Claim 1. (1) If Q* = Q0, then every NMq -path in OD(Q0) of length at least two contains one of v1v4 and v6v9. (2) If Q* = Q4, then every chordless NMq4-path in OD(Q4) of length at least two has length exactly two, one end is either v4v9 or v0v5, and that end does not overlap Mq4 .
Proof. Suppose first that Q* = Q0. Let P be any NMq -path in OD(Q0) that has length at least 2. We may assume P is chordless: otherwise there is a shorter NMq -path P' of length at least 2 and V (P') C V (P); if P' contains either v1 v4 or v6v9, then so does P. By Lemma 8.11 (2), the ends of P are adjacent in OD(Q0). Thus, P together with this edge of OD(Q0) makes an induced cycle. As this cycle has only one vertex in N, Lemma 8.12 implies the cycle has length 3 and contains one of v1v4 and v6v9.
Now suppose that Q* = Q4 and P = (B1, B2,..., Bk, Mq4) is a chordless NMq4-path in OD(Q4) of length at least 2. Then B1 G N. Since P is chordless and Bk G N, Lemma 8.12 (2) implies B1 does not overlap Mq4 . Now Lemma 8.11 (2) implies B1 is either v4v9 or v0v5. Thus, B2 is skew to B1. Since att(B1) C att(MQ4), B2 is also skew to Mq4 . Since P is chordless, k = 2, as required.	□
m	-
If Q* = Q0, then set M to be the set {Mq , v1v4, v6v9}, while if Q* = Q4, then set M to be the set {Mq4, v4v9, v0v5}. In either case, let M- = M \ {Mq* }.
Let N+ be the set of Q*-bridges B so that there is an NB-path in OD(Q*) that is disjoint from AL The next lemma shows that J\f+ consists of the members
a	j
o
CM
CD O
00
CD
00	of N, which have attachments in both r* and r+5, and other Q*-bridges B that
simply extend out along either r* or r+5. This structure is what will allow us to find natural "breaking points" a' and b' in r* and r+5, respectively, to allow us to "flip" the members of N into M, yielding the embedding with H C M and N = 0.
Claim 2. If B g N+, then att(B) C r* U r+5. Furthermore, if B G N+ \N, then either att(B) C r* or att(B) C r+5.
Proof. Let P be a shortest NB-path in OD(Q*) that is disjoint from M. We proceed by induction on the length of P.
If B G N, then the result follows from Lemma 8.8. Otherwise, B G N. The neighbour B' of B in P is closer to N than B is, so att(B') C r* U r+5.
If B overlaps Mq. , then P extends to a chordless NMq.-path in OD(Q*) -M-of length at least 2. This contradicts Claim 1, showing B does not overlap Mq. .
Suppose by way of contradiction that B has an attachment x in the interior of some H-spoke s contained in Q*. As B overlaps B' and att(B') C r* U r+5, not all attachments of B can be in [s]. But any attachment y of B in Q* - [s] combines with x to show that B is skew to the ends of s and, therefore, overlaps MQ. . Therefore, att(B) C r* U r+5.
Next suppose that B has an attachment in (r*). If B also has an attachment in Q* - [r*], then B overlaps Mq. (the two identified attachments of B are skew to the two ends of r*). Thus, if B has an attachment in (r*), then att(B) C r*. Likewise, if B has an attachment in (r+5), then att(B) C r+5.
If B has an attachment in each of r* and r+5, then the preceding paragraph shows that att(B) consists of some of the four H-nodes that comprise the ends of r* and r+5. Because B overlaps B', att(B) cannot be just the two ends of one of CM	the two H-spokes in Q*. In the remaining case, B is skew to Mq. , a contradiction.
Thus, either att(B) C r* or att(B) C r+5.	□
cm	- - +
Let OD-(Qo) = OD(Qo) - {v1v4,v6v9} and let OD-(Q4) = OD(Q4). By Lemma 8.4 or Theorem 5.23 (1), OD-(Q*) is bipartite; let (S, T) be a bipartition of OD-(Q*), with Mq. G T. We briefly treat separately the cases Q* = Qo and
Q* = Q4.
For the former, every element of N overlaps Mq and so N C S. There is an embedding $ of (G - {v1v4,v6v9}) - Nuc(Mq ) in the plane so that all the Qo-bridges in N are on the same side of $[Qo].
In the case of Q* = Q4, N\ {v4v9,vov5} C S. There is an embedding $ of G - Nuc(Mq4) in the plane so that all the Q4-bridges in N\ {v4v9, vov5} are on the same side of $[Q4]. Any of v4v9 and vov5 that is also in S can also be embedded on that same side of $[Q4].
Among the attachments of the elements of N+, let a9 be the one in r* nearest v9 and let a4 be the one in r+5 nearest v4.
Claim 3. No Q*-bridge not in M is skew to {a4, a9}.
Proof. It is clear that, in the case Q* = Q4, neither v4v9 nor vov5 is skew to {a4, a9}. We show that a Q*-bridge not in M that is skew to {a4, a9} must overlap some Q*-bridge in N+; this implies the contradiction that it is in N+.
By the Ordering Lemma 4.8, the elements of N O S occur in order on Q* in $. Thus, there is one element B' of N O S that has both an attachment nearest to v4 (relative to r*) and an attachment nearest to v9 (relative to r+5). Let x'
o
CM
00
o
CO	and y' be the attachments of B' nearest V4 in r* and V9 in r+5, respectively. In
the case Q* = Q°, B° is a candidate for B', so, even in this case, we have that x' G [v4,r4,«5] and y' G [v9,r9,v°].
Suppose by way of contradiction that some Q*-bridge B'' not in M has attachments x'' and y'' in the two components of Q* — {a4, a9}. We note that, when Q* = Q4, B'' = V4V9 and B'' = v°«5.
If one of x'' and y'' is in the component of Q* — {x',y'} that is disjoint from s4 — {x',y'}, then B'' overlaps B'. Since B' G N, Lemma 8.11 implies B'' G N and, therefore, B'' G N+. But this contradicts the definition of either a4 or a9 and, therefore, both x'' and y'' are contained in the component of Q* — {x', y'} that contains s4 — {x',y'}. In particular, we may assume y'' G (a4,r4,x'] U (a9,r9,y']. For the sake of definiteness, we assume y'' G (a9,r9,y'].
Some Q*-bridge B+ in N+ has a9 as an attachment; since y'' is in (a9,r9,y'], y' = a9 and, therefore, B+ is not in N. There is a shortest path P = (B', B1,..., Bn) in OD-(Q*) — Mq* from B' to some element Bn of N+ so that Bn has an attachment yn in [a9, r9, y''); choose yn so that it is as close to a9 in [a9, r9, y'') as possible.
The Q*-bridge Bn-1 is in N+ and so, by minimality of n, does not have an attachment in [a9,r9,y''). Since Bn overlaps Bn-1, there is an attachment zn of Bn in (y'', r9, x']. Since B'' is skew to {a4, a9}, there is an attachment z'' of B'' in (a9,r9,v9]s4[v4,r4,a4). But now zn, y'', yn, and z'' show B'' overlaps Bn. Since B'' G M, B'' is in N +. But this contradicts the definition of a4 or a9.	□
The following is immediate from Claim 3.
Claim 4. Each Q*-bridge not in M has all its attachments in one of the two subpaths of Q*. □
The proof now bifurcates into the two cases. We consider first the case Q* = Q° and that s° is exposed in n. The following is immediate from Claim 4.
Claim 5. The planar embedding $ of (G — {v1v4,v6v9}) — Nuc(Mq ) has the property that there is a simple closed curve in the plane that meets $[(G — {v1v4, v6v9}) — Nuc(Mq*)] precisely at a4 and a9.	□
CO
We are now prepared to describe a representativity 2 embedding of G in RP2 CO	so that all H-spokes are in M.
Let ^ be an embedding of H in RP2 so that all H-spokes are contained in the Mobius band M^ bounded by ^[R] and let be a non-contractible, simple, closed curve that meets H in precisely the points a4 and a9. The claim is that this embedding extends to an embedding of G so that meets G only at a4 and a9.
Claim 4 implies that we can add all the Q°-bridges other than v1 v4, v6v9, and
M^ to ^ so that there is no additional intersection with y$. It remains to show
Q0 _
that we may also add the at most three remaining Q°-bridges.
Claim 6. At most one of v1v4 and v6v9 is in G.
Proof. Suppose both are in G. We consider a 1-drawing D2 of G—(s2). As Q2 must be crossed in D2 (it has BOD and s2 is contained in a planar Q2-bridge; apply Lemma 5.9), we conclude that r° r1 r2 r3 crosses r5 r6 r7 r8 in D2. In particular, s° and s4 cannot be exposed.
In order for v1v4 to be not crossed in D2, we must have the crossing in r°. Likewise, v6v9 implies the crossing is in r5. But then neither r1 r2 nor r6 r7 is crossed, so Q2 is not crossed in D2, a contradiction.	□
a	2
o
CD O
00
CO CD
a
CD CU
00	We note that v1v4 and V6V9 are not symmetric: the embedding n of G in RP2
distinguishes these two cases. However, it is easy to add either of these to ^ so that the newly added edge is in the closed disc D^ bounded by ^[R] in
Finally, it remains to show that we may also add Mq to Here the argument depends slightly on which of v1v4 and v6v9 occurs in G. We will assume, for the sake of definiteness, that it is v1v4 that occurs; the argument in the other case is completely analogous. We shall simply import n[MQ ] in RP2 as its embedding in
To this end, let B be any H-bridge contained in Mq so that n[B] C D. We show that either att(B) C r0 r1 r2 r3 [v4, r4, a4] or att(B) C r5 r6 r7 r8 [v9, r9, a9].
We begin by observing that such a B cannot overlap v1v4 (as R-bridges), as both are are embedded in D by H An analogous discussion applies if v1v4 is replaced by v6v9.
The embedding n shows B cannot have an attachment in each of (r1 r2 r3) and (r5 r6 r7 r8 r9). Likewise, B cannot have an attachment in each of (r6 r7 r8) and r0 r1 r2 r3 r4. The next claim treats the remaining possibilities.
Claim 7. The H-bridge B does not have an attachment in each of (r1 r2 r3) and (a4,r4,v5]. Likewise, B does not have an attachment in each of (r6 r7 r8) and either r5 or (09, r9, V0].
in OD(Q0) — {v1v4, v6v9, Mq } joining some Bn in N to a Q0-bridge Bn+ so that Bn + has an attachment in [v4,r4,x).
If Bn + G N, then Bn + C D. Lemma 8.8 shows Bn + has an attachment in each of r* and r+5; therefore, Bn + is not contained in the closed disc bounded by P and a subpath of r1 r2 r3 r4, Bn + and P must cross in H Therefore, Bn + G N+ \ N.
The neighbour B^- + of Bn + in S does not have an attachment in [v4, r4,x). Since Bn + overlaps B^- +, it follows that Bn + has another attachment in (x,r4, v5, r5, b]. In particular, the edge e of [v4,r4,x] incident with x is H-green because of
Bn +.
On the other hand, if either x = v5 or y G (r1), then P combines with the xy-subpath of r1 r2 r3 [v4,r4,x] to make another H-green cycle containing e, contradicting Theorem 6.7. Therefore, x = v5 and y G (r1). But then att(B) C Q0, contradicting the fact that B C Mq^ .
The "likewise" statement has an analogous proof.	□
Proof. Suppose by way of contradiction that B has an attachment x in (a4,r4,v5] and an attachment y G (ri r2 r3). Let P be an H-avoiding xy-path in B. Since a4 is an attachment of some element of N+, there is a shortest path S
We now see that ^ may be extended to include n[M^], completing the proof when Q* = Q0.
The proof will be completed by now considering the case Q* = Q4. The only difference in how we proceed is to note that the H-bridges v4v9 and v0v5, if they exist, may be transferred to M at the start. To see this, first observe that v4v9 and
CD
v0v5 overlap on R and so cannot both be embedded in D. If v4v9 is not contained in M, then we may consider H' to be (H — (s4)) + v4v9, relabel H' so that v4v9 — the exposed spoke — is s0 and proceed as above to move v4v9 into M.	■
The following notions will be helpful for the duration of the work.
o
m
CD
$H
CD
m
u
a
CD U
00	Definition 8.14. Let G be a graph, V10 = H C G and let B be an H-bridge
in G.
(1)	If there is an i G {0,1, 2, 3,4} so that att(B) C Qj, then B is both a local H-bridge and a Qi-local H-bridge.
(2)	Otherwise, B is a global H-bridge.
Corollary 8.15. Let G G M2 and V10 = H C G. Then there is no i so that Qi has BOD and each edge of ri-2 ri-1 ri ri+1 is in an H-green cycle consisting of a global H-bridge and a path in R having at most two H-nodes other than vi.
Proof. By way of contradiction, suppose there is such an i. By Theorem 8.1, G has a representativity 2 embedding in RP2 so that H C M. Thus, si is in a Qi-bridge other than Mq .
By Lemma 6.6 (10), no edge of ri-2 ri-1 ri ri+1 can be crossed in any 1-drawing D of G — (si). By hypothesis, Qi has BOD, so Lemma 5.9 implies Qi is crossed in D, which further implies that some edge of ri-2 ri-1 ri ri+1 is crossed in D, a contradiction.	■
00 1-H
o
cm
00
u
CD
csr
CM
CO CO
CHAPTER 9
Parallel edges
In this very short chapter, we present some observations on how parallel edges can occur in 2-crossing-critical graphs. This will be used in later sections, especially Section 15, where we determine all the 3-connected, 2-crossing-critical graphs that do not have a subdivision of V8. There are easy generalizations to k-crossing-critical graphs.
Definition 9.1. For an edge e of a graph G, ^(e) denotes the number of edges parallel to e (including e itself).
Observation 9.2. Let G be a 2-crossing-critical graph and let e and e' be
parallel edges of G. Then: , _ _
(1)	if G is the underlying simple graph, then G is not planar;
(2)	the edge e' is crossed in any 1-drawing of G — e;
(3)	Me) < 2;
(4)	if e' is an edge parallel to e, then G — {e, e'} is planar;
(5)	if cr(G) > 2, then G is simple^ and
(6)	if n > 4 and V2n — H C G, then one of e and e' is in the H-rim.
—
Proof. For (1), a planar embedding of G allows us to introduce all the parallel edges of G with no crossings, showing G is planar, a contradiction.
For (2)-(5), let D be a 1-drawing of G — e and suppose e' is not crossed in D Then we may add e alongside D[e'] to obtain a 1-drawing of G, a contradiction. Since D has at most one crossing, it must be of e', which is (2). Adding e alongside D[e'] yields a 2-drawing of G. Thus we have (4) and (5). Also, (3) follows, since any other edge e'' parallel to e does not cross e' in De. Thus, e'' is not crossed in De, which contradicts the second sentence, with e'' in place of e'.
Finally, for (6), we may suppose e is not in H. Lemma 3.6 shows that the only edges that are in every non-planar subgraph of G — e are those in the H-rim. Therefore, e' is in the H-rim.	□
CO CD
$H
CD CO
$H
a
61
$H
00 1-H
o
00
u
CD
O
$H
CD
CHAPTER 10
Tidiness and global H-bridges
In this section, we show that, if G e M3 and V10 = H C G, then there is a V10 = H' C G with many useful additional characteristics that we call "tidiness". The main result is that a tidy subdivision of V10 has only very particular global bridges, each of which is an edge. We start with a slightly milder version of tidiness.
Definition 10.1. Let n be a representativity 2 embedding of G in RP2 and let V10 = H C G. Then H is n-pretidy if:
(1)	all H-spokes are embedded in M; and
(2)	for every H-quad Q and for every Q-bridge B other than Mq, Q U B has no non-contractible cycle in H
The first step in this section is to find an embedding with a pretidy subdivision of V10.
Lemma 10.2. Let G e M3 and V10 = H C G. Then G has a representativity 2 embedding n in RP2 so that H is n-pretidy.
Proof. By Theorem 8.1, G has a representativity 2 embedding n in RP2 so that all the H-spokes are contained in M and so that, for any Q4-bridge B other than Mq4, n[Q4 U B] has no non-contractible cycle. We note that every global H-bridge is contained in D. We describe a particular representativity 2 embedding n* of G in RP2 for which H is n*-pretidy. Let 7 be the non-contractible simple closed curve that meets n(G) at just the two points a and b.
CO
The embedding n* is obtained by adjusting the local H-bridges; we do not adjust those that are Q4-local. We start with n* being the same as n on H and all the Q4-bridges other than Mq4 . Let Q be an H-quad other than Q4. By Theorem 5.23, Q has BOD and all Q-bridges other than Mq are planar. Let (S, T) be a bipartition of OD(Q) labelled so that Mq e T. Let nQ be a planar embedding of Q and all the Q-bridges other than Mq so that all the Q-bridges in T \ {Mq} are on one side of nQ [Q] and all the Q-bridges in S are on the other side of nQ [Q].
Extend n* to include all the Q-bridges other than Mq by placing the Q-bridges in S into the H-face in n* bounded by n*[Q], using nQ. As every Q-bridge in T \ {Mq} does not overlap Mq, each of these has all its attachments on one of the four H-branches in Q and these may be embedded in n* on the other side of n* [Q], and without crossing Mq U 7.
The only concern here is that a local H-bridge can be local for distinct H-quads. Such an H-bridge B must have all its attachments on the same H-spoke sj. We claim it is in T for one of Qj-1 and Qi and in S for the other one of Qj-1 and
Qi.
As G is 3-connected, OD(Qi) is connected (see [6, Thm. 1], where this is proved for binary matroids). There is a shortest MQiB-path P = (B0, B1,..., Bn)
o
CM
CD O
00
o
cm
I
00 cm cm
in OD(Qj) (thus, B0 = Mq and Bn = B). Let k be least so that Bk has an attachment in (sj).
Claim 1. For j > k, att(Bj) Ç s,, and k < 1.
Proof. If, for some j > k, Bj has an attachment not in s,, then j < n. If Bj has an attachment in (sj), then Bj is skew to Mq and P is not a shortest MqB-path, a contradiction. Thus, there is a least j' > j so that Bj/ has an attachment in (sj). Since Bj/ overlaps Bj/-1 and Bj/-1 has no attachment in (sj), Bj/ has an attachment not in s,. Again, Bj/ is skew to Mq , so P is not a shortest MqB-path, a contradiction. Thus, for all j > k, att(Bj) Ç s,.
If k = 0, then obviously k < 1, so we may assume k > 1. As Bk has an attachment in (s,) and Bk-1 does not, it follows that Bk has an attachment not in s,. But then Bk is skew to Mq . Because P is a shortest Mq B-path, we deduce that k < 1.	□
The claim shows that the Qj-bridges Bk+1, Bk+2, ..., Bn are also Qi-1-bridges and, therefore, (Bk+1, Bk+2,..., Bn) is a path in OD(Qj-1). Suppose first that k = 0. Then MQi contains a vertex x in (sj) so that x and vj+1 are skew to B1. There is a shortest Qj-avoiding path P in MQi joining x to a vertex in Nuc(MQi) n H. Since P is not in the face of n[Qj] contained in M, we deduce that P is contained in the face of n[Qj-1 ] contained in M. But then we conclude that P is contained in a Qj-1-local H-bridge B', showing that B' is skew to both MQi-1 and to B1. We deduce that, in OD(Qj), MQi and B1 are on opposite sides of the bipartition of OD(Qj), while MQi-1 and B1 are on the same side of the bipartition of OD(Qj-1). Since B1 and B = Bn have not changed their relative positions, we see that in
one of OD(Qj) and OD(Qj_i), B is on the same side of the bipartition as the
corresponding Mobius bridge, while in the other B and the other corresponding Mobius bridge are on opposite sides of the bipartition.
The argument works exactly in reverse when k = 1. In this case, B1 is skew to MQi and B2. Since B1 C MQi-1, we conclude that B2 is skew to MQi-1, and the result follows analogously to the argument in the preceding paragraph.
Finally, suppose B is a global H-bridge. Then, for each H-quad Q, B C Mq, so B does not overlap any of the Q-local H-bridges already embedded in Dn* and, since n[B] C D, B can also be added to n*.	■
We are now ready to move to tidiness.
Definition 10.3. Let V10 = H C G and let n be a representativity 2 embedding of G. Then H is n-tidy if:
(1)	H C M;
(2)	every local H-bridge is contained in M;
(3)	for each H-quad Q, no two Q-local H-bridges overlap; and
(4)	there is no H-avoiding path P in D and an index i G {0,1, 2,..., 9} so that P has both its ends in (vj, rj, vj+1, rj+1, vj+2, rj+2, vj+3).
If V10 = H C G, then H is tidy if there is a representativity 2 embedding n of G so that H is n-tidy.
Our aim is the following result.
Theorem 10.4. Let G g M2
have a subdivision of V10. Then there exists a representativity 2 embedding n in RP2 of G with a n-tidy subdivision of V10.
o
CM
o» o
cm
i
cm cm
CO
00	The following concept is central to the proof.
Definition 10.5. Let Vi0 = H C G. Then Loc(H) denotes the union of H and all the local H-bridges in G.
n
Proof of Theorem 10.4. For any Vi0 = H C G, Lemma 10.2 implies there is a representativity 2 embedding n of G in RP2 so that H is n-pretidy. Among all H for which Loc(H) is maximal and all n so that H is n-pretidy, we consider the pairs (H, n) so that G n Mn(H) is maximal. Among all these pairs (H, n), we choose one for which the number of edges of G in H-spokes in minimized. We claim that this H is n-tidy. We note that (1) is satisfied by the fact that H is n-pretidy.
If H and n fail to satisfy either (2) or (4), then either there is an H-quad Q so that some Q-local H-bridge B is not embedded in MH, or there is an H-avoiding path P contained in and an index i e {0,1, 2,..., 9} so that P has both ends in (r, r.j+i r.j+2). In the first
case, as Q U B has no non-contractible cycles, the only 1	possibility is that B has all its attachments in one of the H-rim branches of Q.
Thus, the first case is a special case of the second; we now consider the second case.
Let P' be the subpath of (r, r^i r.j+2) joining the ends u and w of P, with the labelling chosen so that u is nearer to v, in P' than w is. Note that the cycle PU P' is an H-green cycle and, therefore, bounds a face of G.
We construct a new subdivision H' of Vi0 in G. The H'-rim is obtained from the H-rim by replacing P' with P. The spokes s,, s.j+3, and s.j+4 of H' are also spokes of H'. The H-spokes s^i and s.j+2 might need extension, using the subpaths of r, r^i r.j+2 joining u and/or w to either v^i or v.j+2 as necessary, to become spokes of H'. Evidently all H'-spokes are contained in , so H' C GnMh' C Loc(H'). Furthermore, if F is the (closed) face of G bounded by PUP', then MH' = MH UF.
Claim 1. Loc(H) C Loc(H').
Proof. Let e be an edge of Loc(H). If e e MH, then e e Loc(H'), so we may assume e e MH'. Let B be the local H-bridge containing e. Since e e MH' and Mh C , we deduce that B C , and so all attachments of B are in some H-rim branch (recall H is n-pretidy). Thus, Corollary 5.15 implies B has precisely two attachments and therefore is just the edge e. Consequently, B is disjoint from P (it is not in Mh'), and so B is an H'-bridge, whence e e Loc(H').	□
If P is not contained in a local H-bridge, then, since P C Loc(H'), we contradict maximality of Loc(H). Therefore, P is contained in, and therefore is, a local H-bridge B. But this implies that H' is n-pretidy and that G has one more edge in MH' than it has in MH, contradicting the maximality of G n MH. Therefore, (2) and (4) hold for (H, n).
It follows that, if H is not n-tidy, then (3) is violated: there exists an H-quad Q and two Q-bridges B and B' in (Mq)# that overlap. As both B and B' are contained in M, one, say B, is Q-interior in n, while B' is Q-exterior. This implies that att(B') C s, for some H-spoke s C Q. Corollary 5.15 implies that B' is just an edge uw. We note that B has an attachment x in (u, s, w) and an attachment y not in [u, s, w].
Let H'' be the subdivision of Vi0 obtained from H by replacing s with (s — (u, s, w))UB'. We note that H'' is n-pretidy, Loc(H') = Loc(H), and MH» = MH, so G n MH" is maximal. However, the H''-spokes have in total at least one fewer edge than the H-spokes, contradicting the choice of H.	■
00	We now turn our attention to the global H-bridges of a tidy H.
Theorem 10.6. Let G G M2 and V10 = H C G. If H is tidy, then any global H-bridge is just an edge, and, in particular, has one of the forms viVj+2, vivi+3, or has vi as one end and the other end is in (ri-3) U (ri+2).
Proof. Let n be a representativity 2 embedding of G for which H is n-tidy. In particular, all H-spokes and all local H-bridges are in M, and, for each i = 0,1, 2 ..., 9, no global H-bridge has two attachments in (ri ri+1 ri+2). Let B be a global H-bridge. We note that B C D.
Claim 1. If there is an i so that att(B) C ri ri+1 ri+2, then either B = vivi+2 or B = vi+1vi+3 or B = vivi+3 or B has vi as one end and the other end is in (ri+2) or B has vi+3 as one end and the other end is in (ri).

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Proof. Because H is tidy, no two attachments of B are in (r ri+i ri+2). Thus, at least one of v and vj+3 is an attachment of B; for the sake of definiteness, let it be Vj. Then tidiness implies no attachment of B can be in (rj rj+1). As tidiness also implies rj+2 has at most one, and therefore exactly one, attachment of B, the result follows.	□
Claim 2. If there is no i so that att(B) C ri ri+1 ri+2, then either att(B) = {v0, v5, z}, with z G (r2) U (r7) or att(B) = {v4, vg, z}, with z G (r1) U (r6).
Proof. We may assume that B is embedded in the (H U Y)-face contained in D and incident with v0, v1, ..., v4. As H is tidy and B is H-global, there exist
i, j € {9, 0,1, 2, 3, 4} so that (taking 9 to be equal to -1) i < j, B has attachments x in rj — vj+i and y in rj — Vj, and j — i > 3; choose such i, j so that j — i is as small as possible. By tidiness, there is no other attachment of B in
00
[rj-i rj rj+i) U (rj_i rj rj+i] .
Subclaim 1. Either i = —1 or j = 4.
Proof. In the alternative, i > 0 and j < 3. As j — i > 3, we conclude that CO	i = 0 and j = 3, so the six H-rim branches ri-1, ri, ri+1, rj-1, rj, and rj+1 are all
distinct and cover the entire a6-subpath in the boundary of (HU 7)-face containing B, with the possible exception of v2, in which case both x = v0 and y = v4.
Let e be an edge in s2 and let D be a 1-drawing of G — e. Theorem 5.23 implies Q2 has BOD; now Lemma 5.9 implies Q2 is crossed in D. In particular, r0 r1 r2 r3 crosses r5 r6 r7 r8 in D.
In the case v2 is an attachment of B, let P and P' be H-avoiding v0v2- and v2v4-paths in B, respectively. Then the cycles r0 r1 [v2,P, v0] and r2 r3 [v4,P', v2] are both H-green. Lemma 7.2 (1) implies neither is crossed in D, yielding the contradiction that r0 r1 r2 r3 is not crossed in D.
Thus, B is the edge xy. Note that B is not a local H-bridge and, therefore, not both v0 and v4 are attachments of B. As B is not crossed in D, we deduce that the xy-subpath of r0 r1 r2 r3 is also not crossed in D. Therefore, either r0 or r3 is crossed in D. From this, we conclude that, since Q2 is crossed in D, r6 r7 is crossed in D. Moreover, either s1 or s3 is exposed in D. By symmetry, we may assume s1 is exposed in D.
If x = v0, then the cycle r1 r2 r3 s3 r8 rg s0 r5 s1 is clean in D and separates x G (r1) from y G (v3, r3, v4], so B must be crossed in D, a contradiction. If y = v4,
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then the cycle r1 r2 s3 r8 s4 r4 r5 s1 is clean in D and separates x G [v0, r1, v1) from y G (r3), and again B is crossed in D, a contradiction.	□
Recall that —1 is equal to 9. The following is immediate from tidiness.
Subclaim 2. (1) If x g [a, r9,v0), then there is no attachment in [vo,ro,vi). (2) If y G (v4, r4, b], then there is no attachment in (v3, r3, v4].	□
The next two subclaims are rather less trivial.
Subclaim 3. (1) If x g [a, r9,v0), then there is no attachment in [v2,r2,v3). (2) If y G (v4, r4, b], then there is no attachment in (v1, r1, v2].
Proof. We prove (1); (2) is symmetric. For (1), suppose there is an at-
tachment y' in [v2,r2,v3). By tidiness, there is no attachment other than y' in (ro r1 r2 r3), and so minimality of j - i implies y' = y.
The only other possible attachment is in [v4, r4, b]. If there is an attachment z in [v4, r4, b], then either y = v2 or z = b = v5. Thus, either z does not exist and B is the edge xy, or z exists, B has exactly three attachments, namely x, y, and z, and Lemma 5.19 shows B is a K13. Let P and P' be the xy- and yz-paths (the latter only if z exists) in B.
Suppose first that y = v2. Then x = v9, as otherwise [y, P, x, r9, vo] ro r1 [v2, r2, y] is an H-green cycle with the three H-nodes vo,v1,v2 in its interior, contradicting Lemma 6.6 (9).
Theorem 5.23 (6a) does not apply, as x = v9 = a implies vo = a. If Theorem
cm
5.23 (6b) applies, then there is a second H-bridge B' attaching at b = v5 and in ro r1. But then B and B' must cross in n, a contradiction. Therefore, Theorem 5.23 (6) shows Q1 has BOD.
Let e be an edge of s1 and let D be a 1-drawing of G - e. Lemma 5.9 implies Q1 is crossed in D. On the other hand, the presence of P and Lemma 7.2 (3a) and (2) imply Q1 cannot be crossed in D, the desired contradiction.
Therefore, y = v2. Since x,y G r9 ro r1, the hypothesis of the claim implies z must exist. The cycles [x, P, v2]r1 ro[vo,r9,x] and [z,P',v2]r2 r3[v4,r4,z] are H-green. Let e be an edge in s2 and let D be a 1-drawing of G - e. Theorem 5.23 implies Q2 has BOD, so Lemma 5.9 implies Q2 is crossed in D. However, Lemma 7.2 (1) shows that ro and r3 are not crossed. If x = v9 , then the same result shows r1 is not crossed and likewise if z = v5, then r2 is not crossed. If, say, x = v9, then Lemma 7.2 (3b) implies r1 can only cross r8. However, if z = v4, then (2) shows r8 cannot be crossed.
In the remaining case, x = v9 and z = v4. In this case, a = x = v9. If Q1 does not have BOD, then Theorem 5.23 (6) implies b = v5 and there is a Q1-bridge
B' different from Mq , having attachments at b and in ro r1, and embedded in D.
Qi
But then B' is an H-bridge different from B that overlaps B on R, while both are embedded in D, a contradiction.	□
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Subclaim 4. (1) If x g [a, r9, vo), then there is no attachment in [v3, r3, v4). (2) If y G (v4, r4, b], then there is no attachment in (vo, ro, v1 ].
Proof. We prove (1); (2) is symmetric. For (1), suppose there is an attachment in [v3, r3, v4). By minimality of j — i, Subclaim 3 and tidiness, this attachment is y. Also by tidiness, there is no other attachment in (r1 r2 r3 r4).
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00	Suppose there is also an attachment z in [v1,r1,v2). The preceding paragraph
shows z = v1. Tidiness now implies that x is v9 and, since a G r9 and x G [a, r9, v0), a = v9. Let P and P' be H-avoiding xz- and yz-paths in B, respectively.
Theorem 5.23 (6) implies Q1 has BOD. If D1 is any 1-drawing of G — (s1), then Lemma 5.9 implies Q1 is crossed in D1. But Lemma 7.2 implies (recall z = v1) the two H-green cycles [z, P, x, r9, v0, r0, z] and [y,P', z, r1, v2, r2, v3, r3, y] are not crossed in D1. Thus, r9 r0 r1 r2 is not crossed in D1 (since x = v9), so Q1 is not crossed in D1, a contradiction.
Therefore, there is no attachment in [v1, r1, v2). Thus, we may assume that the only attachments in [a, r9,v0]r0 r1 r2 r3 are x G [a, r9,v0) and y G [v3,r3,v4). Tidiness further shows there is no attachment in [v4, r4, v5), so the only other possible attachment of B is v5, in which case y = v3.
In each of the two cases x = v9 and x = v9, we show that Q4 has NBOD by showing that B, Mq , and the Q4-bridge B4 containing s4 are mutually overlapping. We remark that B and B4 are in different faces of n[H], so B = B4. Obviously, B4 is skew to Mq .
•
Case 1. x = v9.
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The attachments x and y of B are skew to v4 and v9, so B and B4 overlap. Also, x and y are skew to v8 and v0, so B and Mq overlap, as required.
Case 2. x = v9.
As x,y G Q3 and B is not Q3-local, there is another attachment z of B. Our earlier remarks imply z = v5 and y = v3. Now y and z show B and B4 are skew, while x and y show B and Mq are skew.
00	We now resume our general discussion. Let Pxy be the xy-path in B. Since
cm	x G [a,r9,vo), vo = a. Suppose some Q1-bridge B' has an attachment at b = V5
cm	and an attachment in r0 r1. Since B is not a Q1-bridge and both B and B' are
H-bridges, B = B'. Then Pxy and a v5 [r0 r1]-path in B' would cross in n, which is impossible. Therefore, Theorem 5.23 shows Q1 has BOD.
Let D1 be a 1-drawing of G — (s1). Because Q4 has NBOD, Lemma 5.6 implies D1[Q4] is not clean in D1. Since Q1 has BOD and s1 is contained in a planar Q1-bridge, Lemma 5.9 implies Q1 is crossed in D1. Therefore, s0 is exposed in D1. Thus D1[H — (s1)] is one of two possible 1-drawings, depending on whether r9 crosses r5 r6 or r4 crosses r0 r1.
If x = v9, then Pxy cannot be added to D1[H — (s1)] without introducing a second crossing, which is impossible. If x = v9, then the three attachments of B are not all on the same face of D1 [H — (s1)], so B cannot be added to D1[H — (s1)] without introducing a second crossing, the final contradiction.	□
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We can now complete the proof of Claim 2. Subclaim 1 implies either x G [a, r9,v0) or y G (v4,r4,b]. By symmetry, we may assume the former. Subclaims 3 and 4 imply y G [v4,r4, b]. If y = v4, then Subclaims 2, 3 and 4 (all six statements) show that there is no other attachment of B. But then B is Q4-local, a contradiction. Therefore, y = v4, and, furthermore, there is an attachment z of B in [v1, r1, V2).
If x = v9, then both x and z are in (r9 r0 r1), contradicting tidiness. Thus,
x = v9.
a	9
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00	The claim will be proved once we know z = v1. By way of contradiction,
suppose z = v1. Consider any 1-drawing D2 of G — (s2). By Theorem 5.23, Q2 has BOD. Thus, Lemma 5.9 implies Q2 is crossed in D2. That is, r° r1 r2 r3 crosses r5 r6 r7 r8 in D2. In particular, neither s° nor s4 is exposed in D2.
Since B is global and has attachments at v4 and v9, it must be that D2[B] is in the face of D2 [R U s° U s4] incident with s4 and the crossing. Since v1 is an attachment of B, v1 must be in the subpath of r° r1 r2 r3 between the crossing and v4. But then s3 is not exposed in D2, implying B must cross s3 in D2, a contradiction that shows v1 is not an attachment of B, completing the proof of the claim.	□
To complete the proof of the theorem, by way of contradiction assume there is
00	no i so that att(B) C ri ri+1 ri+2. Claim 2 shows either att(B) = {v°,v5,z}, with
z G (r2) U (r7) or att(B) = {v4,v9,z}, with z G (r1) U (r6). These are all the same up to the labelling of H, a, and b, so we may assume att(B) = {v°,v5,z}, with z G (r2). Let H' be the subdivision of V10 consisting of H — (s°), together with the v°v5-path in B.
In order to apply Theorem 7.1, we show that n is H'-friendly. If n is not H'-friendly, then Lemma 6.5 (1) implies (since H and H' have the same nodes) v6v9 is an edge and n[v6v9] is contained in MH, which is the same as MH. But v6 and v9 are not incident with the same H-face in MH and, therefore, this is impossible. Thus, n is H'-friendly. However, H' violates Theorem 7.1, a contradiction.
Therefore, there is an i so that att(B) C ri ri+1 ri+2. Claim 1 implies B has one of the three desired forms.	■
CM	We can go somewhat further in our analysis of the global H-bridges of a tidy
V10 = H C G.
cm	3
Definition 10.7. Let G g M3 and V10 = H C G, with H tidy. Let B be a global H-bridge with attachments x and y.
(1) The span of B is the xy-subpath R with the fewest H-nodes.
CO	(2) An edge or subpath of R is spanned by B if it is in the span of B.
CO	(3) B is: a 2-jump if, for some i, its attachments are v and Vi+2; a 3-jump if,
for some i, its attachments are vi and vi+3; or else is a 2.5-jump.
We remark that Theorem 10.6 implies that, in the case of a 2.5-jump, there is an i so that vi is one attachment and the other attachment is in (ri-3) U (ri+2). Theorem 10.6 further implies a global H-bridge has precisely two attachments and its span has at most four H-nodes. It follows from Definition 6.2 that every global H-bridge combines with its span to form an H-green cycle.
Lemma 10.8. Let G G M3 and V10 = H C G, with H tidy. For each i G {0, 1, 2, 3, 4}, either Qi has BOD or one of vi_1vi_4 and vi+1 vi+4 is a global H-bridge.
Proof. Let n be an embedding of G in RP2 so that H is n-tidy. Suppose neither of the edges vi-1vi_4 and vi+1vi+4 occurs in G. The Qi-bridges that are Qi-exterior consist of Mq , those that are contained in M and, therefore, attach along either si-1 or si+1, and those that are contained in D. Since H is n-tidy, these latter must be global. By Theorem 10.6 they are 2-, 2.5-, and 3-jumps.
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00	Consider any global H-bridge. It is embedded in D so that it, together with
its spanned path in R, bounds a face of G. In particular, if we are considering a 2-jump B that is a Qrbridge, the 2-jump is either Vj_1vi+1 or vj+4Vj+6. In this CD	case, Qi U B has no non-contractible cycle in RP2 and so, by Lemma 5.16, B does
not overlap any other Qrexterior Qrbridge.
It is not possible for a 2.5-jump to be a Qrbridge. The only 3-jumps that can be a Qrbridge are vi+1vi+4 and vi-4vi-1, and these are assumed not to be in G. We conclude that the Qrexterior Qrbridges do not overlap and, therefore, Q4 has BOD.	■
Lemma 10.9. Let G G M3 and V10 = H C G, with H tidy. Then:
(1)	no two global H-bridges have an H-node in common;
(2)	at most one global H-bridge is a 3-jump;
(3)	there is no i so that VjVj+3 is a 3-jump and some 2.5-jump has an end in (vj_1,rj_1, Vj];
(4)	if B1 and B2 are global H-bridges, then, for every i G {0, 1, 2, 3, 4}, there is some edge of Qj O R that is not spanned by either B1 or B2; and
(5)	for each i G {0, 1, 2, 3, 4}, at most one of (rj) and (^+5) can contain an end of a 2.5-jump.
Claim 1. No two global H-bridges have an H-node in common.
Proof. suppose by way of contradiction that the two global H-bridges B1 and B2 have the H-node Vj in common. For j = 1, 2, let Pj be the subpath of R spanned by Bj. Then each of Bj U Pj is a green cycle; therefore, Theorem 6.7 implies P1 and P2 are edge disjoint. We choose the labelling so that rj U ri+1 C P1 and rj-2 U rj_1 C P2. We treat various cases.
Proof. We start with (1).
Subclaim 1. At least one of B and B2 is not a 3-jump.
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Proof. Suppose to the contrary that Bi and B2 are both 3-jumps, so Bi = VjVi+3 and B2 = vj-3Vj, respectively. Then there is a 1-drawing Dj of (H — Sj) U Bi U B2; Lemma 10.8 implies Qj has BOD, so Lemma 5.9 implies Qj is crossed in
Dj.
Because of Bi, Lemma 7.2 (3a) implies rj+i and rj+2 are not crossed in Dj, while (3b) of the same lemma implies that if rj were crossed, it would cross rj+3. However, (2) shows rj+3 is not crossed. Therefore, no edge of rj rj+i rj+2 is crossed in Dj. Analogously, no edge of rj-3 rj-2 rj-i is crossed in Dj. These two assertions show Qj cannot be crossed in Dj, a contradiction.	□
Subclaim 2. Neither B1 nor B2 is a 3-jump.
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Proof. By Claim 1, not both B1 and B2 are 3-jumps. So suppose for sake of definiteness that B1 is the 3-jump VjV^ and B2 is a global H-bridge with one end at Vj and one end in (vj-3,rj_3,vj_2].
The embedding in RP2 shows that	is not an edge of G (it would cross
B1) and Claim 1 shows vj_3vj is not an edge of G. Therefore, Lemma 10.8 implies Qj+1 has BOD. Thus, in any 1-drawing Dj+1 of G — (sj+1), Lemma 5.9 implies Qj+1 is crossed in Dj+1.
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00	By Lemma 7.2 (3a) (when B2 is a 2.5-jump) or (1) (when B2 is a 2-jump), rj_i
is not crossed in Di+1. Likewise, (1) shows that none of ri; ri+1, and ri+2 is crossed in D. But then Qi+1 is not crossed in Di+1, a contradiction.	□
CD
By Claim 2, we know that neither B1 nor B2 is a 3-jump. By Theorem 6.7, neither vi_1vi_4 nor vi+1vi+4 can occur in G; Lemma 10.8 implies Qi has BOD. Let Di be a 1-drawing of G — (sj). By Lemma 5.9, Qi is crossed in Dj.
Lemma 7.2 (1) shows that P1 and P2 are both not crossed in Di. This implies that ri_2 ri_1 ri ri+1 is not crossed in D and, therefore, Qi is not crossed in Di, a contradiction that completes the proof of the claim.	□
We move on to (2).
00
Claim 2. There is at most one global H-bridge that is a 3-jump.
Proof. Suppose there are distinct 3-jumps. Claim 1 implies that, up to relabelling, they are either vivi+3 and vj+4vj+7 or vivi+3 and vj+5vj+8. Theorem 6.7 and Claim 1 imply that there cannot be a third 3-jump. Thus, Lemma 10.8 implies Qj+1 has BOD.
Let C1 and C2 be the two H-green cycles containing these 3-jumps. Lemma 5.9 implies Qj+1 is crossed in a 1-drawing Dj+1 of G — (sj+1). But Lemma 7.2 (1) implies that neither ri rj+1 nor rj+5 rj+6 is crossed in Dj+1, a contradiction proving the claim.	□
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We next turn to (3).
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Claim 3. There is no i so that vivi+3 is a 3-jump and some 2.5-jump has an end in (vj_1, r-j_1, vj].
Proof. Suppose to the contrary that there is such an i. From Claim 1, the 2.5-jump has an end w e (vj_1, rj_1, vj). Its other end is vj-3. Lemma 10.8 and Claim 2 imply that Qj+2 has BOD. Let Dj+2 be a 1-drawing of G — (sj+2). Lemma 5.9 implies Qj+2 is crossed in Dj+2.
By Lemma 7.2 (2), rj+3 is not crossed in Dj+2. The same lemma (1) implies ri rj+1 rj+2 is not crossed in Dj+2. Consequently, Qj+2 is not crossed in Dj+2, contradicting the preceding paragraph and proving the claim.	□
Now we prove (4).
Claim 4. If B1 and B2 are global H-bridges, then, for every i e {0,1, 2,3,4}, some edge of Qj n R is not spanned by either B1 or B2.
Proof. Suppose by way of contradiction that the global H-bridge B1 spans the side rj U rj+1 of Qj+1 and a second global H-bridge B2 spans rj+5 U rj+6. To see that Qj+1 has BOD, by Lemma 10.8 it suffices to show that neither of the 3-jumps vjvj_3 and vj+2vj+5 is in G. For the former, Theorem 6.7 implies vj is an
attachment of B1, contradicting Claim 1. For the latter, vi+2 is an attachment of B2, with the same contradiction. Therefore Qi+1 has BOD.
Lemma 5.9 implies that, for any 1-drawing Di+1 of G—(si+1), Qi+1 is crossed in Dj+i. However, Lemma 7.2 (1) implies that neither rj ri+1 nor ri+5 ri+6 is crossed in Dj+1, showing Qi+1 is not crossed in Dj+1, a contradiction proving the claim. □
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00	Finally, we prove (5). Suppose, for j G {i, i + 5}, (rj} contains an end of the
2.5-jump Bj. We may use the symmetry to assume that Bj = wvj-2. If Bj+5 has Vj+3 as an end, then we contradict Claim 4. Therefore, Bj+5 has vj+8 = vj-2 as an end, contradicting Claim 1.	■
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We conclude this section with two observations about local bridges of a tidy subdivision of V10.
Lemma 10.10. Let G g Mf and V1o = H C G, with H tidy. Then no H-bridge has all its attachments in one H-spoke.
Proof. By way of contradiction, suppose B is an H-bridge and s is an H-spoke so that att(B) C s. By Corollary 5.15, B has precisely two attachments, so B is just an edge uw. Choose B so that no other H-bridge has all its attachments in a proper subpath of [u, s, w]. If [u, s, w] has no interior vertex, then B and [u, s, w] are parallel edges not in the H-rim, contradicting Observation 9.2 (6). Thus, some H-bridge B' has an attachment x in (u, s, w}.
Let n be an embedding of G in RP2 for which H is n-tidy. Since H C M, B' is a local H-bridge. Moreover, Corollary 5.15 and the choice of B show that not all attachments of B' can be in [u, s, w], so B has an attachment y not in [u, s, w]. But then, for at least one of the two H-quads Q containing s, B and B' are overlapping Q-bridges, contradicting the definition of tidiness.	■
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CM	Lemma 10.11. Let G G M3, V10 = H C G, with H tidy. For any H-spoke s, if
B is an H-bridge having an attachment in (s}, then B has no other attachment in [s].
CO
Proof. Suppose B is an H-bridge and s an H-spoke so that B has attachments x,y in s, with x G (s}. Let n be an embedding of G in RP2 for which H is n-tidy. Then n shows B is not a global H-bridge. By Lemma 10.10, B has a third attachment z not in [s]. Let Q be the unique H-quad containing all of x, y, and z.
CO	If y is not an H-node, then let r be an H-rim branch of Q not containing z.
CO	Then x, y, and z are all contained in Q — [r], contradicting Corollary 5.15. Thus,
y is an H-node Vj. We choose the labelling so that r C Q. Corollary 5.15 shows that z is not in Q — [rj+5] and, therefore, z is in rj+5. Furthermore, Corollary 5.15 now shows that B can have no other attachment, so Theorem 8.2 implies B is isomorphic to K13. Let w be the vertex in Nuc(B).
Claim 1. The cycles [y, B, w, B, x, s, y] and [z, B, w, B, x, Q — y, z] bound faces of n[G].
Proof. For the latter, [z, B, w, B, x, Q — y, z] is an H-green cycle, so the result follows from Lemma 6.6. The former, call it C, has just one vertex in R, so Lemma 5.20 implies it has BOD and every one of its bridges other than the one containing H — (s} is planar. If it has a second bridge B', then C is clean in any 1-drawing of contradicting Lemma 5.9.	□
The chosen labelling shows that Qj-1 is the other H-quad containing s.
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Claim 2. There
is no Qj_1-local H-bridge that has an attachment in (s}.
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00	Proof. Suppose B'' is a Qj_1-local H-bridge having an attachment x' in (s).
Lemma 10.10 implies B'' has an attachment z' not in [s]. If z' is in the same H-rim branch rj_1 contained in Qj-1 as y, then [x', B'', z', r, y, B, w, x, s, x'] is an H-green cycle C. As the edge of s incident with y is C-interior, C does not bound a face of n[G]. If z' is not in rj_1, then [z', B'', x', s, x, B, w, B, z, Q — y, z^ is a non-facial H-green cycle. Both conclusions contradict Lemma 6.6 (8).	□
We conclude that s has length 2 and that B is the only H-bridge attaching in o	(s). Let D be a 1-drawing of G — wy. Then D[s U (B — wy)] is clean in D and we
may extend D to a 1-drawing of G by adding in wy alongside [w, B, x, s, y]. ■
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CHAPTER 11
Every rim edge has a colour
In this section we introduce, for a tidy subdivision H of V10 in G, H-yellow edges. The main result is that every H-rim edge has a colour: H-green, H-yellow, or red. This is a major step on the route. In the next section, we will analyze red edges, with the main result being that there are red edges.
Definition 11.1. Let H be a subdivision of V10 in a graph G.
(1)	A 3-rim path is a path contained in the union of three consecutive H-rim branches.
(2)	The closure cl(Q) of an H-quad Q is the union of Q and all Q-local H-bridges.
(3)	Let H be tidy in G. A cycle C in G is H-yellow if C may be expressed as the composition P1P2P3P4 of four paths so that:
(a)	P2 and P4 are R-avoiding (recall R is the H-rim) and have length at least 1;
(b)	P1 and P3 are 3-rim paths and P1 U P3 is not contained in a 3-rim path; and
(c)	there is an H-green cycle C' so that P1 C (C' n R).
(4)	An H-rim edge e is H-yellow if it is not H-green and is in an H-yellow cycle.
We remark that the H-rim edges that are H-yellow are those in P3. The next result elucidates the nature of an H-yellow cycle.
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Lemma 11.2. Let G G M3, V10 = H C G, with H tidy. Let C be an H-yellow cycle, with decomposition P1P2P3P4 into paths as in Definition 11.1, and let C' be the witnessing H-green cycle. Then:
(1)	C' — (C' n R) is a global H-bridge;
(2)	for i G {2, 4}, Pi is either H-avoiding or decomposes as Pi1 P2, where Pi1 is contained in some H-spoke, including an incident H-node, and P2 is H-avoiding;
(3)	there is only one C-bridge in G; and
(4)	there is an i G {0,1, 2, 3, 4} so that C C cl(Qi). ^ 2
Proof. Let n be an embedding of G in RP2 for which H is n-tidy; in particular, every H-green cycle bounds a face of n[G].
For (1), the alternative is that C' is contained in cl(Q), for some H-quad Q. Lemma 6.6 (8) shows that C' bounds a face of G in RP2, so P2 and P4 are contained in global H-bridges. Each of P2 and P4 is in an H-green cycle (as is every global H-bridge) and, since P2 has an end in (C' n R), some edge of C' n R is in two H-green cycles, contradicting Theorem 6.7.
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For (2), let i G {2,4}. Since Pi has positive length, the end ui of Pi in P1 is distinct from the end wi of Pi in P3. Because C' bounds a face of G and is contained in D, we see that the edges of Pi incident with ui is in M. Since Pi is R-avoiding, Pi is contained in M, with only its ends in R.
Now suppose Pi has an edge e not in H. Choose e to be as close to ui in Pi as possible. As wi is in H, there is a first vertex y of Pi after e that is in H. If y = wi, then we are done, so we may assume y = wi. Since Pi is R-avoiding, we see that y must be in the interior of some spoke s. Let z be the vertex of Pi incident to e so that e is in [z, Pi, y].
As Pi is contained in M, we see that [ui, Pi, y] is contained in a closed n[H]-face bounded by some H-quad Q. Also, [z,Pi,y] is H-avoiding and so is contained in some Q-local H-bridge Bi. By Lemma 10.11, y is the only attachment of Bi in [s]. Since z = y and both are attachments of Bi, we have that z G [s].
The path [ui, Pi, z] is R-avoiding and contained in H. Therefore, either it is trivial or it is contained in some H-spoke s'. In the latter case, z = y implies s' = s. In the former case, ui = z, so ui G s. In both cases, [ui,Pi,y] U Q contains an H-green cycle that contains an H-rim edge incident with ui, contradicting Theorem 6.7 and completing the proof of (2).
For (3), we start by noting that there exist i and j so that P1 C ri ri+1,..., rj and i — 1 < j < i + 2; we assume P1 has one end in [vi,ri, vi+1), one end in (vj ,rj ,vj+1], and that j = i — 1 only if P1 is just the single H-node vi. Item 2 implies P2 is contained in cl(Qi-1) U cl(Qi) and that P4 is contained in cl(Qj) U cl(Qj+1). It follows that P3 has its ends in ri+4 ri+5 and rj+5 rj+6. There are at most (j + 6) — (i + 3) < 5 H-rim branches ri+4 ri+5 ... rj+6, so P3, being a 3-rim path, must be contained in this path. It follows that C is disjoint from either si-2 or si+2.
Let s be an H-spoke disjoint from C and let MC denote the C-bridge containing
s.
Set R' = (R — (C' n R)) U (C' — (C' n R)). Then R' U s contains a non-contractible cycle C'' disjoint from C. Lemma 5.20 shows C is contractible, has BOD, and every C-bridge other than MC is planar.
Suppose there is a C-bridge B other than MC; let D be a 1-drawing of B#.
CO	Lemma 5.9 implies D[C] is crossed. Let s, s', and s'' be the three H-spokes disjoint
from (C' n R). Then RU s U s' U s'' is a subdivision of V6 in B# that is edge-disjoint from both P2 and P4; this shows that some edge of P1 U P3 is crossed in D.
But now R' U s U s' U s'' is another subdivision of V6 in Therefore, the crossing in D must involve two edges of R' U s U s' U s''. In particular it does not involve an edge of C' n R, and, since P1 C C' n R, no edge of P1 is crossed in D.
Likewise, let R'' be obtained from R' by replacing P3 with P2P1P4. Now R'' U s U s' U s'' is a third subdivision of V6 in B# that is disjoint from P3. Thus, the crossing in D does not involve an edge of P3. Thus, none of P1, P2, P3, and P4 is crossed in D, contradicting the fact that C is crossed in D. We conclude that there is no C-bridge other than MC, as claimed.
Finally, for (4), suppose first that P1 is not contained in a single H-rim branch. Then there is an H-node vi in the interior of P1. However, P1 is incident on one side with the face bounded by C', so the edge of si incident with vi is on the other side of P1. Since C is contractible, we conclude that there are at least two C-bridges, contradicting (3). Therefore, there is an i G {0,1, 2, 3,4} so that P1 C ri.
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00	If both P2 and P4 are contained in cl(Qj), then so is P3, as it is a 3-rim path.
Therefore, by symmetry, we may assume that P2 has some edge not in cl(Qj). As we traverse P2 from its end in P1, we come to a first edge e that is not in cl(Qj). One end of e is the vertex u that is in either s» or si+1; for the sake of definiteness, we assume the former. Then (2) implies [v^s^u] C P2 and that the remainder of P2 consists of an H-avoiding uw-path, with w an end of P3. It follows that w G rj+4. Let e be the edge of s» incident with u and not in P2.
Switching paths, we know that P4 has an end x in r». If x = vj+1, then (2) implies P4 C cl(Qj). In this case, e is in a C-bridge other than Mq, contradicting (3). Otherwise x = vi+1, in which case P1P2 [w, ri+4, vi+5, ri+5, vi+6, si+1, vi+1] is an H-yellow cycle G. There is a G-bridge other than Mq containing e, also contradicting (3) for C.	■
We now turn our attention to the all-important red edges. We comment that, if n > 4 and V2n = H C G, then any red edge of G is in the H-rim. The remainder of this section is devoted to proving the following.
Lemma 11.4. Let G G M2 and let V1o = H C G. If H is tidy and the H-rim edge e is either H-green or H-yellow, then e is not red.
Theorem 11.3. Let G gM2 and let V1o = H C G. If H is tidy, then every H-rim edge is one of H-green, H-yellow, and red..
We start with an easy observation.
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Proof. Suppose first that e is H-green and let C be the H-green cycle containing e. There are three H-spokes s, s', and s'' disjoint from (C n R). Thus, (R- (C n R)) U
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(C - (C n R)) together with s, s', and s'' is a subdivision of V6 contained in G - e, 00	showing e is not red.
Now suppose e is H-yellow and let C be the H-yellow cycle containing e. Let
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C' be the H-green cycle and P1P2P3P4 the decomposition of C as in Definition 11.1. Then e is in P3 and there are three H-spokes s, s', and s'' disjoint from C U (C' n R). In this case, (R - ((C' n R) U (P3))) U (C' - (C' n R)) U P2P1P4, together with s, s', and s'' is a subdivision of V6 contained in G - e, showing e is
CO	not red.
The following concepts and lemma play a central role in the proof of Theorem
Definition 11.5. Let V1o = H C G. Let e and f be two edges of the H-rim R. Then e and f are R-separated in G if G has a subdivision H' of V8 so that the H'-rim is R and e and f are in disjoint H'-quads.
The following two observations are immediate from the definition.
CD	Observation 11.6. Let V1o — H C G and suppose e and f are two edges of the
H-rim R that are R-separated in G.
(1)	If D is a 1-drawing of G, then e and f do not cross each other in D.
(2)	If H' is a V8 in G witnessing the R-separation of e and f, then there are two H'-spokes that have all their ends in the same component of R-{e, f}.
The following is a kind of converse of Observation 11.6 (1).
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00	LEMMA 11.7. Let G0 G be a graph and let V10 = H C G0, with H tidy.
Suppose G C Go with H C G. Let e G r and f G rj+4 U rj+5 U rj+6 be edges that are both neither H-green nor H-yellow. If e and f are not R-separated in G, then there is a 1-drawing of G in which e crosses f.
S	Proof. Let n be an embedding of G in RP2 so that H is n-tidy.
We may write r = [vj,..., xe, e, ye,..., vj+1] and, by symmetry, we may assume f is in
rj+5 U rj+6 = [vj+5, rj+5, ... ,x/, f, y/,. .. ,rj+6, Vj+7] .
If f G rj+5, then let Je,/ = cl(Qj) and Q = Qj, while if f G rj+6, then let Je,/ = cl(Qj) U cl(Qj+1) and Q = Qj+1. The two H-spokes contained in Q are sj and se,/, which is either sj+1 or sj+2.
Claim 1. There are not totally disjoint sjse,/-paths in Je,/ — e.
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Proof. Because H is n-tidy, n[Je,/] is contained in the closed disc bounded by n[Q]. Therefore, one of a pair of totally disjoint sjse,/-paths in Je,/ would be disjoint from rj+5 rj+6 and it, together with a subpath of Q — rj+5 rj+6 yields the contradiction that e is H-green.	□
Let we be a cut-vertex in Je,/ — e separating sj from se,/. Then Je,/ — e has a separation (He, Ke) with sj C He, the other H-spoke se,/ contained in Q is contained in Ke, and He n Ke = ||we||. Clearly, we G rj+5 rj+6.
There is also a separation (H/, K/) of Je,/ — f, so that H/ n K/ is a single vertex w/, sj C H/, and se,/ C K/. For x G {e, f}, there is a face Fx of n[Je,/] incident with both x and wx. If Fe = F/, then any vertex of rj rj+1 in the boundary cycle C of Fe may be selected as w/. Similarly, we may be any vertex of 00	rj+5 rj+6 that is in C. We choose we and w/ so that they are in different com-
CM	ponents of C — {e, f}. Thus, whether Fe = F/ or not, the cycle Q has the form
cm	[we,..., e,..., w/,..., f,...]. In particular, e and we are in the same component
of Q — {w/, f}, while f and w/ are in the same component of Q — {we, e}. By interchanging the roles of e and f and exchanging the labels of Vj and Vj+5, for j = 0,1, 2, 3,4, we may assume Q has the form
[we, . . . , Vj+5, Sj, Vj . . . , e, . . . , w/, . . . , Se,/, . . . , f, . . .] .
vj+5
For technical reasons, we choose we as close as possible to f in rj+5 rj+6 and w/ as close as possible to e in rj+5 rj+6, while respecting the ordering that was just described of these four elements of Q.
Set N = Ke n H/. Then Je,/ — {e, f} = He U N U K/, He n N = ||we|, and K/ n N = ||w/1|. See Figure 11.1.
Claim 2. N does not have disjoint paths both with ends in the two components of N n R.
Proof. Such paths, together with the H-rim and the H-spokes sj-1 and sj+3, would show e and f are R-separated.	□
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Let w be a cut-vertex in N separating the two components of N n R, and let (Nj, Nj+5) be a separation of N so that, for j G {i, i + 5}, Nj n R is contained in rj U rj+1 and Nj n Nj+5 = ||w|. We proceed to describe a new 2-representative embedding of G in RP2 that shows that G has a 1-drawing.
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Figure 11.1. The locations of e, f, we, wf, He, N, and Kf.
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Let G' be the subgraph of G obtained by deleting all the vertices and edges of N that are not in N O R. There is a face of n[G'] contained in M and incident with both e and f.
Claim 3. No global H-bridge has a vertex in (N O R) U (Nj+5 O R) in its span.
Proof. For sake of definiteness, suppose some vertex of (N O R) is in the span of the global H-bridge B. If the H-node ve f in r rj+1 incident with se f is in the interior of the span of B, then the cycle bounding Ff is H-yellow, contradicting the fact that f is not H-yellow. Letting z be the vertex of N nearest e in r rj+1, we conclude that B has an attachment in (z, r rj+1, ve f ], and B does not span any edge of ri+2.
By Theorem 10.6, B is either a 2-, 2.5-, or 3-jump. It follows from the preceding paragraph that e is contained in the span of B, yielding the contradiction that e is H-green.	□
We can now easily complete the proof of the lemma. By Claim 3, we can separately embed N and Nj+5 in the face outside of M. As no global H-bridge can attach on both paths in R — {e, f} without making at least one of e and f H-green, we can join the two copies of w together to obtain a representativity 2 embedding n' of G in RP2 having a non-contractible simple closed curve meeting n'[G] only in the interiors of e and f. This implies that G has a 1-drawing, as required. ■
We further investigate the detailed structure of H-rim edges.
Lemma 11.8. Let G G M3 and V10 = H C G, with H tidy. If VjV^ is a global H-bridge, then, for j G {i — 1, i + 3} there is an edge ej G rj that is neither H-yellow nor H-green.
Proof. The two sides are symmetric, so it suffices to prove the existence of ej_1. Lemmas 10.8 and 10.9 (2) imply that Qj+1 has BOD. Let D be a 1-drawing of G — (sj+1). Lemma 5.9 implies Qj+1 is crossed in D.
However, the cycle C consisting of VjV^ and the path it spans is H1-close, for H1 = R U sj_1 U Sj U sj+3. Therefore, Lemmas 5.3 and 5.4 imply that C is not crossed in D. We conclude from the nature of 1-drawings of V8 that rj_1 crosses rj+5 U rj+6; let e be the edge in rj_1 that is crossed in D.
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00	Suppose, by way of contradiction, that there is a global H-bridge B spanning e.
Theorem 10.6 implies B is either a 2-, 2.5- or 3-jump, while Theorem 6.7 implies B does not span any edge of rj (such an edge is already spanned by vjvj+3). Lemma 10.9 (1) implies vj is not an attachment of B, so B must be a 2.5-jump with one end in (rj-1), contradicting Lemma 10.9 (3). Thus, e is not spanned by a global H-bridge.
It follows that, if e is in an H-green cycle C', then C' C cl(Qj-1). But such a C' is H2-close, for H2 = R U sj U sj+2 U sj+3. By Lemmas 5.3 and 5.4, C' is not crossed in any 1-drawing of G — (sj+1). This contradicts the fact that e is crossed in D. We conclude that e is not H-green.
So now we suppose e is in the H-yellow cycle C' and that C'' is a witnessing H-green cycle. Then C' C cl(Qj-1) and C'' contains a global H-bridge B that spans an edge in rj+4. This implies B = vjvj+3, so Lemma 10.9 (2) shows that B is not a 3-jump.
Moreover, (3) of the same lemma shows B cannot have an attachment in [vj+3, rj+3, vj+4), while (4) shows B cannot have vj+7 as an attachment. Therefore, B is a 2- or 2.5-jump vj+4w, with w G [vj+6, rj+6, vj+7).
The cycle (R — (C'' n R)) U B, together with the H-spokes sj-1, sj+2, and sj+3 is a subdivision H3 of V6 for which C' is H3-close, showing that e is not crossed in any 1-drawing of G — (sj+1). This contradicts the fact that e is crossed in D and, therefore, e is not H-yellow.	■
Definition 11.9. Let G be a graph and let V10 = H C G, with H tidy. Let n be an embedding of G in RP2 so that H is n-tidy and has the standard labelling relative to 7. For i G {0,1, 2,..., 9}:
(1) Pj = rj-2 rj-1, Pj = rj+3 rj+4, Pj = rj+1 rj+2, and Pj = rj+6 rj+7.
The proof of Theorem 11.3 will also depend on the following new concepts.
(2)	the spines □j and j d of Qj consist of the paths Pj U sj U Pj and Pj U sj+1 U Pj, respectively (see Figure 11.2);
(3)	the scope Kj of Qj consists of cl(Qj) U □j U jd U Bj, where Bj consists
of all global H-bridges having both attachments either in Pj U Pj or in P U P ,•; and
(4)	the complement K? of Kj is obtained from MQi by deleting the edges (but not their incident vertices) that comprise the H-bridges in Bj.
(5)	The two vertices vj-2 and vj+3 are the trivial □j j d-paths in Kj. Any other □j jd-path in Kj is non-trivial.
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We note that □j n jd is equal to ||{vj_2, vj+3}y. For our purposes, these are
not "useful" □jjd-paths.
We observe that, for each i G {0,1, 2, 3,4}, G = Kj U Kj1.
The following lemma plays an important role in the rest of this section.
Lemma 11.10. Let G G M2 and V10 = H C G, with H tidy. Let e be an edge of R and let i be such that e G rj. Then G — e has a subdivision of V5 if and only if there are disjoint non-trivial □j j^d-paths in Kj — e.
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Figure 11.2. The paths with small dashes are P 1; P 1; P 1; and P1. The spine □1 is the path r9 r0 s1 r5 r4, while 1C is r3 r2 s2 r7 r8.
There is some subtlety here; 2-criticality is important. Suppose we have a subdivision H of V1o embedded in RP2 with representativity 2 so that all the H-spokes are in M. Give H the usual labelling relative to 7. Now delete (r1) and (re), and then add the 2.5-jump av2 and the 3-jump v6v9. Then there are disjoint non-trivial □11 C-paths in the union H' of (H — (r1)) — (r6) and the two jumps, but H' is planar.
We shall need the following.
LEMMA 11.11. Let G G M2 and V10 = H C G, with H tidy. Let e be an edge of R and let i be such that e G rj. If there are disjoint non-trivial □a C-paths in Ki — e, then there are two such paths so that at least one of them is contained in cl(Qj) and the other contains at most one global H-bridge'..
In the proof, we consider many possibilities for the two disjoint □ii C-paths. For a given i, some possibilities might not occur because of limitations imposed by n. In principle, for i = 2, all of the considered possibilities can occur, while for i = 4, several of the considered possibilities cannot occur.
Proof. Let P1 and P2 be the hypothesized disjoint paths.
Claim 1. If there is a □ii C-path in Ki — e disjoint from ri+5, then there are disjoint □ii C-paths so that one of them is contained in cl(Qi) and the other contains at most one global H-bridge.
Proof. Suppose that P and ri+5 are disjoint paths. If P contains two (or more) global H-bridges, then they must be 2.5-jumps having an end in (ri). By Theorem 6.7, they must be of the form vi-2w1 and w2vi+3, with w1 being no further from vi in ri than w2 is. By symmetry, we may assume e is not in [vi, ri, w1]. Now [vi, ri, w1] (P — vi-2) and ri+5 are the desired disjoint □i iC-paths in Ki — e. □
Thus, we may assume both P1 and P2 intersect ri+5.
Claim 2. If either of P1 and P2 contains two global H-bridges, then there are disjoint □ii C-paths in Ki — e so that one of them is contained in cl(Qi) and the other contains at most one global H-bridge.
Proof. We may assume P1 contains two global H-bridges B1 and B2. Both B1 and B2 are 2.5-jumps. Both have ends in (ri) U (ri+5). By Lemma 10.9 (5), they
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00	both have an end in the same one of (r) and (r+5). We choose the labelling so
that (B1,B2) is either (vi-2w1, w2vi+3) or (vi+3w1, w2vi+8). We treat these cases separately.
Suppose (B1,B2) = (vi-2w1, w2vi+3). Assume first that e G [w1,r1,w2]. Then B1 U [w1, r1, w2] U B2 is disjoint from ri+5, and we are done by Claim 1. Therefore, we may assume e G [w1, r1, w2].
In this case, P1 consists of B1, B2, and a w1w2-path P1 contained in cl(Qj). We know that P1 contains a vertex in ri+5. Lemma 10.9 (5) implies that P2 consists of a global H-bridge with no vertex in (r+5). Therefore, we may choose [vj, r, w1] U P1 U [w2, r, vi+1] and P2 as the desired paths.
We conclude the proof of this claim by considering the case (B1, B2) = (vj+3w1, w2 vj+8). First, by way of contradiction suppose P2 is not contained in cl(Qj). Lemma 10.9
(5) implies that P2 consists of a global H-bridge having both ends in Pj U Pj. But then P2 is disjoint from rj+5 and we are done by Claim 1. Thus, we may assume P2 C cl(Qj).
If P2 is disjoint from either [vj+5, rj+5, w1) or (w2, rj+5, vj+6], then we may replace either B1 with the former or B2 with the latter, and we are done again. Otherwise, there is a [vj+5, rj+5, w1) (w2, rj+5, vj+6]-path P2' contained in P2 that is rj+5-avoiding; let its ends be W3 G [vj+5, rj+5, W1) and W4 G (W2, rj+5, Vj+6].
If P2' is rj-avoiding, then P2' U [w3,rj+5,w4] is an H-green cycle. Since B1 together with the subpath of R it spans is also H-green, the edge of [vj+5, rj+5, w1] incident with w1 is in two H-green cycles, contradicting Theorem 6.7.
Therefore, P2' is not rj-avoiding and so contains two subpaths, one being a w3rj-
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path P'1 and the other being an rw-path P„2
path P21 and the other being an rjw4-path P2'2. For k =1, 2, let uk be the vertex of P2k in rj. If e G [vj, rj, u1], then the paths [vj+5, rj+5, w3] UP2'1 U [u1, rj, vj+1] and B1 U [w1,rj+5, vj+6] constitute the required disjoint paths. Otherwise, [vj,rj,u1] U
[«1, P2', w4, rj+5, vj+6] and [vj+5, rj+5, w2]UB2 constitute the required disjoint paths.
□
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CO	To complete the proof of the lemma, we may now assume that, for each j = 1, 2,
Pj contains a unique global H-bridge Bj.
We first suppose, by way of contradiction, that both B1 and B2 have an end in (rj) U (rj+5). Lemma 10.9 (5) shows that such ends are in the same one of (rj) and (rj+5); let i' G {i, i + 5} be such that, for j = 1, 2, Bj has an end Wj G (r+). We may assume B1 = vj/_2w1 and B2 = w2vj/+3.
Theorem 6.7 implies w1 is closer to v+ in r+ than w2 is. The paths P1 — vj/_2 and P2 — vj/+3 are both in cl(Qj); the former is a w1sj+1-path, with end x1 G sj+1, and the latter is a w2sj-path, with end x2 G sj.
Recall that n[cl(Qj)] is a planar embedding of cl(Qj) with Qj bounding a face. The vertices w1,w2,x1, x2 occur in this cyclic order in Qj, so the disjoint paths P1 — vj/_2 and P2 — vj/+3 must cross in n[cl(Qj)], a contradiction. Therefore, at most one of B1 and B2 has an end in (rj) U (rj+5), while the other is equal to the path among P1 and P2 that contains it.
We may choose the labelling so that P2 consists only of B2. Theorem 6.7 implies no edge of rj U rj+5 is spanned by both B1 and B2; since B2 spans one of rj and rj+5 completely, one of B1 and B2 spans edges in rj and the other spans edges in rj+5. If either Bj spans all of rj, then, as it is disjoint from rj+5, we are done by
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CO CD
CD CO
u
a
CD U
00	Claim 1. In particular, B2 spans rj+5, edges spanned by B1 are in ri; and B1 does
not span all of rj.
Therefore, B1 is a 2.5-jump with one end w1 in (rj). We may assume the other end of B1 is vj+3. If e G [vj,rj,w1], then [v, rj,w1] U B1 is disjoint from rj+5, and we are done by Claim 1. If e G [vj, rj, w1], then (P1 — vj+3) [w1, rj, vj+1] and P2 are the desired paths.	■
o
Proof of Lemma 11.10. The following claim settles one direction.
Claim 1. If there are not disjoint non-trivial □j jC-paths in Kj — e, then G — e is planar.
Proof. For this proof, we need to apply Menger's Theorem; in order to do so, we treat the copies of vj-2 and v.¿+3 in Zk as different from their copies in jC.
Let u be a cut-vertex of K — e separating Zk and j C. Let Kj be the union of
the ||u||-bridges in Kj — e that have an edge in □j and let Kj be the union of the
remaining ||u||-bridges in Kj — e. Then Kj — e = Kj U Kj and Kj n Kj is just ||u||.
Let n be an embedding of G in RP2 so that H is n-tidy. Since rj+s C Kj — e, u € rj+5. Because Kj — {e, u} is not connected, there is a non-contractible, simple closed curve in RP2 that meets n[G —e] only at u. Thus, there is no non-contractible cycle in G — {e, u}, showing that G — e is planar.	□
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For the converse, Lemma 11.11 shows there are disjoint non-trivial □j jC-paths Pi and P2 in Kj — e so that Pi C cl(Qj). In particular, Pi is an sjsj+i-path. It follows from the embedding n[Kj] that P2 is disjoint from either rj or rj+5.
In every case, we find our V6 by adding three spokes to the cycle contained in (R — ((rj) U (rj+5>)) U Pi UP2 U s j U sj+i and containing (R — ((rj) U (rj+5>)) UPi UP2.
If P2 contains no global H-bridges, then sj+2, sj+3, and sj+4 may be chosen as the spokes.
If P2 contains precisely one global H-bridge B2, then B2 is one of:
(1)	Vj-2vj+i (symmetrically, VjVj+3);
(2)	vj-iVj+2;
(3)	vj-2w and w is in (rj) (symmetrically, wvj+3);
(4)	wvj+i and w is in (rj-2) (symmetrically, vjw, with w € (rj+2));
(5)	vj-iw and w is in (rj+i) (symmetrically, wvj+2, with w € (rj-i));
(6)	vj-ivj+i (symmetrically, VjVj+2);
(7)	and the comparable jumps with ends in rj+3 rj+4 rj+5 rj+6 rj+7.
We choose, in all cases, sj-2 and sj+2 as two of the spokes, with third spoke (taking the cases in the same order):
(1)	the Pj^-subpath of sj+i (symmetrically, the PiP2-subpath of sj);
(2)	sj-i;
(3)	the P^iP2-subpath of sj+i (symmetrically, the PiP2-subpath of sj
(4)	the P^iP2-subpath of sj+i (symmetrically, the PiP2-subpath of sj
(5)	sj-i (symmetrically, the same);
(6)	sj-i (symmetrically, the same); and
(7)	these cases are symmetric to the preceding ones. In every case, we have found a V6 in G — e, as required.
s¿);
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00 cm cm
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00	We conclude this section by proving that every rim edge is either red, H-green,
or H-yellow.
Proof of Theorem 11.3. Let e be an edge in the H-rim. There is an i so that e G ri. By Lemma 11.10, G is red if and only if there are no disjoint non-trivial
SZk iC-paths in Ki — e.
Now suppose there are disjoint non-trivial □i iC-paths P1 and P2 in Ki — e. By
CD
Lemma 11.11, we may assume P1 is contained in cl(Qi), while P2 contains at most one global H-bridge. If P1 is disjoint from ri+5, then every maximal ri-avoiding subpath of P1 is contained in an H-green cycle. The edge e is in one of these H-green cycles, as required.
Thus, we may assume P1 contains a vertex in ri+5. If P2 C cl(Qi), then
00
the planar embedding of cl(Qi) shows P2 is disjoint from ri+5 and the preceding paragraph, with P2 in place of P1, shows e is H-green. Consequently, we may further assume P2 contains a global H-bridge B2.
Case 1: B2 has its ends in Pi U ri U Pi.
In this case, if e is spanned by B2, then there is an H-green cycle containing e, namely the cycle consisting of B2 and the subpath of R that it spans. The only other possibility in this case is that B2 is a 2.5-jump with an end w2 in (ri) and that e is in the one of [vi, ri, w2] and [w2, ri, vi+1] not spanned by B2. For the sake of definiteness, we suppose B2 = vi-2w2 and that e is in [w2, ri, vi+1]. Since P1
C cl(Qi), we see that, in this case, P2 is disjoint from ri+5 and, therefore, we may assume P1 = ri+5. We replace P2 with [vi,ri, w2] (P2 — vi-2) so that there are disjoint □iiC-paths contained in cl(Qi) — e; a situation resolved in the paragraph preceding this case.
cm
Case 2: B2 has its ends in Pi U ri+5 U Pi.
In this case, either P2 is B2 or, up to symmetry, B2 is a 2.5-jump wvi+8, with w G (ri+5), and P2 is [vi+5, ri+5, w] U B2. On the other hand, P1 is an sisi+1-path in cl(Qi) intersecting ri+5.
Let x be the first vertex in rj+5 as we traverse P1 from Sj and let P1 be the sjx-subpath of P1. We note that P2 prevents x from being in [vj+5, rj+5, w], so x G (w,ri+5, vi+6]. Let y be the end of P{ in Sj. The cycle P1 [x, ri+5, vi+6] Sj+1 r [vj,Sj,y] is H-yellow, as witnessed by the H-green cycle containing B2. Therefore, e is either H-yellow or H-green (in Definition 11.1, an H-yellow edge is not H-green). ■
00 1-H
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CHAPTER 12
Existence of a red edge and its structure
In this section, we prove that if G is a 3-connected, 2-crossing-critical graph containing a tidy subdivision H of V10, then some edge of the H-rim is red. Furthermore, we prove that each red edge e has an associated special cycle we call Ae. These "deltas" will be the glue that hold successive tiles together and so form a vital element of the tile structure.
1	The argument for proving the existence of a red edge depends on whether or
not there is a global H-bridge that is either a 2.5- or 3-jump. Once these cases are disposed of, matters become simple. However, with the knowledge of the A's, it turns out we can show that there is no 3-jump. This will be our first aim and so, since we need the A's to complete the elimination of 3-jumps, we shall begin by determining the structure of the A of a red edge.
Theorem 12.1. Let G g M2, V10 = H C G, with H tidy. Let e = uw be a red edge of G and let i G {0,1, 2, .. ., 9} be such that e G rj. Then there exists a vertex xe G [^+5] and internally disjoint xeu- and xew-paths and , respectively, in cl(Qi) so that, letting Ae = (Au U ) + e:
(1) there are at 'most two Ae-bridges in G; 00	(2) there is a Ae-bridge Ma so that H C Ma U Ae, while the other Ae-
CM	bridge, if it exists, is one of two edges in a digon incident with xe; and
CM	(3) when there are two Ae-bridges, let ue and we be the attachments of the
one-edge Ae-bridge, labelled so that ue G and we G ; otherwise let ue = we = xe. In both cases, Ae — e contains unique uue- and wwe-paths Pu and Pw, each containing at most one H-rim edge, which, if it exists, is in the span of a global H-bridge and, therefore, is H-green.
Proof. Let n be an embedding of G in RP2 for which H is n-tidy. We may assume r = [vj, r», u, e, w, r», vj+1]. Lemma 11.10 implies K — e has a cut vertex xe separating Zj and j d (again adopting the perspective that vj_2 and vj+3 are split into different copies in Z and ¿c). As rj+5 is a Zj id-path in K — e, xe is in rj+5.
Because cl(Qj) is 2-connected and n[cl(Qj)] has Qj bounding the exterior face, there is a face Fe of G in RP2 that is in the interior of Qj and incident with both e and xe. As G is 3-connected and non-planar, Fe is bounded by a cycle Ce and Ce — e consists of a uxe-path and a wxe-path .
For (1) and (2), we begin by noticing that Ce C cl(Qj). Thus, there is a Ce-bridge MCe containing the three H-spokes not in Qj.
Claim 1. Each of Ce n si; Ce n sj+i, and Ce nr is connected. Either Ce nri+5 is connected or it has two components that are joined by an edge e' of ri+5 and Ce has an edge parallel to e'. In particular, each of Sj, sj+1, r, and ri+5 — e' is contained in Ce U MC .
Jh
a
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CM
CD O
00
00	Proof. Suppose by way of contradiction that Ce n sj is not connected. As
Ce bounds a face of n, it follows that there is a Qj-local H-bridge having all its attachments in sj, contradicting Lemma 10.10. Thus, Ce n sj is connected.
It follows that any part of sj that is not in Ce is in the same Ce-bridge as either rj_1 or rj+4. That is, it is in MCe, and therefore, sj C MCe U Ce.
Symmetry shows that this also holds for sj+1.
Now suppose Ce n rj is not connected. Then there is a Qj-local H-bridge B having all attachments in rj. Corollary 5.15 implies B has precisely two attachments x and y, and so Lemma 5.19 implies B is just the edge xy. Thus, [x,rj,y]B is an H-green cycle C. Lemma 6.6 (8) shows C bounds a face of n[G].
By symmetry, we may assume that x and y are both in [vj, rj, u]. Suppose that z is any vertex in (x, rj,u].
Suppose first that z = u. As G is 3-connected, z has a neighbour z' not in [x, rj, y]. If zz' is in the interior of Qj, it must be parallel to an edge in rj, as any other edge would go into one of the faces of n[G] bounded by Ce and C. Therefore, zz' is outside M and, so is an edge of another H-green cycle. But then one of the edges of [x,rj,y] incident with z is in two H-green cycles, contradicting Theorem 6.7.
This same argument, however, also applies if z = u, with the small variation that, by Lemma 11.4, zz' cannot span the red edge e, giving the contradiction that the edge of [x, rj,u] incident with u is in two H-green cycles. Thus, Ce n rj is connected. As it did for sj, this implies that rj C MCe U Ce.
Finally, we consider Ce n rj+5. Proceeding as we did for rj, if Ce n rj+5 is not connected, there is (up to symmetry) a Qj-local H-bridge B having all attachments in [vj+5, rj+5, xe]; B is a single edge and is in an H-green cycle. One end of B is CM	xe, and the H-green cycle containing B consists of two parallel edges.
00	Thus, there are at most two such H-bridges B, each of which is an edge parallel
cm	to an edge in rj+5. If they both exist, then the 3-connection of G implies xe has
cm	another neighbour, which, as above, is adjacent to xe by an edge not in M, showing
one of the edges of rj+5 incident with xe is in two H-green cycles, contradicting Theorem 6.7.	□
CO
We can now define Ae. If Ce n rj+5 is connected, then Ae = Ce. Otherwise, Ae is obtained from Ce by replacing the edge of Ce incident with xe and not in rj+5 with its parallel mate that is in rj+5. Notice that the Ae- and Ce-bridges are the same, except for these exchanged edges incident with xe. Set Mab to be MCe. The following is evident from what has just preceded.
Claim 2. H C Mac U Ae and Ae n rj+5 is connected.	□
Consider again rj n Ae. It is connected, so if it is more than just [u, e, w], the symmetry shows we may assume it contains an edge xu other than e. The 3-connection of G implies that u is adjacent with a vertex y other than x and w. The edge uy is not interior to Qj, as then it would be in the face of G bounded by Ce.
Thus, uy is not in M, and, as uw is red, Lemma 11.4 implies uy spans xu. The vertex x is seen to be H-green by the H-green cycle Cy containing uy. Since x has at least three neighbours in G, there is a neighbour of x different from the two neighbours of x in rj_1 rj. Because Cy bounds a face of G (Lemma 6.6 (8)), every edge incident with x and not in rj_1 rj is in M. There is a unique neighbour z of
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x so that z is not in ri-1 r» and xz is an edge of Ag. This shows that x is one end of r n Ae. These observations easily yield the following claim.
Claim 3. Each of Au n r and n r has at most one edge.	□
We now turn our attention to rj+5.
Claim 4. (1) No edge of rj+5 n Ae is H-yellow.
(2) No global H-bridge has xe in the interior of its span.
Proof. For (1), suppose by way of contradiction there were an H-yellow edge in rj+5 n Ae. Then Lemma 11.2 (3) shows the witnessing H-yellow cycle must be Ae. However, the witnessing H-green cycle must have Ae n rj in the interior of its span, yielding the contradiction that e is H-green.
For (2), suppose by way of contradiction that there is a global H-bridge xy with xe in the interior of the span of xy. Then xy U (rj+5 — xe) contains a □j jC-path in Kj — {e,xe}, contradicting Lemma 11.10.	□
Claim 5. (1) If [vj+5, rj+5,xe] n Ae contains three vertices x, y, and xe of rj+5, then (choosing the labelling of x and y appropriately) [vj+5, rj+5, xe]n Ae = [x, xy, y,yxe, xe], y and xe are joined by a digon, and y is incident with a global H-bridge that spans x.
(2) If [vj+5, rj+5, xe] n Ae does not contain three consecutive vertices of rj+5, but has a vertex x other than xg, then either x and xg are joined by a digon, or xe is incident with a global H-bridge that spans x.
CM
The symmetric statements also hold for [xg, ri+5, vi+6] n Ag.
Proof. For (1), the fact that Agnrj+5 is connected implies that there are vertices x and y so that [x, xy, y, yxg, xg] C [vj+5, rj+5, xg]. Because G is 3-connected, y is adjacent to a vertex z other than x and xg. The edge yz cannot be in M, as then it would be in the face of G bounded by Cg, a contradiction. Therefore, it is a 2.5-jump. Claim 4 (2) shows yz does not span xg.
As G is 3-connected, x has a neighbour x' different from the two neighbours of
CO
x in R. If the edge xx' is in D, then it is in the face bounded by the H-green cycle containing yz, a contradiction. Therefore, xx' is in M and, in particular, for that x' giving the edge nearest to xy in the cyclic rotation about x, xx' is in Ag and, therefore, no other vertex of [vj+5, rj+5, xg] is in Ag.
Since yxg is not R-separated from e in G, Lemma 11.7 implies yxg is either H-yellow or H-green. Claim 4(1) implies it is not H-yellow; we conclude that yxg is H-green and let CyXe be the witnessing H-green cycle.
As pointed out in the first paragraph of the proof, CyXe cannot contain a global H-bridge that spans xg. On the other hand, xy is H-green by the global H-bridge yz. By Theorem 6.7, this is the only H-green cycle containing xy. Thus, the only H-rim edge contained in CyXe is yxg. It follows that CyXe is contained in cl(Qj). Claim 1 implies CyXe is a digon.
For (2), the fact that [vj+5, rj+5, xg]nAg is connected implies that [vj+5, rj+5, xg]n Ag = [x,xxg,xg]. Lemma 11.7 implies that xxg is either H-yellow or H-green, and Claim 4(1) shows it is not H-yellow. Therefore, it is H-green.
Claim 4 (2) shows any global H-bridge spanning xxg has
xg as an attachment.
Otherwise, the H-green cycle Cxxe containing xxg is contained in cl(Qj). Again, Claim 1 shows Cxxe is a digon.	□
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00	There is one more observation to make before we complete the proof of the
theorem. From Claim 5 (1), it seems possible that both [vj+5, rj+5, xe] n Ae and [vj+5, rj+5, xe] n Ae have three vertices. However, this is not possible, as xe must have a neighbour z different from its neighbours in R. But now xez cannot be in M, as then it would be in the face bounded by Ce, and it cannot be in D, as then it is a global H-bridge and one of the digons incident with xe is also spanned by xez, contradicting Theorem 6.7. Therefore, rj+5 n Ae has at most three edges, and all such edges are H-green.
If there are no edges, then rj+5 n Ae is just xe. If no edge of rj+5 n Ae is in a digon, then ue and we are defined in (3) of the statement to be xe. In this case, Claim 5(1) implies there can be at most one edge of rj+5 n Ae on each side of xe, but any such edge is spanned by a global H-bridge. If there is a digon, then it is uewe, each of ue and we is incident with at most one other edge in rj+5 n Ae, and any such edge is spanned by a global H-bridge.
Finally, By Lemma 10.9 (4), not both u and ue, for example, can be incident with such global H-bridges, so Pu has at most one H-rim edge.	■
Definition 12.2. Let G g M2, V10 = H C G, with H tidy, and e a red edge
10
of G with ends u and w. With ue and we as in the statement of Theorem 12.1, the peak of Ae is the subgraph of G induced by ue and we. If the peak has just one vertex, then Ae is sharp.
Corollary 12.3. Let G G M2, V10 = H C G, with H tidy, and e a red edge
cm
of G. Then the peak of Ae is either a single vertex or a digon and no edge of the CO	peak is in the interior of the span of a global H-bridge.
The following observations are given to summarize important points from Theorem 12.1.
Proof. That the peak is either a single vertex or a digon is a rephrasing of Theorem 12.1 (2) and (3). In the case the peak is a digon, neither ue nor we can be in the interior of the span of a global H-bridge, since then the H-rim edge in the digon is in two H-green cycles, contradicting Theorem 6.7.
So suppose the peak is just the vertex ue = we, let B be a global H-bridge with ue in the interior of its span, and let i be such that e G rj. If Ae n rj+5 has an edge e', then e' is incident with ue and, moreover, is H-green by a global H-bridge B' incident with ue. But then B provides a second H-green cycle containing e', contradicting Theorem 6.7. So Ae n rj+5 is just ue, in which case B provides a witnessing H-green cycle that shows Ae is H-yellow. But then e is H-yellow, contradicting Lemma 11.4.	■
Our next goal is to eliminate 3-jumps. For this the next two lemmas are helpful.
Lemma 12.4. Let G e M2 and V10 = H Ç G, with H tidy. Suppose C is an H-yellow cycle and C is the witnessing H-green cycle. Let e be an edge of G not in C U C' U R. Suppose either C' does not contain a 3-jump or e is in one of the CD	four spokes containing an H-node spanned by C. Then no H-yellow edge in C is
crossed in any 1-drawing of G — e.
Proof. There are at least four H-spokes contained in G — e. By hypothesis, at least one of these has no end in C' and, therefore, no end in C U C'. Therefore, Lemma 7.2 (2) applies.	□
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Figure 12.1. One of several examples of a A.
LEMMA 12.5. Let G G M2 and V10 = H C G, with H tidy. Suppose C is an H-green cycle in G. Suppose that C does not contain a 3-jump, e is an edge of G not in R U C and D is a 1-drawing of G — e. If an edge e' of C is crossed in D, then C contains a 2.5-jump with an end in (ri), for some i, and e' is in ri.
Proof. This is a straightforward consequence of Lemma 7.2 (3a and 3b). ■
Theorem 12.6. Let G G M3 and V10 = H C G, with H tidy. Then no global H-bridge is a 3-jump.
Proof. The proof begins by showing that if vi-3vi is a global H-bridge that is a 3-jump, then there is a red edge in ri. The next step is to show that the edge of ri incident with vi is red. The final step is to show that, if e* is the edge of si incident with vi, then cr(G — e*) > 2, contradicting the criticality of G. Let n be an embedding of G in RP2 so that H is n-tidy.
Claim 1. There is a red edge of G in ri.
Proof. Lemma 10.9 (2) implies neither vi+5vi-2 nor vivi+3 is in G. Thus, Lemma 10.8 implies Q^ 1 has BOD.
Let Di_1 be a 1-drawing of G — (si-1). Lemma 5.9 implies Q^ 1 is crossed in Di-1. Let H' be the subdivision of V6 consisting of the H-rim R and the three spokes si, si_3, and si+1. Lemma 7.2 implies the cycle ri_3 ri-2 ri_1 [vi, vi_3vi, vi_3] is clean in Di_1. In particular, the crossing must be of an edge in ri+3 U ri+4 and an edge e in ri.
We prove e is red in G by proving it is neither H-green nor H-yellow. Lemma 10.9 (1) and (3) imply that no global H-bridge other than vi_3vi has an end in [vi, ri, vi+1). Therefore, no H-green cycle containing e can contain a global H-bridge. Thus, any H-green cycle C containing e is contained in cl(Qi). Lemma 12.5 implies C is not crossed in Di_1, contradicting the fact that the edge e is in C and is crossed in Di_1. We conclude that e is not H-green.
So suppose C is an H-yellow cycle containing e and let P1P2P3P4 be the decomposition of C into paths as in Definition 11.1. By Lemma 11.2, there is a global H-bridge B so that the interior of the span of B contains P^ Lemma 10.9 (2) says there is at most one 3-jump in G, so B is either a 2- or 2.5-jump.
That e is not H-yellow is an immediate consequence of Lemma 12.4.	□
e
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00 CSI CSI
CO
CO CD
5H
CD CO
00	We now aim to show that the edge of rj incident with vj is red. By Claim 1,
there is a red edge in r; let e1 be the red edge nearest to vj in r. Let r' be the component of r — e1 containing vj and let u be the end of e1 in r'.
Claim 2. No edge of r^ is H-yellow.
Proof. Suppose some edge e' of r' is H-yellow and let C and C' be the witnessing H-yellow and H-green cycles, respectively. Lemma 11.2 (1) implies C' contains a global H-bridge B. We note that Lemma 10.9 (1) and (3) imply (because v.j_3v.j is present and v.j_3 = v.j+7) that B has no vertex in (vj+6,r+6,vj+7]. On the other hand, to make C H-yellow, B must have one end in (vj+5,r+5,vj+6].
Due to the presence of vj_3vj, Lemma 10.9 (4) implies vj+3 is not in B. Therefore, Theorem 10.6 implies B has vj+6 as one end and its other end is in (vj+3,r+3, vj+4]. Theorem 12.1 (3) implies the edge e of Aei — e1 incident with u is not in H; by Theorem 12.1, it is in cl(Qj).
Let D be a 1-drawing of G — e. By Theorem 5.23, Qj has BOD, so Lemma 5.9 implies Qj is crossed in D. Lemma 7.2 implies no edge in r+4 r+5 is crossed in D, so the crossing in D is of r with r+6.
Lemmas 12.4 and 12.5 combine with Theorem 11.3 to show that the edge e'' of r+6 crossed in D is red in G. Lemma 11.7 implies e'' and e1 are R-separated in G and we conclude that they are also R-separated in G — e'; in fact, e'' is R-separated from r'[u, e1, w]. It follows that the edge f of r crossed in D is in [w, r, vj+1 ].
Lemmas 12.4 and 12.5 combine with Theorem 11.3 to show that f is red in G; however, e1 and f are not R-separated in G — e' and, therefore, not separated in G, contradicting Lemma 11.7. It follows that no edge of r' is H-yellow, as required. □
I
Claim 3. « = vj.
Proof. By way of contradiction, suppose that u = v. By definition of e1, no edge of r^ is red, and Claim 2 shows no edge of is H-yellow. Theorem 11.3 shows that every edge of r^ is H-green. Because of v-3v, Lemma 10.9 (1) and (3) shows no edge of r^ is H-green by a global H-bridge.
Let e be the edge of Aei — e1 incident with u; Theorem 12.1 and the fact that e1 is not incident with v imply that e is not in H. Let D be a 1-drawing of G — e. Note that e is in a Qj-local H-bridge. Since Qj has BOD (Theorem 5.23), it is crossed in D (Lemma 5.9). Every edge of r-1 is H-green in G — e; thus, Lemma 6.6 (10) implies the following.
Subclaim 1. No edge in r-1 is crossed in D.	□
We next rule out another possibility.
Subclaim 2. No edge in r+1 is crossed in D.
Proof. Suppose some edge eD of r +1 is crossed in D. Since Qj is crossed in D, the other crossed edge e'D is in r+5. By Lemma 12.4, no H-yellow edge in
rj+1 U r+5 can be crossed in D. Since H C G — e, Lemma 6.6 (10) implies no H-green cycle not containing e can be crossed in D; in particular, no H-green edge in r +1 U r+5 can be crossed in D. Now Theorem 11.3 implies eD and e'D are both red in G.
Suppose first that e'D is in [v.j+5, r+5, uei]. (Recall that uei is the vertex in the peak of Aei nearest u in Aei — e1.) Lemma 11.7 implies e'D and e1 are R-separated
o
CM

o
CD
00	in G; this implies that A£,d is disjoint from Aei. One of the rjrj+5-paths in A£,d ,
si+:b si+2, and si+3 combine with R to show that e'D is R-separated in G — e from every edge in ri+1, a contradiction.
If, on the other hand, e'D is not in [vi+5, ri+5, uei ], then Lemma 11.7 shows eD and e'D are R-separated in G and there is a subdivision of V8 that both witnesses
this separation and does not contain e (the spokes are Sj+2, Sj+3, and the "nearer" (r ri+1)(rj+5 ri+6)-paths in AeD and Ae,n ). This shows that eD and e'D are R-separated in G — e, a contradiction.	□
Since Qj is crossed in D, Subclaims 1 and 2 imply that some edge eD of r is crossed in D.
Subclaim 3. eD g r'.
c^	j j
Proof. If eD is not in rj, then let e'D be the edge of rj+4 rj+5 rj+6 that is crossed in D. Then eD and e'i are not R-separated in G — e. Observe that Aei shows no H-green or H-yellow cycle containing eD can also contain e. Therefore, eD is red in G and, consequently is R-separated from e'D in G. In particular, e is in every subdivision of Vg that contains R and witnesses the R-separation of eD and e'D. This implies that e'D is in [vj+5, rj+5, uei ].
As ei and ejD are both red in G, by Lemma 11.7 there is a subdivision K of Vg containing R and witnessing the R-separation of e1 and e^ . There is an rjrj+5-path P in K that is disjoint from Aei. Moreover, P C cl(Qj). But now, P
CM
together with the rjri+5-path in Aei — u, Sj+2, and sj+3 make the four spokes of a subdivision of Vg containing R and witnessing the R-separation of eD and ej4 in G — e, a contradiction.	□
00
cm	We now locate the edge ej4 . To this end, let e be the edge of si-1 incident with
Vj_i and let D be a 1-drawing of G — e. By Lemmas 10.8, 10.9 (2), and 5.9, Q^ 1 must be crossed in D. However, Lemma 7.2 shows that none of rj_3 ri-2 rj_1 can be crossed in D. Since the edges in rj are all H-green and none of the witnessing H-green cycles contains a global H-bridge, Lemma 12.5 implies that no edge of rj is crossed in D. Thus, some edge of rj+3 rj+4 crosses an edge of rj — (rj} in D.
Subclaim 4. Every edge in rj+4 is H-green in G and no edge in rj+4 is crossed in D.
Proof. If e' G rj+4 is H-yellow, then vj_3vj is in the witnessing H-green cycle and, therefore, the edge of sj_1 incident with vj+4 is in the interior of an H-yellow cycle containing sj_2; this contradicts Lemma 11.2, so e' is not H-yellow.
Now we eliminate the possibility that e' is red. To do this, it will be helpful to know that no H-green edge in rj+4 is crossed in D: fortunately, this is just Lemma 12.5, combined with Lemma 10.9 (1) and (4) to eliminate the possibility of a 2.5-jump.
Choose e' to be the red (in G) edge in rj+4 that is nearest in rj+4 to vj+5. Lemma 11.7 implies e' is R-separated from e1 in G; we may choose the witnessing subdivision K of Vg to contain sj_2 and sj+2; in particular, K avoids e. Therefore, e' is R-separated from e1 in G — e. Since the edges in rj+4 between e' and vj+5 are neither red (choice of e') nor H-yellow (two paragraphs preceding), they are
o
CM
00 Gï
O
Qï
o
cm
i
00	H-green (Theorem 11.3), we know they are not crossed in D (preceding paragraph).
The subgraph K shows that none of the rest of ri+3 ri+4 can be crossed in D, which is a contradiction. Therefore, no edge of ri+4 is red in G; since none is H-yellow CD	by the preceding paragraph, Theorem 11.3 shows they are all H-green.	□
rO
It follows that an edge of ri+3 is crossed in D and it must cross some edge in [u, ri, vi+1 ]. This further implies that the uuei -subpath Pu of Aei — e1 intersects si as otherwise each edge of [u, ri, vi+1 ] is R-separated from ri+3 in G — e.
We now return to consideration of D. No edge in ri+4 is red in G and, because Pu intersects si, every edge (if there are any) of [vi+5, ri+5, uei] is H-green. This combines with Lemma 10.9 (1) and (4) to show that no edge in ri+4[vi+5, ri+5, uei] is in the span of a global H-bridge; therefore, Lemmas 12.4 and 12.5 imply that no edge of ri+4 [vi+5, ri+5, uei ] is crossed in D. Thus, the edge e'D that crosses eD in D is in [uei,ri+5, vi+6]ri+6.
Because of vi-3vi, no edge in [uei, ri+5, vi+6]ri+6 is in the span of a global H-bridge. Therefore, Lemmas 12.4 and 12.5 imply e'D is red in G. But now Lemma 11.7 implies e^ is R-separated in G from e1; there is a witnessing subdivision K of V8 that contains si-1, si, and the nearer (ri ri+1)(ri+5 ri+6)-paths in Aei and Ae,D. Note that the path taken from Aei does not contain e. Therefore, K is also
contained in G — e; Observation 11.6 (1) shows that these edges cannot be crossed in D, the final contradiction that proves the claim.	□
We now move into the final phase of the proof that there is no 3-jump. Let e* be the edge of si incident with vi and let D* be a 1-drawing of G — e*. Lemma 10.9 (2) implies vi-3vi is the only 3-jump of G, so Lemma 10.8 implies Qi has BOD. Lemma 5.9 implies Qi is crossed in D*. In particular, there is an edge e in ri+3 ri+4 ri+5 ri+6 that is crossed in D*. Lemma 7.2 shows that ri+3 is not crossed in D*.
Claim 4. e is red in G.
Proof. If e is H-yellow in G, then Lemma 12.4 shows that e is not crossed in D*. Thus, e is not H-yellow.
Suppose e is H-green in G, and let C be the witnessing H-green cycle. Lemma 10.9 (2) implies C does not contain a 3-jump and Lemma 7.2 implies both that it does not contain a 2-jump and is not contained in the union of some Qj together with a Qj-local H-bridge. Therefore, C contains a 2.5-jump b and Lemma 7.2 implies e is in the H-rim branch that contains the end x of b that is not an H-node.
The edge e has already been shown to be in ri+4 ri+5 ri+6. Suppose e is in ri+4. If b = vi+2x, then we contradict Lemma 10.9 (4) — vi-3vi and b span the opposite sides of Qi-2, a contradiction. The other alternative is that b = xvi-3, which violates Lemma 10.9 (1). Thus, e G ri+4.
If e G ri+5, then either b = xvi-2 or b = xvi+3. The former does not occur, as otherwise the edges of ri-3 are all in two H-green cycles, contradicting Theorem 6.7. If the latter occurs, then we contradict Lemma 10.9 (4) — vi-3vi and b span the opposite sides of Qi-1. Thus, e G ri+5.
So e G ri+6. In this case b is either xvi-1 or xvi+4. For the former, the edges of ri-3 ri-2 are all in two H-green cycles, contradicting Theorem 6.7. For the latter, the edge e1 of ri incident with vi is red by Claim 3. The existence of b shows Qi is
o
CM
u
CD
CO CO
J-H
CD
00	H-yellow, contradicting the fact that e1 is red. This is the final contradiction that
shows e is red.	□
Recall that the edge e is in rj+3 rj+4 rj+5 rj+6, since it is involved in a crossing with Q4. We have already observed that e is not in rj+3.
S	Suppose first that e G rj+4. Lemma 11.7 implies e and e1 are R-separated in
G; in particular, v is not in Ae. But then Vj_3Vj shows Ae C cl(Qi_1) — v to be an H-yellow cycle, contradicting the fact that e is red. o	Therefore, e G rj+5 rj+6. Let e be the edge crossed by e in D*. Since Lemma
7.2 implies rj_2 is not crossed in D*, e G rj_2. Since e and e1 are both red in G, Lemma 11.7 implies they are R-separated in G; there is a witnessing subdivision K of V8 that contains sj_1 and sj_2. This K does not contain e*, and so is contained in G — e*. Therefore, K separates e from rj_1 in G — e*, and so, in D*, e does not cross rj_1. Thus, e is not in rj_1.
Therefore, e G r rj+1. Lemma 10.9 (4) implies there is no 2.5-jump xvj+4 — it and vj_3vj would span the opposite sides of Q4_2. Also, Lemma 10.9 (3) implies there is no 2.5-jump xvj+3 with x G (rj).
It follows from Lemmas 12.4 and 12.5 (the preceding pararaph is used here) that the edge e crossed by e in D* is red in G. This implies that e and e are R-separated in G and this in turn implies that e and e are R-separated in G — e*,
the final contradiction.	■
^ 3
Corollary 12.7. Let G G M2 and V10 = H C G, with H tidy. Then every H-hyperquad has BOD.
Proof. By Theorem 12.6, no global H-bridge is a 3-jump. By Lemma 10.8, every H-hyperquad has BOD.	■
CM	We are now prepared for the main result of this section.
cm
Theorem 12.8. Let G	and V10 = H C G, with H tidy. Then there is a
red edge in the H-rim.
Proof. We prove this by first considering the case there is a global H-bridge. By Theorem 12.6, there is no 3-jump. By Theorem 10.6, a global H-bridge is either a 2.5- or a 2-jump.
Claim 1. If G has a 2.5-jump, then G has a red edge.
Proof. By symmetry, we may assume wvi+2 is a 2.5-jump with w G (ri-1}. By way of contradiction, we assume that G has no red edge. We first treat two special cases.
Case 1: there is a 2.5-jump Vj_3w', with w' G (rj_1}.
In this case, let D be a 1-drawing of G — (sj+2}. Corollary 12.7 and Lemma 5.9 show that Qi+2 is crossed in D. Lemma 7.2 implies each of the cycles consisting of one of these two 2.5-jumps and the subpath of R it spans is clean in D. The same lemma implies that neither ri+3 nor ri+s is crossed in D. The combination of facts imply that some edge e2 in ri+2 crosses some edge e6 in ri+6.
Since G has no red edge, Theorem 11.3 implies each of e2 and e6 is either H-yellow or H-green in G. There is complete symmetry between them (relative to the two 2.5-jumps), so we treat e6. If e6 is H-yellow in G, then it is in some
o
CM
CD O
00
o» o
cm
i
cm cm
CO CO
co
00	witnessing H-yellow cycle C for which there is a witnessing H-green cycle C'. The
only possibility is that C' contains wvj+2.
We have that C C cl(Qj+1) — vj+2. Let C = P1P2P3P4 be the composition of paths showing C is H-yellow, as in Definition 11.1. Since P1 C (C' n R), we have P1 C rj+1 — vi+2. Choose the labelling of P2 and P4 so the rj+1-end of P2 is nearer vj+2 in rj+1 than the rj+1-end of P4 is.
If P2 is not disjoint from (sj+2), then the edge of rj+1 incident with v»+2 is in two H-green cycles, contradicting Theorem 6.7. Therefore, C U C' is disjoint from (si+2). But then Lemma 12.4 implies e6 is not crossed in D and, therefore, e6 is not H-yellow. Likewise, e2 is not H-yellow.
Therefore, e6 is H-green, so Lemma 12.5 implies e6 is spanned by some 2.5-jump J6 and, moreover, is not in either H-rim branch fully contained in the span of J6. By Theorem 6.7, no H-rim edge is in two H-green cycles. Thus, the only possibility for the 2.5-jump J6 spanning e6 is v»+4w6, with w6 G (rj+6). An analogous argument applies to e2, so e2 is spanned by the 2.5-jump J2 w2v»+5, with w2 G (r»+2). But now we have that every edge of r»+4 is in the distinct H-green cycles containing J2 and J6, contradicting Theorem 6.7, completing the proof in Case 1.
o
Case 2: There is a 2.5-jump vj_4w', with w' G r»_2.
Let D1 be a 1-drawing of G — (sj+1). Corollary 12.7 and Lemma 5.9 imply Qi+1 is crossed in D. Lemma 7.2 (1) shows none of [w,ri-1vi], r», rj+1, and r»+6 is crossed in D, while (2) of the same lemma shows r»+2 is not crossed. It follows that some edge e5 G r»+5 crosses an edge e9 G [v»+9, r»+9, w].
Since e9 is not red, Theorem 11.3 shows it is either H-yellow or H-green. If e9 is H-yellow as witnessed by the H-yellow cycle C and the H-green cycle C', then the global H-bridge J in C' is a 2- or 2.5-jump (Theorems 10.6 and 12.6) and C C cl(Q-1) (Lemma 11.2 (4)). Lemma 12.4 implies that e9 is not crossed in D, a contradiction.
Likewise, if e9 is H-green, the Lemma 12.5 shows it is not crossed in D, the final contradiction completing the proof in Case 2.
Case 3: All the remaining cases.
Let e» be the edge of s» incident with v» and let D» be a 1-drawing of G — e». Corollary 12.7 and Lemma 5.9 imply Q» is crossed in D».
Since G (in particular, rj_2) has no red edge, Lemma 12.4 shows any H-yellow edge in rj_2 is not crossed in D», while Lemma 12.5 implies that, as we are not in Case 2, no H-green edge of rj_2 is crossed in D». Lemma 7.2 (1) implies no edge of [w^j-^vj]r» rj+1 is crossed in D». Therefore, it must be that some edge ej_1 of [vj_1, rj_1, w] is crossed in D».
As ej_1 is not red in G, Theorem 11.3 implies ej_1 is either H-green or H-yellow. If it is H-green, then, because we are not in Case 1, Lemma 12.5 implies ej_1 is in an H-green cycle C contained in cl(Qj_1) and e» G C. But then every edge in [w, rj_1,vj] is in two H-green cycles, contradicting Theorem 6.7.
We conclude that ej_1 is H-yellow. Let C and C' be the witnessing H-yellow and H-green cycles, respectively, and let B be the global H-bridge contained in C'. Lemma 12.4 implies e» G C. Moreover, v»+5 is in the span of B, as otherwise B attaches at v»+2, contradicting Lemma 10.9 (1). By Lemma 10.9 (4), vj+7 is not in the span of B.
o
CM
00

CM
i
CM
CO
$H
a
CD Jh
00	If B has an end in (rj+2), then the other end of B is Vj+5. The R-avoiding
path (one of P2 and P4 in the decomposition of the H-yellow cycle as in Definition 11.1) in C containing ej contains a positive-length H-avoiding subpath joining a vertex of (sj) to a vertex of [vj+4,rj+5,vj+5). This yields the contradiction that the edge of rj+4 incident with vj+5 is in two H-green cycles. Therefore, B has one attachment in [rj+5 r.j+6) and one attachment in rj+3.
Let Dj+1 be a 1-drawing of G — (sj+1). Lemma 12.4 implies no H-yellow edge in either rj_1 or rj+2 is crossed in Dj+1. An H-green edge of rj+2 is not spanned by a global H-bridge (there is no room for such a jump between B and wvj+2), so Lemma 12.5 implies no H-green edge of rj+2 is crossed in Dj+1. Because we are not in Case 1 and there is no 3-jump, Lemma 12.5 implies no H-green edge of either rj_1 or rj+2 is crossed in Dj+1.
Lemma 7.2 (1) implies no edge of r rj+1 is crossed in Dj+1. Thus, none of rj_1 r rj+1 rj+2 is crossed in Dj+1, and therefore Qj+1 cannot be crossed in Dj+1. However, Corollary 12.7 and Lemma 5.9 imply that Qj+1 is crossed in Dj+1. This contradiction completes the proof that G has a red edge when there is a 2.5-jump.
o	n
At this point, we may assume G has no 2.5-jump and no 3-jump. Claim 2. If G has a 2-jump vjvj+2, then either rj_1 or rj+2 has a red edge.
Proof. In this case, let Dj+1 be a 1-drawing of G — (sj+1). Corollary 12.7
and Lemma 5.9 imply that Qj+1 is crossed in Dj+1. Lemma 7.2 (1) shows no
edge of r rj+1 is crossed in Dj+1. Therefore, some edge of rj_1 U rj+2 must be crossed in Dj+1. Lemmas 12.4 and 12.5 imply that no H-yellow or H-green edge in rj_1 U rj+2 is crossed in Dj+1. Therefore, Theorem 11.3 shows some edge in rj_1 U rj+2 is red.	□
In the final case, there are no global H-bridges. Therefore, there are no H-
CM	yellow cycles and every H-green cycle is contained in cl(Qj), for some i. For j G
{0,1, 2, 3,4}, let ej be the edge in Sj incident with Vj and let Dj be a 1-drawing of
CO
G — ej. Corollary 12.7 and Lemma 5.9 imply that Qj is crossed in Dj, so some edge in rj+3 rj+4 rj+5 rj+6 is crossed in Dj. Since ej cannot be in any H-green cycle containing an edge in rj+3 rj+4 rj+5 rj+6, Lemma 12.5 implies no H-green edge in rj+3 rj+4 rj+5 rj+6 can be crossed in Dj. Therefore the edge in rj+3 rj+4 rj+5 rj+6 crossed in Dj is red in G.	■
We conclude this section with the technical lemma (12.14) below that will be used in the next section. We start with four lemmas leading to a more refined understanding of R-separation in cases of interest for us. The first three are primarily used in the proof of the fourth. (Recall that an RR-path is an R-avoiding path with both ends in R.)
Lemma 12.9. Let G G M3 and let V10 = H C G, with H tidy, 'witnessed, by the embedding n. Let P be an RR-path in G. If B is a global H-bridge so that one CD	end of P is in the interior of the span of B, then there is an H-quad Q so that
P C cl(Q) and the two cycles in R U P containing P are non-contractible in RP2.
Proof. As P is R-avoiding, Theorem 6.7 implies P is not contained in D. If P is just an H-spoke, then both conclusions are obvious. Otherwise, as we traverse P from an end u in the interior of the span of B, there is a first edge e that is not in
o
CM
00 Gï
O
cm
i
cm
m
u
a
CD U
00	H. Since P Ç M, there is an H-quad Q so that e e cl(Q). Let P' be the H-bridge
in H U P containing e. Then P' is an H-avoiding path with both ends in H, so P' Ç cl(Q).
Since P is R-avoiding, if both ends of P' are in R, then P' = P and P Ç cl(Q), as claimed. Otherwise, one end w of P' is in the interior of some H-spoke Sj. Our two claims eliminate many possibilities for the other end x of P'. We choose the labelling so that u e ri-1 rj.
o	Claim 1. x is not in (sj_1 ri-1 rj sj+1).
CD
Proof. Suppose first that u is an end of P'. The choice of e implies P' = [u, P, w] is just the edge e. If u is an end of sj, then e is an H-bridge having all its attachments in sj, contradicting Lemma 10.10. If u is not an end of sj, then there is an H-green cycle that contains an edge f of R incident with u. But then f is in two H-green cycles, contradicting Theorem 6.7. Thus, u is not an end of P'.
If x e (sj_1 rj_1 rj sj+1) — u, then u = vj and P' = [w, P, x] is contained in either cl(Qj_1) — rj+4 or cl(Qj) — rj+5. In this case, we again have the contradiction that some edge of R incident with u is in two H-green cycles.	□
Claim 1 implies u = vj and [u, P, w] Ç sj. Moreover, x is in Qj_1 U Qj and either P' Ç cl(Qj_1) or P' Ç cl(Qj). The next claim eliminates another possibility for x.
Claim 2. x e (sj).
Proof. Suppose by way of contradiction that x e (sj). Let B' be the H-bridge containing P'. Observe that B' is H-local and that w and x are both attachments of B' in (sj). Corollary 5.15 implies that these are the only attachments of B', contradicting Lemma 10.10.	□
We conclude from Claims 1 and 2 that x is in rj+4 rj+5. Evidently, P is in
1)
cl(Qi-1) or cl(Qj), respectively, as required for the first conclusion. Furthermore, both cycles in n[R U P] that contain P are non-contractible in RP2.	■
Lemma 12.10. Let G G M2 and let V10 = H C G, with H tidy, witnessed by the embedding n. For i G {0, 1, 2, 3, 4} and j G {i + 3, i + 4, i + 5}, let e G rj and f G rj be edges that are not H-green. Suppose P is an RR-path in M having both ends in the component R' of R — {e, f} containing rj+6 rj+7 rj+8 rj+g and so that the cycle in n[R' U P] is non-contractible. Then
P C ^cl(Qj) — [vj,sj, vj-5} ^ U ^ (J cl(QfcU ^ cl(Qj) — (vj+e, Sj+1,vj+1] ^ .
Proof. Choose the labelling u and w of the ends of P so that u is nearer in R' to the end incident with f than w is.
Let y be a non-contractible curve meeting n[G] in just the two points a and b;
CD
we note that u and w are on different ab-subpaths of R (allowing a or b to be an end of P). We may choose the labelling of a and b so that a G R', and if both a and b are in R', then a is closer to the end of R' incident with f than b is.
Claim 1. (1) If vj and w are on the same ab-subpath of R, then P n (sj} is empty.
(2) If vi+1 and u are in the same ab-subpath of R, then P n (si+1) is empty.
i—l
Proof. The statements are symmetric, so it suffices to prove the first. Suppose to the contrary that P n (sj) is not empty. As we traverse P from w (which is not incident with Sj), let x be the first vertex in (sj) and let P' denote the wx-subpath
S	of P. Evidently, P' is contained in one component M' of M \ (7 U Sj). On the other
hand, f is between vj and w, and so f is in M'. If w G rj, then P' and f are in an CD	H-green cycle, a contradiction.
Otherwise, vj+1 G M' and P' intersects (sj+1). In this case, some (sj+1) (sj)-subpath of P' is in an H-green cycle with f, also a contradiction.	□
If both vi+1 and w are in the same ab-subpath of R and both vi+4 and u are in the same ab-subpath of R, then Claim 1 implies P is trapped between Sj and si+1, as required. By symmetry, we may assume that vi+1 is not in the same ab-subpath of R as w. Let Rw denote the ab-subpath of R containing w and let Ri+1 denote the other ab-subpath of R, so vi+1 G Ri+1.
This implies that vi+1, vi+2, ..., vj are all in Ri+1. We noted above that u G Rw, so u is also in Ri+1. From Claim 1 (1), we conclude that P is disjoint from (sj). Thus, P is contained in the component of M — Sj disjoint from vi+2.
It follows from the fact that all the H-spokes are in M that vj_5 is on the same ab-subpath as w. This combines with the fact that vi+1 is not in that ab-subpath and the fact that P meets 7 at most in a to tell us that
Gï
P Ç ( cl(Qj) - (Si_i) ) U ( (J cl(Qfc) J U
<N 00
CO CO
CO
Jh
a
CD U
P Ç (cl(Qj) -(si-i^u I U cl(Qk)J ^cl(Qi) -(si+i))
as required.
The additional fact that P cannot include vj and vi+1 follows from the knowledge that these vertices are not in R'.	■
In a similar vein, we have the following.
LEMMA 12.11. Let G G M2 and let V10 = H C G, with H tidy as witnessed
by the embedding n. Suppose e G ri; f G ri+3 ri+4 and P is an RR-path with both ends in the component of R — {e, f} containing ri+i ri+2. If e is not H-green, then both cycles in n[R U P] containing P are contractible.
Proof. Let R' be the component of R — {e, f} containing ri+1 ri+2 and let C be the cycle in R U P that contains P and is contained in R' U P. Since R is contractible, the other cycle in R U P containing P is homotopic to C; thus, it suffices to show C is contractible.
Let ue be the end of R' incident with e. Suppose there is a ([ue, ri, vi+1] si+1)si-path P' in P contained in cl(Qi). Since C is disjoint from ri+1, P' is contained in an H-green cycle containing e, a contradiction.
Thus, there is no ([ue, ri, vi+1] si+1)si-path in P contained in cl(Qi). Since C is disjoint from ri+5, there is an arc in the disc bounded by n[Qi] joining a point of [vi,ri, ue) to ri+5 that is disjoint from C; this shows that C is contractible, as required.	■
Our next lemma takes us one step closer to the useful description of R-separation.
o


00
CO
$H
a
CD Jh
00	Lemma 12.12. Let G gM2 and let V10 = H C G, with H tidy. Suppose e G r
and f G r+3 r+4 are R-separated as witnessed by the subdivision H' of V8. If e is not H-green, then the component of R — {e, f} containing both ends of some H'-spoke is the one containing r+5 r+6 r+7 r+8 r+9.
Proof. Let n be an embedding of G in RP2 so that H is n-tidy. Recall that R is also the H'-rim. Observation 11.6 (2) shows that two of the four H' spokes have all their ends in the same component of R — {e, f}. Of the four H'-spokes, at most one can be in D. Thus, of the two that have both ends in the same component R' of R — {e, f}, there is at least one, call it s, that is in M.
In particular, the two cycles in R U s containing s are non-contractible. Now Lemma 12.11 shows the two ends of the RR-path s are not in the component of R — {e, f} containing r +1 r+2 and so must be in the component containing rj+5 r+6 ... rj+9, as claimed.	■
Our next lemma in the series gives a quite refined description of R-separation.
Lemma 12.13. Let G G M3 and let V10 = H C G, with H tidy. Let e G r and f G rj+4 rj+5 be edges that are both not H-green. If e and f are R-separated in G, then there is a witnessing subdivision H' of V8 having sj+2 and sj+3 as H'-spokes and the other two H'-spokes are in cl(Qj_1) U cl(Qj).
Proof. Let n be an embedding of G in RP2 for which H is n-tidy. Let H1 be a subdivision of V8 witnessing the R-separation of e and f. Let s be an H1 spoke having both ends in the same component R' of R — {e, f}.
Claim 1. The cycles in n[R U s] containing s are non-contractible.
Proof. Suppose first by way of contradiction that n[s] in not contained in M. Since H is n-tidy, s is a global H-bridge. Theorems 10.6 and 12.6 show s is either a 2- or a 2.5-jump. By hypothesis, it is not possible for both e and f to be in the span of s and, therefore, neither is. On the other hand, each of the other three H1-spokes has precisely 1 end in the span of s, and is contained in M. Let these spokes appear in the order t1,t2,t3 in the span of s.
We claim that the tj imply the existence of an H-yellow cycle that does not bound a face of n[G], contradicting Lemma 11.2 (3). Let P be the span of s and, for i = 1, 2, 3, let uj be the end of tj that is not in P. Because n[s U P] bounds a closed disc, both cycles in n[RU tj] containing tj are non-contractible. Thus, tj has an end in each of the a6-subpaths of R.
Lemma 12.9 implies that each tj is contained in an H-quad. Thus t1 U t2 U t3 is contained the the union of the closures of the H-quads that have an edge in P. In particular, u1, u2, and u3 occur in a 3-rim path P1 having u1 and u3 as ends. Letting P3 be the minimal subpath of P containing the ends of the tj, we see that P111 P3 ¿3 is an H-yellow cycle C. However, n[C] bounds a face of n[G]; the contradiction is that t2 and s are on different sides of n[C].
Thus, s is contained in M. Since s is one of four H1-spokes, the two cycles in n[R U s] that contain s are non-contractible.	□
In particular, s has an end in each of the two a6-subpaths of R determined by the standard labelling of n[G].
In the case f G r+5, we may, if necessary, use the reflective symmetry j ^ 4 — j (for 0 < j < 4), to arrange that the end sf of s is, in n[R'], between the end uf
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of f in R' and a, say, while the other end se of s is between a and the end ue of e. In particular, vi+1, vi+2, vi+3, and vi+4 are not in R'. Lemma 12.12 shows this always holds when f G ri+4.
Let s' be the other H'-spoke having both ends in R'. The arguments above for s apply equally well to s'. Lemma 12.10 shows that (s U s') C cl(Qi-1) U cl(Qi). In particular, s and s' are disjoint from si+2 and si+3, so these H-spokes may replace the two H1-spokes having ends in both components of R — {e, f}, as required. ■
The final technical lemma of this section will be used in the next.
Lemma 12.14. Let G G M2 and V10 = H C G, 'with H tidy. If e and e' are red, edges in the same H-rim branch, then Ae and Ae/ are disjoint.
Proof. We may choose the labelling of e and e' so that e = uw and e' = xy are such that ri = [vi, ri, u, w, ri, x, y, ri, vi+1]. As we follow Ae — e from w, there is a first edge f that is not in R. In fact, Theorem 12.1 (3) implies f is incident with w, as there can be no global H-bridge spanning e'.
Observe that f is not in H, so H C G — f. Moreover, if f is in an H-yellow cycle, then either e or e' is H-yellow, a contradiction. Thus, Lemmas 12.4 and 12.5 imply the colours of an edge of R are the same in G and G — f, unless the edge is in an H-green cycle in G that contains f. Such an edge is necessarily in [w,ri, x].
Let D be a 1-drawing of G — f and let e1 and e2 be the edges of G — f crossed in D. Since f is incident with w G (ri), Theorem 5.23 and Lemma 5.9 imply that Qi is crossed in D, so we may assume e1 G ri-1 ri ri+1 and e2 G ri+4 ri+5 ri+6. Moreover, no H-green cycle containing e2 contains f, so e2 is red in G. In particular, Lemma 11.7 implies e2 is R-separated from both e and e'.
Let ue and we be the first vertices in ri+5 as we traverse Ae — e from u and w, respectively. Likewise, we have xe and ye in ri+5 n (Ae/ — e').
CM
Claim 1. e2 g
ue,ri+5,ye
Proof. Suppose by way of contradiction that e2 g ri+4 [vi+5, ri+5, ue]; a sim-
y , ri+5, Vj+6
ri+6.
ilar argument will treat the case e2 G
If e1 G ri-1 [vi,ri,u], then e1 is red in G, so e1 and e2 are R-separated in G. Note that either e1 G r or e2 G rj+5. Lemma 12.13 implies there is a witnessing subdivision H' of V8 that contains Sj+2 and Si+3, while the other two spokes are in cl(Qi-1) U cl(Qj). Furthermore, Ae shows that f G H'; therefore, H' Ç G — f shows that e1 and e2 are R-separated in G — f, and therefore cannot cross in D, a contradiction.
The other possibility is that e1 G [u, rj, vi+1] ri+1. Since e and e2 are both red in G, Lemma 11.7 implies e2 is R-separated from e in G — f. As in the preceding paragraph, we may choose the witnessing subdivision H' of V8 to contain Sj+2 and Sj+3, while the other two spokes are in cl(Qi-1) U (cl(Qj) — f ). Again H' witnesses the R-separation of e1 and e2 in G — f, a contradiction.	□
rH
Theorem 12.1 (2) shows that any edge in either Ae n rj+5 or Ae/ n rj+5 is in a digon in G and so is not e2. Thus, e2 is further restricted to be in we,rj+5,xe . Lemma 11.7 implies Ae and Ae2 are disjoint, as are Ae2 and Ae/, which further implies that Ae and Ae/ are disjoint, as required.	■
00 1-H
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CHAPTER 13
The next red edge and the tile structure
We now know that there are red edges and every red edge comes equipped with a A. The tiles are determined by what is between "consecutive" red edges. In this section, we explain what "consecutive" means, show that consecutive red edges determine one of the tiles, and complete the proof of our main result, Theorem 2.14, by demonstrating that every red edge has a consecutive red edge on each side.
Definition 13.1. Let G g M3 and V10 = H C G, with H tidy. Let e = uw be a red edge in rj, labelled so that rj = [vj, rj, u, e, w, rj, vj+1j. A red edge ew is w-consecutive for e if:
(1)	ew G [we, rj+5, vj+6]rj+6 rj+7 (recall that we is the vertex in the peak of Ae nearest w in Ae — e);
(2)	there is no red edge in [we, rj+5, vj+6] rj+6 rj+7 between we and ew;
(3)	there is no red edge in [w, rj,vj+1]rj+1 rj+2 between w and the peak of Ae„ ;
(4)	if ew is the edge of Pw nearest w that is not in R, then there is a 1-drawing D of G — ew in which e crosses ew.
(5)	There is an analogous definition for u-consecutive.
Our first main goal is, therefore, the following.
Theorem 13.2. Let G g Mf and V10 = H C G, with H tidy. Let e = uw be red in G. Then there is a w-consecutive red edge and a u-consecutive red edge for e.
The next lemma will be helpful in the proof.
Lemma 13.3. Let G G M3 and V10 = H C G, with H tidy. Let e = uw and e be red edges in G, with e G rj and the labelling chosen so that rj = [vj, rj, u, e, w, rj, Vj+1] and e G [we, rj+5, vj+6]rj+6 rj+7. If ew is the w-nearest edge of Pw that is not in R and e and e are not R-separated in G — ew, then e has a w-consecutive red edge.
-H
Proof. Suppose there is a red edge e' in rj rj+1 rj+2 between w and the peak of Ae. Then e' is R-separated from e in both G and G — ew, showing that e and e are R-separated in G — ew, a contradiction. Thus, no such red edge exists.
Let e' be the we-nearest red edge in [we, rj+5, vj+6]rj+6 rj+7. Lemma 11.7
CD
implies e' is R-separated from e in G; if e' were also R-separated from e in G — ew, then so would e, which contradicts the hypothesis. But now Lemma 11.7 implies there is a 1-drawing of G — ew in which e crosses e', as required.	■
And now the final major proof needed to prove Theorem 2.14.
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00	Proof of Theorem 13.2. It obviously suffices to prove the existence of a w-
consecutive red edge for e. Let rj be the H-rimbranch containing e. Let ew be the edge of Pw nearest w and not in R. There are two principal cases.
Case 1: ew is incident 'with w.
We note that ew is contained in a Qj+1-bridge that is not M^ ^. Let D be a
1-drawing of G — ew. Corollary 12.7 and Lemma 5.9 show that Qj+1 is crossed in D.
Let
•	f be the edge of rj+4 rj+5 rj+6 rj+7 that is crossed in D and
•	f' be the other edge crossed in D; thus, f' G rj_1 rj rj+1 rj+2.
Claim 1. If f is not red in G, then there is a w-consecutive red edge for e.
Proof. Because we are in Case 1, no global H-bridge has w in the interior of its span and, therefore, ew is not in any H-yellow cycle that could witness the H-yellowness of any edge in rj+4 rj+5 rj+6 rj+7, (in particular, the H-yellowness of f). Therefore, Lemma 12.4 shows f is not H-yellow. Since f is not red, Theorem 11.3 implies f is H-green. Lemma 12.5 implies there is a 2.5-jump J that spans f and so that f is in the H-rim branch whose interior contains an end of J. We note that if vj+6 is in the span of J, then Lemma 7.2 (1) shows no edge in the span of J is crossed in D. Therefore, vj+6 is not in the span of J. Furthermore, if ew is not in sj+1, then H C G — ew and, therefore Lemma 6.6 (10) implies f is not crossed in D, a contradiction. This implies w = vj+1. We summarize these remarks as follows.
Subclaim 1.	• w = vj+1 and
• there is a 2.5-jump J so that:
—	f is spanned by J;
—	f is in the H-rim branch whose interior contains an end of J; and
—	vj+6 is not in the span of J.	□
Subclaim 2. Let j G {i+ 4, i + 5, i + 6, i + 7} so that f is in the H-rim branch CO	rj. Then no edge of rj is H-yellow.
Proof. Suppose some edge e' of rj is H-yellow. This implies e' is not H-green and, therefore, is not spanned by J. Let C and C' be the witnessing H-yellow and H-green cycles, respectively.
Suppose first that j G {i + 4, i + 5}. Then rj = [vj, rj, f, rj, e',rj,vj+1]. Because e G rj is not H-green, Vj+5 G {vj_1, vj} is in the interior of C' n R. This implies there is an H-yellow cycle containing Sj and the portion of rj from Vj to e'. By Lemma 11.2 (3), this H-yellow cycle must be C and, therefore, f G C. Now the fact that f is crossed in D contradicts Lemma 12.4. A completely analogous argument holds for j G {i + 6, i + 7}.	□
Let w be the vertex in rj+5 that is nearest w in Pw. Observe that w is not necessarily in the peak of Ae. (See Figure 12.1, where w is the vertex of Ae at the top right hand corner of Ae.) The following claim will be helpful in completing the proof of Case 1.
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Subclaim 3. If w = vj+6, then [w, rj+5, vj+6] is in an H-green cycle contained in cl(Qj).
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$H
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00	Proof. Let PW be the wsj+1-subpath of Pw. Since ew G Sj+1, PW C Pw — w.
Let we be the end of P^ in sj+1. Since w G sj+1 and we G sj+1, w = we. By definition of w, P^ — w is disjoint from rj+1. Therefore, P^ [we, sj+1, v»+6, r»+5, w] is an H-green cycle containing [w, r»+5,v»+6], as required.	□
The proof of Claim 1 is completed now by treating separately each of the four possibilities for f: f G rj+4, f G rj+5, f G r»+6, and f G rj+7.
Subcase 1: f G r»+4.
CD
In this case, J has an end x' G (r»+4) and the other end of J is v»+2. Lemma 7.2 (3b) implies f' G r» rj+1. We claim that if f' G rj+1, then there is another 1-drawing of G — ew in which f crosses e.
Since f G r»+4 and f' G rj+1, we see that s» is exposed in the 1-drawing D of G — ew. Note that D[Qj_1] consists of a simple closed curve crossed by D[f'], with D[r»] on one side (the inside of D[Qj_1]) and most of D[H] on the other side (this is the outside of D[Qj_1]).
We claim that we may reroute f inside D[Qj_1] so that it crosses e instead of f'. If this fails, then there is an (H — (sj+1))-avoiding path P having one end in the component of r»+1 — f' that contains v»+1 , and having its other end in Qj_1 U [vj,rj,u].
We note that D[sj+1 — vj+1] (which is possibly just v»+6) is completely outside D[Qj_1]. Therefore, P is H-avoiding. In RP2, we conclude that P cannot start inside Qj+1. Thus, P is contained in a global H-bridge. Therefore, P is a global H-bridge; we note that P has one end in the component of rj+1 — f containing vj+1. No edge of r»+2 can be spanned by P, as such an edge is already spanned by J and therefore would contradict Theorem 6.7. In the other direction, P cannot span e, as e is red and not H-green. This contradiction shows that f may be redrawn as claimed. Consequently, we may assume f' G r».
Observe that no global H-bridge can have an end y in (r»), since yv»+3 shows e is H-green, a contradiction, and yv»_2 shows f is H-yellow and, therefore, by Lemma 12.4 cannot be crossed in D. It follows from this, using Lemmas 12.4 and 12.5 and Theorem 11.3, that f' is red in G.
Suppose first that some edge e' of [x', r»+4, v»+5] is red in G. Then Ae and Ae/ are R-separated in G as witnessed by a subdivision H' of Vg consisting of R, s»_3, s»_2, and two RR-paths P1 and P2, contained in Ae and Ae/, respectively. The paths P1 and P2 are disjoint from sj+1 except that, possibly P1 contains v»+6. Thus, H' and Lemma 7.2 show that f cannot be crossed in D, a contradiction. Therefore, there is no red edge in [x', r»+4, v»+5].
Furthermore, no global H-bridge other than J has an end in [x', r»+4, v»+5), as otherwise either e is H-yellow, or f is in two H-green cycles, both contradictions, the latter of Theorem 6.7. We conclude that each edge of [x',rj+4,vj+5] is either H-yellow or contained in an H-green cycle in cl(Qj_1). Subclaim 2 shows the following.
Jh
Subcase 1 Observation: Each edge of [x , r»+4, v»+5] is in an H-green cycle contained in cl(Qj_1).
Suppose there is a red edge e' in ri+s. By Lemma 11.7, e' is R-separated from e in G. Therefore, Pu is disjoint from Sj and now we see that G — ew contains the subdivision H' of V10 consisting of (H — (sj+1)) U Pu. But J is in an H'-green
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00	cycle C and so, by Lemma 6.6 (10), C, and in particular, f, is not crossed in D, a
contradiction.
Thus, no edge of rj+5 is red in G. We consider next a 1-drawing Dj_1 of G — (sj_1). By Corollary 12.7 and Lemma 5.9, Q4_ 1 is crossed in Dj_1. From Lemmas 12.4, 12.5, and 7.2 (1), no edge in rj+2 rj+3 rj+4 is crossed in Dj_1. Therefore, it is some edge f'' in rj+5 that is crossed in Dj_1. Since no edge of rj+5 is red in G, Lemmas 12.4 and 12.5 imply that f'' is spanned by a 2.5-jump J'' = x"vj_2, with
x'' G (rj+5).	_
Now consider a 1-drawing Dj+3 of G — (sj+3). As for Qj_1 in the preceding paragraph, Qj+3 is crossed in Dj+3. In this case, rj+1 is contained in the H-yellow cycle Qj+1 (with witnessing H-green cycle containing J"). Therefore, rj+1 is not crossed in Dj+3. Lemma 7.2 (1) implies no edge in the span of J is crossed in Dj+3. Subcase 1 Observation combines with Lemma 12.5 to show that no edge in [x',rj+4, Vj+5] is crossed in Dj+3. But now we see that Qj+3 cannot be crossed in A+3, a contradiction that shows Subcase 1 cannot occur.
Subcase 2: f G rj+5.
In this case, J has an end x' G (rj+5). Subclaim 1 implies that Vj+6 is not spanned by J, so the other end of J is Vj+3. Lemma 7.2 implies the edge f' (crossed by f in D) is in rj+2.
We first show that there is no global H-bridge spanning any edge in r rj+1 rj+2. For if J' is a global H-bridge that spans such an edge, then J' does not span e, while Lemma 10.9 (1) shows it cannot be the 2-jump v^v^. Theorem 6.7 shows J' cannot span any edge in rj+3, so no edge of rj+1 rj+2 is spanned by a global H-bridge. On the other side, J' would have to span rj_2 rj_1. In that case, J and J' contradict Lemma 10.9 (4).
We also conclude that no edge of rj+5 rj+6 r.j+7 is H-yellow.
Our next principal aim is to show that each edge of [x', rj+5, Vj+6] is H-green, witnessed by a cycle in cl(Qj). We have already seen that none of the edges in [x',rj+5, Vj+6] is H-yellow; to see they are H-green, it suffices by Theorem 11.3 to CO	show none is red.
If e' is one of these edges that is red, then Lemma 11.7 implies it is R-separated from e. We note that Ae and Ae/ are disjoint, both are in cl(Qj), and w = vj+1. Therefore, e' is in rj+5, between x' and the peak of Ae. However, this shows e' and e are R-separated in G — ew and, therefore, f and rj+2 are R-separated in G — e', showing that f cannot cross anything in D, a contradiction. Therefore, no edge of [x',rj+5, Vj+6] is red, and so they are all H-green.
We next show they are not spanned by a global H-bridge. Recall that w is the vertex in rj+5 that is nearest w in Pw.
If w = Vj+6, then (Pw — ew) U (sj+1 — ew) U [w, rj+5, Vj+6] contains an H-green cycle that contains [w, rj+5, Vj+6] and is contained in cl(Qj). Theorem 6.7 shows no edge of [w, rj+5, Vj+6] is spanned by a global H-bridge, so no edge of [x', rj+5, Vj+6] is H-green by a global H-bridge. In this case, every edge of [x', rj+5, Vj+6] is H-green by a local cycle.
So suppose w = Vj+6. By way of contradiction, we suppose there is a global H-bridge J'' spanning the edge of rj+5 incident with Vj+6. Then J'' must be x''vj+8, for some x'' G [x', rj+5, Vj+6]. All edges in [x', rj+5, x''] are H-green by local cycles. For j G {i + 3, i + 8}, let ej be the edge of Sj+3 incident with Vj and let Dj be a
o
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00	1-drawing of G — ej. Corollary 12.7 implies Qj+3 has BOD and Lemma 5.9 implies
Qj+3 is crossed in Dj. Lemma 7.2 (3a) implies neither rj+6 rj+7 nor rj+3 rj+4 is crossed in Dj, while (2) of the same lemma implies neither rj+g nor rj+1 is crossed
CD	in Dj. Therefore, rj+8 crosses rj+2.
If the edge ej+8 of rj+8 that is crossed in Dj+3 is H-green because of some 2.5-jump, then Lemma 7.2 implies ej+8 can cross only rj+1 in Dj+3. Therefore, Theorem 11.3 and Lemmas 12.4 and (because no H-green cycle containing ej+8 can contain ej+3) 12.5 imply ej+8 is red in G. Likewise the edge ej+2 of rj+2 that is crossed in Dj+8 is red in G.
By Lemma 11.7, ej+2 and ej+8 are R-separated in G. Moreover, the nearer of the (rj+7 rj+8)(rj+2 rj+3)-paths P2 in A^+2 and P8 in Aei+g, along with Sj and sj+1 witness their R-separation. We now show that P8 is contained in cl(Qj+8) and must be disjoint from sj+4.
If P8 intersects sj+4 at a vertex other than vj+4, then P8 U sj+4 U rj+8 contains an H-green cycle that includes ej+8. Otherwise, P8 and sj+4 intersect just at vj+4, in which case P8 U sj+4 U rj+8 contains a cycle C that includes ej+8. The H-green cycle containing J shows C is H-yellow. Both possibilities contradict the fact that ej+8 is red.
Symmetrically, we use J'' to show that P2 is disjoint from sj+2. Thus, G contains a subdivison of V12 consisting of R, P2, P8, sj-1, sj, sj+1 and sj+2. But then G — ew contains a subdivision of V10, yielding the contradiction that f cannot be crossed in D. Therefore, there is no global H-bridge J'' spanning the edge of rj+5 incident with vj+6.
CM
We conclude that every edge of [x', rj+5, vj+6] is in an H-green cycle contained in cl(Qj).
We are now in a position to show that rj+6 has a red edge. By way of contradiction, we suppose rj+6 has no red edge. If there were a global H-bridge having an end in (rj+6}, then rj+2 is H-yellow; Lemma 12.4 shows rj+2 is not crossed in D, a contradiction. Thus, no global H-bridge has an end in (rj+6}.
Let Dj be a 1-drawing of G — (sj}. Then Corollary 12.7 and Lemma 5.9 imply Qj is crossed in Dj. However, Lemma 7.2 shows none of rj+3 rj+4 rj+5 rj+6 can be crossed in Dj, a contradiction.
Thus, rj+6 has a red edge e'. Then e is R-separated from e' in G. If e is R-separated from e' in G — ew, then f is R-separated from rj+2 in G — ew and so f cannot be crossed in D, a contradiction. Therefore, e is not R-separated from e' in G — ew, so Lemma 13.3 implies there is w-consecutive red edge for e, completing the proof in Subcase 2.
Subcase 3: f G rj+6.
CO
In this case, J has an end x' G (rj+6} and the other end is vj+g. Also, Lemma 7.2 implies f' (crossed by f in D) is in rj+g.
Suppose by way of contradiction that no edge of rj+6 is red in G. We show that no edge of rj+6 is H-yellow. As every edge in [x', rj+6, vj+7] is H-green (because of J), we assume by way of contradiction that there is an H-yellow edge in [vj+6, rj+6, x']. Let C and C' be the witnessing H-yellow and H-green cycles, respectively. Lemma 11.2 (1) implies there is a global H-bridge B contained in C', while (4) shows C C cl(Qj+1).
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The edges of the span PB of B are all H-green, so PB does not contain the red edge e. One end of B is in [w, rj, vi+1, ri+1, vi+2] and the other end is in ri+3. Furthermore, Lemma 10.9 (4) and the presence of J shows vi+4 is not the other end of B.
Write C = pP2P3P4 as in Definition 11.1 (H-yellow). Because C bounds a face n[G], C C cl(Qj+1), so that P1 = rj+1 n C. In particular, ew G C.
Choose the labelling of P2 and P4 so that the end of P2 in ri+6 is nearer to vi+6 than is the corresponding end of P4. Since there is an H-yellow cycle containing P2 and sj+2, Lemma 11.2 (3) shows this must be C. It follows that P4 = Sj+2.
Consider the subdivision H' of V6 whose rim consists of (R—(PB)) — (x', ri+6, vi+6), B, C — (ri+6 n C), and whose spokes are sj-1, sj, and Sj+3, [vi+3, ri+3, z]. Then H' does not contain ew and so must contain the unique crossing in D. Since f is not in H', this is a contradiction, showing that no edge of [vi+6, ri+6, x'] is H-yellow.
Because of J, a global H-bridge spanning an edge in [vj+6, ri+6, x'] would have to be a 2.5-jump having vi+4 as an end. But then e is in an H-yellow cycle, which is impossible. Thus, for each edge e of [vi+6, ri+6, x'], e is in an H-green cycle Cg contained in cl(Qi+1). Theorem 6.7 implies Cg is disjoint from Sj+2.
Let Di+2 be a 1-drawing of G — (sj+2). We know that Qi+2 is crossed in Di+2 (Corollary 12.7 and Lemma 5.9). Lemma 12.5 shows no edge in [vj+6,ri+6,x'] is crossed in Di+2, while J and Lemma 7.2 show no edge in [x', ri+6, v.j+7] ri+7 ri+8 is crossed in Di+2. Therefore, the crossing in Di+2 must be of an edge f'' in ri+5 crossing rj+1 rj+2.
If f'' is red in G, then Lemma 11.7 implies f'' and e are R-separated in G. Since ew G sj+1, f'' is between (in ri+5) vi+5 and the peak of Ae. Thus, f'' and e are R-separated in G — (sj+2) (using si+3 and Sj+4 as two of the four spokes). In CM	turn, this implies f'' cannot cross ri+1 ri+2 in Di+2, a contradiction that shows f''
00	is not red. Therefore, Lemmas 12.4 and 12.5 imply f'' is spanned by a 2.5-jump
vi+3x'', with x'' G (ri+5).
cm	Now let Dj+3 be a 1-drawing of G — (sj+3). We know that Qj+3 is crossed in
Dj+3. However:
•	Lemma 12.5 implies [vj+6, rj+6, x'] is not crossed in Dj+3;
•	Lemma 7.2 (1) implies [x', rj+6, «¿+7]r.j+7 rj+8 is not crossed in Dj+3; and
•	Lemma 12.4 implies rj+9 is not crossed in Dj+3. These three observations imply the contradiction that Qj+3 cannot be crossed in Dj+3, showing that some edge e' in rj+6 is red in G.
Obviously, e' G [vj+6, rj+6, x']. By way of contradiction, suppose e and e' are R-separated in G — ew. Because e G rj and e' G rj+6, Lemmas 12.12 and 12.13 imply that there is a a witnessing subdivision H' of V8 with two H'-spokes in cl(Qj) U cl(Qj+1) and the other two H'-spokes are sj+3 and sj+4. Furthermore, six of the eight ends of the H'-spokes are in the component R' of R — {e, e'} containing rj+1 rj+2 rj+3 rj+4.
Let y be the end of e' in R'. Because w = vj+1 and x' G [vj+6,rj+6, x), R' is contained in
rj+1 rj+2 rj+3 rj+4 rj+5 [vj+6, rj+6, x) . In particular, J is not an H'-spoke and at most two of the H'-spokes have ends in the span of J. Lemma 7.2 (1) implies the contradiction that the span of J, which includes f, cannot be crossed in D. We conclude that e and e' are not R-separated in G — ew. Lemma 13.3 implies that e has a w-consecutive edge, as required.
u
CD
CO	Subcase 4: f G r+7.
I
In this case, J has an end x' G (r+7). If the other end of J is v+5, then Lemma 7.2 (3b) implies f' is in r+3. The contradiction is that Qj+1 is not crossed in D. Therefore, the other end of J is v. Lemma 7.2 (3b) implies f' is in r r+1.
Suppose there is no red edge in r+6 r+7. Let ej+8 be the edge of sj+3 incident with v+8 and let Dj+8 be a 1-drawing of G — ej+8. Corollary 12.7 and Lemma 5.9 imply Qj+3 is crossed in Dj+8. No edge in r+6 is spanned by a 2.5-jump o	having an end in (r+6), as otherwise e is H-yellow. Therefore, Lemmas 12.4 and
12.5 imply no edge of r+6 is crossed in Dj+8. Lemma 7.2 (1) shows that no edge of [x',r+7,v+8]r+8 r+9 is crossed in Dj+8. We conclude that some edge f of [v+7, r+7, x'] is crossed in D.¿+8.
Lemmas 12.4 and 12.5 imply that there is a 2.5-jump v+5x'', with x'' G (v+7,r+7,x'], and, furthermore, that f G [v+7,r+7,x'']. Lemma 7.2 (3b) implies f crosses an edge e' in r+4. Lemmas 12.4 and 12.5 imply e' is red in G.
Let y be the end of e' nearest v+5 in r+4. The rr+5-path P0 contained in the uue-subpath of Ae — e must have vj+5 as an end, since otherwise e is either H-green or H-yellow. Symmetrically, the r+4r+9-path P4 contained in the yye -subpath of Ae' — e' has v as an end.
Lemma 11.7 implies e' is R-separated from e in G. Therefore, P0 and P4 are disjoint. This implies that R U P0 U P4 U sj+2 U sj+3 U sj+4 is a subdivision V10 in G — ew, showing that f cannot be crossed in D, a contradiction that proves there is a red edge e'' in r+6 r+7.
Suppose e and e'' are R-separated in G — ew. Lemma 12.12 implies that a witnessing subdivision H' of V8 is such that the component R' of R — {e, f} containing six of the eight ends of H'-spokes contains r +1 r+2 r+3 r+4 r+5.
However, J spans [x', r+7, v+8] r+8 r+9, so at most two H'-spokes have ends that are in the span of J. Lemma 7.2 (1) combines with H' to yield the contradiction that the span of J, including f, cannot be crossed in D. It follows that e and e'' are not R-separated in G — ew, and now Lemma 13.3 implies e has a w-consecutive red edge, completing the proof of Claim 1.	□
With Claim 1 in hand, we may assume f is red. Recall that f and f' are the edges crossed in D, with f G rj+4 rj+5 rj+6 r+7 and f' G rj_1 rr+1 rj+2. The proof in Case 1 is completed by finding a w-consecutive red edge for e. We proceed in four cases, basically depending on which side of Ae each of f and f' is on.
Subcase 1: f is in r+4[v+5,r+5,ue] and f' is in r_1 [vj,rj,u].
Since f and f' are not R-separated in G — ew and, therefore, not R-separated in G, f' cannot be red (Lemma 11.7). If f' is H-yellow in G, then Lemma 12.4 shows it is not crossed in D. Therefore, Theorem 11.3 implies f' is H-green in G. Lemma 12.5 says there is a 2.5-jump J spanning f' so that f' is in the partial H-rim branch spanned by J. As J cannot span e (e is not H-green), Lemma 7.2 (3b) and our current context (f in r+4 r+5 and f' in r_1 r) implies this is possible only if f' G r_1 and f G r+5. However, the red edges f and e are R-separated in G, implying that G — ew still has five spokes (we may replace sj+1 with the rr+5 subpath of Pu). Thus, f' is H'-green in G — ew, for some H' = V10. This is impossible, as f' is crossed in D (Lemma 6.6 (10)).
Subcase 2: f G rj+4[v+5,rj+5,ue] and f' G [u, rj, v+1]rj+1 rj+2.
CD
■ 1 J-H
CD C/3
a
CD U
o
CM
00
o
I
cm 00 cm cm
00	In this subcase, f is R-separated in G from e. The witnessing subdivision H' of
V8 can be chosen to contain the "nearer" (ri-1 rj)(ri+4 ri+5)-paths, one from each of Af and Ae, together with the H-spokes si+2 and si+3 to construct H'.
We claim that this H' also shows that f is R-separated from f ' in G — ew. If f € ri+4, then, since Qi+1 is crossed in D, f ' G r ri+1. In this case, H' contains the spokes si+2 and Sj+3, so indeed f and f' are in disjoint H'-quads, as required. If f G ri+5, then f' G ri+2 by Lemma 7.2 (3b), and again f and f' are in disjoint H '-quads, showing f and f ' are R-separated in G — ew. Observation 11.6 (1) yields the contradiction that f and f ' do not cross each other in D.
Subcase 3: f G [we, ri+5, vi+6]ri+6 ri+r and f ' G rj_1 [vj,rj,w].
If f is R-separated from e in G—ew, then it cannot cross f ' in D, a contradiction. Otherwise, Lemma 13.3 implies there is a w-consecutive red edge for e.
Subcase 4: f G [we, ri+5, vi+6]ri+6 ri+r and f' G [u,rj,vi+1]ri+1 ri+2.
If f ' = e, then we are done: Lemma 13.3 implies e has a w-consecutive edge. So we assume f ' = e. If f ' is red in G, then Lemma 11.7 implies it is R-separated from f in G. Therefore, f ' is R-separated from f in G — ew, a contradiction; so f ' is not red in G.
Suppose by way of contradiction that f ' is H-yellow, with witnessing H-yellow and H-green cycles C and C', respectively. If ew is not in C, then Lemma 12.4 yields the contradiction that f' is not crossed in D.
If ew is in C, then let P2 be the RR-subpath of C containing ew, let P' be the
cm
RR-subpath of Ae — e that contains ew, and let J be the global H-bridge contained in C'. The end of P' in ri+5 cannot be in the interior of the span of J, as then either the peak of Ae is a vertex, in which case we have that Ae is H-yellow, yielding the contradiction that e is H-yellow, or the peak of Ae consists of parallel edges, both in the span of J, contradicting Theorem 6.7.
It follows that P' has its end in ri+5, but not in the interior of the span of J.
On the other hand, P2 has, by Definition 11.1, one end in the interior of the span of J. But now (P2 U P') — ew contains an R-avoiding subpath that intersects at most the one spoke sj+1. Therefore, this subpath is in an H-green cycle and contains an edge spanned by J, contradicting Theorem 6.7. It follows that f' is H-green.
Theorem 12.6 implies that H has no 3-jumps. If f' is H-green by a 2.5-jump J, then, because J cannot span e, Lemma 7.2 (3b) implies f G [we, rj+5, vj+6]rj+6 and f' G rj+2. Let x be the end of f closest to we in rj+5 rj+6. Let H' be the subdivision of V8 obtained from H — (sj+1) by replacing sj+2 with Px (recall this is defined in Theorem 12.1 (3)). Now f and f' violate Lemma 7.2 (3b) relative to H'. Therefore, f' is not H-green by a 2.5-jump.
Lemma 7.2 implies f' is not H-green by a 2-jump, as then it is not crossed in D. Thus, f' is H-green by a local H-green cycle C. Lemma 12.5 implies ew is in C.
CD	Since f cannot be R-separated from f' in G — ew, we see that f is not R-separated
from e in G — ew. Now Lemma 13.3 implies there is a w-consecutive red edge for e, concluding the proof for Case 1.
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Case 2: ew not incident with w.
£
By Theorem 12.1 (3), w is incident with a global H-bridge Jw. Since w is not incident with ew, w = vj+1, and therefore Jw is the 2.5-jump wvj+3.
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CM
00 Gï
00	We observe that, since ew is not incident with w, its incident vertex in r is
in the interior of the span of Jw. Moreover, ew is the first edge of an R-avoiding riri+5-path P in Ae — e, which, together with a subpath of r ri+1, Si+2, and a subpath of rj+5 rj+6 makes an H-yellow cycle C. By Lemma 11.2 (3), there is only one C-bridge in G and, therefore, P = si+1. In particular, ew G si+1.
Claim 2. No edge in rj+7 r.j+s is H-yellow.
CD
Proof. Suppose some edge e' in rj+7 r.j+s is H-yellow. Let C and C' be the witnessing H-yellow and H-green cycles, respectively. By Lemma 11.2 (1), C' contains a global H-bridge J'.
In the case e' is in rj+7, the span of J' contains a vertex of rj+2 in its interior. Theorem 6.7 implies J' = Jw. But now C U Qj+1 contains an H-yellow cycle C'' for which there is a C''-interior C''-bridge containing an edge of Si+2, contradicting Lemma 11.2 (3). Therefore, no edge in rj+7 is H-yellow.
Now we suppose e' is in rj+s. Lemma 10.9 (1) shows J' does not have Vj+3 as an end, so J' has one end x G (rj+3) and its other end is Vj+6. But now C U Qj+4 contains an H-yellow cycle C'' having a C''-interior C''-bridge containing an edge of Sj+4, contradicting Lemma 11.2 (3).	□
Claim 3. Some edge of ri+7 is red.
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Subclaim 1. If there is a red edge in either ri+3 ri+4 or ri+8 ri+g, then there is a red edge in ri+8 ri+g. Furthermore, among all such red edges, the one e'' with an end x'' nearest vi+8 in ri+8 ri+g is such that (e'')x is not incident with x'' (that is, Case 1 does not apply to e'' and x'').
Proof. Suppose no edge of ri+7 is red. By Theorem 11.3 and Claim 2, every edge in ri+7 is H-green.
CM
Proof. We first suppose no edge of ri+3 ri+4 is red. Then there is a red edge in ri+8 ri+g. For any such red edge e'', if the end x'' of e'' nearest to vi+8 is incident with (e'')x , then Case 1 shows there is an x''-consecutive red edge e for e''. By Definition 13.1 (1), e G ri+1 ri+2 ri+3 ri+4. Since the edges in ri+1 ri+2 are H-green, e G ri+1 ri+2. But then e is a red edge in ri+3ri+4, a contradiction. Therefore, x'' is not incident with (e'')x , as required.
The alternative is that there is a red edge in ri+3 ri+4. Among all such edges, let e' be the one having an incident vertex x' nearest vi+3 in ri+3 ri+4. Because of Theorem 6.7 and Jw, x' is not incident with a 2.5-jump x'vi+1 or x'vi+2. Therefore, x' is incident with (e')x , and we conclude from Case 1 that there is an x'-consecutive red edge e'' for e'. Because of Jw, every edge in ri+6 is either H-yellow or H-green and so, in particular, is not red. By assumption, no edge of ri+7 is red. By Definition 13.1 (1), e'' G ri+8 ri+g. Also, Ae» separates si+3 from Ae' in cl(Qi+3) U cl(Qi+4).
Let x'' be the end of e'' nearest vi+8 in ri+8 ri+g. By way of contradiction, suppose x'' is incident with (e'')x . Then Case 1 shows there is an x''-consecutive red edge e for e''. But e is not in ri+1 ri+2 because Jw makes every one of those edges H-green. Therefore, e is in ri+3 ri+4. Since Ag separates si+3 from Ae» in cl(Qi+3) U cl(Qi+4), we see that e is nearer to vi+3 than e' is, contradicting the choice of e'. Therefore x'' is not incident with (e'')x , as required.	□
• 1
Subclaim 2. No edge in either ri+3 ri+4 or ri+8 ri+g is red.
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CM 00
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00
0
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£ CO CO
Proof. Suppose by way of contradiction that there is a red edge in either r»+3 r»+4 or rj+8 rj+9. By Subclaim 1, there is a red edge e'' in rj+8 rj+9 so that the end x'' of e'' nearest vj+8 in r»+8 r»+9 is not incident with (e'')x . Therefore, Theorem 12.1 (3) implies x'' is incident with a 2.5-jump that is either x''v»+6 or x''vj+7. It cannot be the former, as the 2.5-jumps x''v»+6 and Jw contradict Lemma 10.9 (4). Therefore, x'' is in the interior of r»+9 and the 2.5-jump is x''vj+7. The contradiction is obtained by showing that cr(G) < 1.
Let D be a 1-drawing of G — (rj+7). There is still a subdivision H' of V8 in G — (r»+7) consisting of the rim (R — (r»+7)) U x''v»+7 and the four spokes s», sj+1, s»+2 and s»+3rj+8[vj+9,r»+9,x'']. We note that x''v»+7 is an H'-rim branch, contained in an H'-quad Q consisting of s»+2, r»+2, s»+3 r»+8 [v»+9, r»+9, x''], and x''vj+7.
We aim to show D[Q] is clean, so by way of contradiction, we assume D[Q] is not clean. The H'-rim branches of Q are r»+2 and x''v»+7. Since rj+1 r»+2 is not crossed in D (Lemma 7.2 (3a)), we deduce that x''v»+7 is crossed in D. Furthermore, the cycle r»+3 s»+4 r»+8 s»+3 (which is Q»+3 in G) is H'-close and, therefore Lemmas 5.3 and 5.4 imply Q»+3 is not crossed in D. It follows that x''v»+7 crosses r»+4 in D, so s»+3 r»+8 [v»+9, r»+9, x''] is exposed in D, from which D[H'] is completely determined. (See Figure 13.1.)
Vi+1
Vi+2
e Vi
Vi+3 \ ■ / x
Vi+8	Vi+9
m
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Figure 13.1. D[H']
Our contradiction is obtained from a detailed consideration of Ae». We first show that v»+4 is in the peak of Ae». To see this, we note that the rj+9rj+4-subpath of Ae" — e'' that starts nearest x'' is simply s»+4, as otherwise there is an H-yellow cycle C with more than one C-bridge. Theorem 12.1 (3) implies the subpath of Ae» — e'' from x'' to the peak of Ae» has at most one edge in R; therefore, there is no edge of r»+4 between v»+4 and the peak of Ae». That is, v»+4 is in the peak of Ae" .
Let y'' be the end of e'' different from x''. Because y'' is too close to Jw, it is not incident with a global H-bridge. Thus, the edge of Ae» — e'' incident with y'' is not in R and, therefore, is the first edge of an rj+9rj+4-subpath P of Ae» — e''. Let z'' be the other end of P.
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00 Gï
CSI CSI
u
a
CD U
00	We note that z'' = vi+4, as D[P] cannot cross D[H']. Therefore, z'' G (vi+4, ri+4,
vi+5]. If z'' is in the peak of Ae», then z'' and vi+4 are joined by parallel edges, one of which is not in H'. That one must cross D[H'], which is a contradiction. Therefore, z'' is not in the peak of Ae». But now Theorem 12.1 (3) implies z'' is in the interior of the span of a global H-bridge J'' that has an end in the peak of Ae»; therefore, this end of J'' is in ri+4.
The end of J'' in ri+4 must be vi+4, as otherwise J'' is a 2.5-jump with one end being vi+7, which, together with x''vi+7, contradicts Lemma 10.9 (1). Therefore, J'' is either vi+4vi+6 or vi+4u'', with u'' G (ri+6). However, Lemma 7.2 (1) or (3a) and J'' show that ri+4 cannot be crossed in D, a contradiction that finally shows D[Q] is clean.
We can now obtain the claimed 1-drawing of G. Observe that x''vi+7 is in an H-green cycle that, by Lemma 6.6 (8), has only one bridge. Also, if there is a Qi+2-bridge other than Mq+2, then cl(Qi+2) has an edge f not in Qi+2. But Theorem 5.23 and Lemma 5.9 imply Qi+2 would be crossed in any 1-drawing of G — f; however, both ri+2 and ri+7 are H-green courtesy of Jw and x''vi+7. Therefore, Qi+2 has only one bridge. It follows that there are only two Q-bridges in G, one of which is ri+7. Since D[Q] is clean, it bounds a face of D[G — (ri+7)] and it is easy to put ri+7 into this face so as to obtain a 1-drawing of G. That is, cr(G) < 1, a contradiction completing the proof of the subclaim.	□
We are now in a position to finish the proof of Claim 3. Let e3 be the edge of si+3 incident with vi+3 and let D be a 1-drawing of G — e3. Corollary 12.7 and Lemma 5.9 imply Q3 is crossed in D. It follows that there is an edge e in ri+6 ri+7 ri+8 ri+9 that is crossed in D.
The H-yellow cycle Qi+1 contains ri+6, so Lemma 12.4 implies ri+6 is not crossed in D. By assumption for ri+7 and by Subclaim 2 for ri+8 ri+9, no edge of ri+7 ri+8 ri+9 is red. Lemmas 12.4 and 12.5 imply that e is spanned by some 2.5-jump J', and, moreover, e is in the H-rim branch whose interior contains the end x' of J'.
If e G ri+7, then J' is either x'vi+5 or x'vi. Suppose first that J' = x'vi+5. Lemma 7.2 (3b) implies e crosses an edge in ri+4. But Theorem 6.7 shows ri+4 cannot be in the span of a 2.5-jump, so Lemmas 12.4 and 12.5 imply no edge of ri+4 is crossed in D. Thus, J' = x'vi+5.
Now we suppose J' = x'vi. In this case, Lemma 7.2 (3b) implies e crosses an edge in ri+1, while (1) of the same lemma implies no edge in the span of J, which includes ri+1, is crossed in D. We conclude that e G ri+7.
If e G ri+8, then J' is either x'vi+6 or x'vi+1. Theorem 6.7 shows the latter does not happen. Lemma 10.9 (4) shows the former does not happen. Therefore,
e G ri+8.
The last possibility is that e G ri+9. In this instance, J' is either x'vi+7 or x'vi+2. Theorem 6.7 precludes the latter possibility, so we assume J' = x'vi+7. However, in this case, Lemma 7.2 (3b) implies e crosses an edge e in ri+5, in which case neither e nor e is in Qi+3, contradicting the fact that Qi+3 is crossed in D. □
We now finish the proof of Case 2 and, therefore, Theorem 13.2. By Claim 3, we may let e' = xy be the red edge in ri+7 that is nearest vi+7 in ri+7, labelled so that x is nearer vi+7 in ri+7 than y is. We look for the x-consecutive red edge
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00
m
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00	for e'. As the edges spanned by Jw are H-green, e is the only possibility for the
x-consecutive red edge for e'.
Suppose first that e' and x satisfy the condition for Case 1. We have proved there is an x-consecutive red edge for e' and, as just mentioned, this can only be e. This implies that x = Vj+7. To see that e' is the w-consecutive red edge for e, it remains to show that e and e' can be crossed in G — ew. (This is the only asymmetric condition in the definition of consecutive.)
The H-quad Qj+1 is also an H-yellow cycle and so (Lemma 11.2 (3)) bounds a face of G. It follows that e and e' are not R-separated in G — ew and, therefore Lemma 11.7 implies there is a 1-drawing of G — ew in which e and e' are crossed, as required.
The alternative is that e' and x do not satisfy the condition for Case 1. Then, just as for w above, there is a 2.5-jump Jx = xvj+5 incident with x. Also, the edge ex of Ae/ — e' that is nearest x and not in R is in Sj+7. Since Qj+1 bounds a face of G, e and e' are not R-separated in G — ew and, therefore, Lemma 11.7 implies there is a 1-drawing of G — ew in which they are crossed.	■
The following is a consequence of Definition 13.1 and Theorem 13.2.
Lemma 13.4. Let G G M3 and V10 = H C G, with H tidy. With the labelling of e = uw and ew as in Definition 13.1, if x is the end of ew nearest we in [we, rj+5, Vj+6]rj+6 rj+7, then e is x-consecutive for ew.
Proof. By Theorem 13.2, there is an x-consecutive red edge e'' for ew. Conditions (2) and (3) of Definition 13.1 applied to ew being w-consecutive for e and the same conditions applied to e'' being x-consecutive for ew imply that e = e''.	■
CM
CM CM
Lemma 13.5. Let G G M2, Vlo = H C G, and let n be an embedding of G in RP2 so that H is n-tidy. Let C be a contractible cycle contained in M so that C is the union of a 3-rim path C O R (recall Definition 11.1 (1)) and an R-avoiding
CO
path P. Then, for every edge e of C O R, there is an H-green cycle containing e CO	and contained in H P.
The main goal of this work is to prove Theorem 2.14. The following lemma will be very helpful.
Proof. The graph H U P is 2-connected and not planar, so every face of n[H U P] is bounded by a cycle. There is a face F of H U P contained in M and incident with e; by the preceding remark, F is bounded by a cycle C'.
Let j be the index so that e G rj; thus, F is Qj-interior. Since F is also C-interior, C' O H C (sj rj sj+i). In particular, there is at least one edge of C' that is in P but not in H.
Observe that (sj rj sj+i) — e has two components and K2. Since C' contains a vertex in each of and K2 (namely the ends of e), C' contains an (sj rj sj+L)-avoiding KLK2-path P'. Thus, P' C P.
Let C'' be the cycle in (sj rj sj+i) UP'. Then C'' is evidently an H-green cycle containing e, as required.	■
Now for the main result.
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Theorem 2.14 If G is a 3-connected, 2-crossing-critical graph containing a subdivision of Vlo, then G G T(S).
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00
00	Proof. By Theorem 10.4, G contains a tidy subdivision H of V10; let n be
an embedding of G in RP2 so that H is n-tidy. The strategy is to show that, between every red edge e = uw and its w-consecutive red edge ew, there is one of the thirteen pictures (as defined just before Lemma 2.11). This is accomplished by showing that e produces "one side" of the picture and ew produces the other. Let i G {0,1, 2,..., 9} be such that e G rj; we choose the labelling so that rj = [vj,rj,u, e, w,rj, vj+1]. Thus, e^ G rj+5 rj+6 rj+7.
Let x be the end of ew so that e is the x-consecutive red edge for ew. Let P1 be the wex-subpath of R that is a 3-rim path (Definition 11.1 (1)); likewise P2 is the xeww-subpath of R that is a 3-rim path.
Claim 1. Let B be a global H-bridge spanning an edge of p. Then:
(a)	B has ends we and x;
(b)	we = vj+5; and
(c)	ew G Sj+1 and (eM)x G Sj+2.
The analogous claims holds for P2.
Proof. We remark that the span of B does not include in its interior a peak vertex of Ae, and does not include ew. Therefore, B has both its attachments in
P1.
Consequently, the attachments of B are contained in rj+5 rj+6 [vj+7, rj+7, vj+8}. Theorem 10.6 implies one end of B is vj+5 and the other end is in [vj+7, rj+7, vj+8}.
It follows that we = vj+5. At the other end, we claim x is in B. We note that ew is in rj+7, so that H — (sj+4} shows that e and ew are R-separated. Let x' be the end of B in rj+7.
If (ew)x is not in sj+2, then let ej+7 be the edge of sj+2 incident with vj+7 and let D be a 1-drawing of G — ej+7. Corollary 12.7 and Lemma 5.9 imply Qj+2 is crossed in D. The presence of J and B combine with Lemma 7.2 (1) to show that neither [w,rj,vj+1] rj+1 rj+2 nor rj+6 [vj+7, rj+7, x'], respectively, is crossed in D.
CO
It follows that some edge e' in [x', rj+7, vj+8] is crossed in D. Let Pw be the path in Ae described in Theorem 12.1 (3). Since Pw does not have vj+1 as one end, and its other end is vj+5, its only intersection with sj+1 can be in (sj+1}. Such an intersection produces an H-green cycle that shows the edge of rj incident with vj+1 is in two H-green cycles, contradicting Theorem 6.7. Therefore, Pw is disjoint from Sj+1.
Using Pw, sj+1, sj+3 and sj+4 as spokes and R as the rim yields a V8 that shows rj+7 is R-separated in G — ej+7 from [vj,rj,w]; thus, Observation 11.6 (1) shows e' crosses an edge e'' of rj+3 in D. Lemmas 12.4 and 12.5 shows e' and e'' are red in G. Lemma 11.7 shows that e' and e'' are R-separated in G. Lemma 12.13 shows that a witnessing V8 can be chosen to avoid ej+7. But now D contradicts Observation 11.6 (1). Therefore, (ew)x is in sj+2 and incident with vj+7.
If B has an end in (rj+7}, then Theorem 12.1 (3) implies Px n rj+7 has just one edge, namely xvj+7 and, consequently, x is in B.
If, on the other hand, vj+7 is an end of B, then Theorem 12.1 (3) implies x must be incident with ex and, therefore x = vj+7. Again, we see that x is in B.
Observe that Jw and B are now seen to be completely symmetric with respect to (e, w) and (x, ew); in particular, we conclude that ew G sj+1.	□
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If there is a global H-bridge B spanning an edge of P1, then we let P1 = B. Otherwise, we let P1 = p. A completely analogous discussion holds for P2 to yield the wxWe-path P2'.
Our next claim identifies the cycle that is the boundary of our picture.
Claim 2. The closed walk PwP>1'PxP2 is a cycle.
Proof. If the edge of Pw incident with w is in R, then Theorem 12.1 (3) shows that w is incident with a global H-bridge B. Claim 1 implies B = P2. Thus, w has degree 2 in the closed walk PwP1 PxP2. Otherwise, P2 = P2, w is incident with ew, and again w has degree 2 in Pw P1 Px P2.
The other "corners" we, x, and xew are treated similarly.	□
Definition 13.6. Let e and e' be red edges and let w and x be the ends of e and e', respectively, so that e' is the w-consecutive red edge for e and e is the x-consecutive red edge for e'. Let P1 be the xAe-path in R that is a 3-rim path and let P2 be the wAe/-path in R that is a 3-rim path. Let Pw be the wwe-path in Ae — e and let Px be the xxe -path in Ae/ — e'. For i = 1, 2, let Pj' be Pj unless there is a global H-bridge Bi spanning an edge of Pi, in which case Pi' = Bi.
The cycle Ce is the composition Pw P1 PxP2.
We will see that Ce is the outer boundary of the one of the thirteen pictures that occurs. We observe that Ce is in the boundary of the closed disc in RP2 consisting of the union of the closed discs bounded by rj rj+1 rj+2 sj+3 rj+7 rj+6 rj+5 sj, P1 P1 (if P1 = p), and P2'P2 (if P2' = P2). Therefore, Ce is the boundary of a closed disc De in RP2.
We now prove three claims that will be useful for finding the various parts of the picture.
Claim 3. Let C be a cycle contained in De. If either C n P1 or Cn P2 is empty, then C bounds a face of n[G].
Proof. By symmetry, we may suppose CnP1 is empty. Let M be the C-bridge containing sj+4.
Subclaim 1. If B is a C-bridge different from M, then n[CUB] is contractible in RP2.
Proof. We start by noting that n[B] C M, since P2 is either just an edge that is a global H-bridge (and so in D and forcing B to be in M) or P2' = P2 and there is no global H-bridge having an attachment in (P2). In the latter case, any global H-bridge having an attachment at an end of P2 (say w), has its other attachment in the H-rim R — (P2). Such an attachment is in Nuc(M), contradicting the assumption that B = M.
CD	It follows that n[C U B] is contained in M and totally disjoint from Sj+4.
Therefore, n[C U B] is contractible, as claimed.	□
CD
Let H' be the subgraph of HUP{UP2' consisting of (R — ((Pi)U(P2)))U(PiUP2') and the three H-spokes sj+3, si+4, and Sj. The following claim shows that H' is a subdivision of V6. (The notation ||y|| is in Definition 4.1 (1).)
Subclaim 2. Ce n Sj C ||vj+5|| and Ce n Sj+3 C ||vj+3|.
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00	Proof. Recall that Pw is contained in Ae. Theorem 12.1 (the existence of
and , together with (3)) implies Pw is internally disjoint from Pu and, therefore, cannot intersect Sj, except possibly at their common end point vj+5. The analogous argument using Ae applies for Sj+3.	□
rO	^
S	If C does not bound a face of n[G], then let e' be any edge of any C-interior
C-bridge and let D be a 1-drawing of G — e. Subclaim 2 implies that C is H'-close (Definition 5.2). Lemmas 5.3 and 5.4 imply C is clean in D. Therefore, D contains a 1-drawing of C U M in which C is clean and Lemma 5.6 implies C has BOD. It now follows from Corollary 4.7 that cr(G) < 1, the final contradiction.	□
We find structures in the Ce-interior that lead to the pictures. Our discussion will be w-centric; there is a completely analogous discussion for x.
A useful observation is the following. Recall that Pw is the wwe-path in Ae — e (Theorem 12.1 (3)) and Px is the analogous xxew-path in Aew.
Claim 4. (1) No Ce-interior Ce-bridge has an attachment in each of the components of (Ce — Px) — ew. (2) No Ce-interior Ce-bridge has an attachment in each of the components of
ex.
(Ce — Pw ) —
Proof. Let H' be a subdivision of V8 witnessing the R-separation of e and
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As e and ew are R-separated in neither G — ew nor G — ex, ew and ex are both in H'. Since e and ew are in disjoint H'-quads, ew and ex are in disjoint H'-spokes, which we denote as Pw and Px, respectively; Pw and Px are contained in the closed disc bounded by n[Ce].
Subclaim 1. There is such an H' so that Px = Px.
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Proof. As a first case, suppose Ce n sj = 0. Then we may choose H' to be R, sj, sj+4, Pw, and Px, and we are done. In the second case, Ce n sj+3 = 0; replace sj with Sj+3.
In the final case, Censj and Censj+3 are not empty. In this instance, ew G rj+7.
CO	We may choose H' to consist of R, sj+4, sj, sj+1, and Px, the latter being contained
in cl(Qj+2).	□
By symmetry, it suffices to prove (1). Suppose by way of contradiction that there is a Ce-interior Ce-bridge B having an attachment in each component of (Ce — Px) — ew. Subclaim 1 implies there is a subdivision H' witnessing the R-separation of e and ew so that Px C H'. Let Pw be the other H'-spoke contained in the interior of Ce.
Let C' be the cycle bounding the Ce-interior face of Ce U Pw that is incident with ew. The Ce-bridge B contains a subpath P' joining the two components of (C' — Px) — ew. Now ((C' — Px) — ew) U P' contains an R-avoiding path P'' that can replace Pw in H' to get another subdivision of Vg that witnesses the R-separation of e and ew in G — ew. However, this contradicts the fact that e and ew are not R-separated in G — ew.	□
Here is our final preliminary claim.
Claim 5. Let B be a Ce-interior Ce-bridge. Then B is just an edge and its ends.
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three attachments.
Subclaim 1. B has at most two attachments in each of Ce — P{ and Ce — P2.
Proof. By symmetry, it suffices to prove the first of these. Suppose B has at least two attachments in Ce — P{. Let y and z be the ones nearest the two ends of Ce — P{. There is a cycle in B U Ce consisting of a Ce-avoiding yz-path in B and the yz-subpath of Ce — P{. Claim 3 implies this cycle bounds a face of n[G] and, therefore, B can have no other attachment in Ce — P{.	□
Subclaim 2. att(B) n P[ C {x, we} and att(B) n P2' C {w,xe™}.
Proof. By symmetry, it suffices to prove the first of these. By way of contradiction, suppose B has an attachment y in (P{}. Because B has at least three attachments, Subclaim 1 implies B has an attachment z in P2'. Any Ce-avoiding yz-path in B contradicts Claim 4.	□
From these two subclaims, we easily deduce that:
•	B has at most four attachments;
•	one of w and xew is an attachment of B; and
•	one of x and we is an attachment of B.
Observe that Claim 4 (1) implies that not both w and we are attachments of B, while (2) implies that not both x and xew are attachments of B. Therefore, att(B) n (P{ U P2') is either {w,x} or {we,xe»}.
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Subclaim 3. att(B) n (P{ U P2') = {we,xe™}.
Proof. Suppose by way of contradiction that att(B) n (P{ U P2') = {w,x}. As B has at least three attachments, there is an attachment y in (Pw} U (Px). By symmetry, we may assume y G (Pw}. Let Pbe a Ce-avoiding yw-path in B. Then the union of Pand the yw-subpath of Pw is a cycle Cin De

Since y and w are in Pw — we, Cis disjoint from P{. Claim 3 implies C bounds a face of n[G]. On the other hand, Pw is contained in the boundary of the face bounded by Ae and, therefore, Cyw n Pw is in the boundary of two faces of n[G]. We deduce that Cn Pw is just the edge wy.
Furthermore, Claim 4 implies w and y are in the same component of Pw — ew. Therefore, the definition of ew implies wy is in R, and consequently P2' is a global H-bridge spanning wy. However, any edge of B incident with w — and there is at least one such — must be in the interior of the face of n[G] bounded by the H-green cycle containing P2' (Lemma 6.6 (8)). This contradiction proves the subclaim. □
We are now ready to complete the proof of the claim. Any vertex in att(B) \ {we, xew } is in (Pw) U (Px). Subclaim 1 implies there is at most one of these. Since B has at least three attachments, there is at least one of these. We conclude there is exactly one such attachment y. We may choose the labelling so that y G (Pw).
Lemma 5.19 implies B is isomorphic to K13.
The vertex y is in the interior of Pw. Thus, both edges of Pw incident with y are in the boundary of the face bounded by n[Ae]. Consequently, any edge of G incident with y is in De.
Let c be the vertex of degree 3 in B. Claim 3 implies that the cycles [y, c, we, y] and [y, c, xew, P2', w, Pw, y] both bound faces in De. Therefore, y has degree 3 in G.
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Let e' be the edge cwe of B and let D' be a 1-drawing of G — e'. Consider the subdivision H' of V6 consisting of (R — ((Pi) U (P2)) U (P{ U P2'), Px, si; and si+4. Then H' shows that Pw U (B — e') is not crossed in D'.
The path P' = [ c, cy, y, Pw, we] is not crossed in D'. Since y has degree 3 in D', we may add the edge wec to D' alongside P' without crossing to obtain a 1-drawing of G. This is the final contradiction that shows B has only two attachments. Lemma 5.19 shows B is just an edge and its ends.	□
We now have our preliminary lemmas in hand and proceed to complete the proof of Theorem 2.14.
Definition 13.7. Let Ce be decomposed as PwP'PxP2' as in Definition 13.6.
(1)	If f is an edge not in Ce with ends w and xew and P2 has length 1, then f is a w-chord.
(2)	If f is an edge not in Ce joining w to a vertex y G (Px) and the yxew-subpath of Px has length 1, then f is a w-slope.
(3)	If f and f' are edges not in Ce, with f joining w with z G (P2') and f' joining z to z' G (Px), and if P2 has length 2, while the z'xew-subpath of Px has length 1, then {f, f'} is a w-chord+w-slope.
(4)	If f is an edge not in Ce joining xew to a vertex y in (Pw), and both P2 and the yw-subpath of Pw have length 1, then f is a w-backslope.
(5)	If f is an edge not in Ce joining y G (Pw) and z G (Px), and the paths Pw and Px have length 2, while Pi' and P2' have length 1, then f is a crossbar.
The five situations in Definition 13.7 are illustrated in Figure 13.2.
P2
Pi
Pi
Pi
Pi
	f		f		f n»				
P 1 w		Px						y*	f
		P Pw		P P Px Pw		PP Px Pw		PP Px Pw	
P1
P1
P1
P1
P1
Figure 13.2. Definition 13.7.
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Claim 6. If ew is in neither an H-yellow nor an H-green cycle, then every edge of P2 is H-green. If C is the set of H-green cycles containing edges of P2, then Ce U (UCeC)C contains either:
(a)	Ce plus a w-chord;
(b)	Ce plus a w-slope; or
(c)	Ce plus a w-chord+w-slope.
Proof. Because ew is not in an H-yellow cycle, Theorem 12.1 (3) implies w is incident with ew.
Case 1: some edge of P2 is spanned by a global H-bridge'..
Let B be a global H-bridge spanning an edge of P2. Claim 1 implies B has ends xew and w, xew = vi+3, ew G si+1, and ex G Sj+2. Since w is incident with ew, we have w = vi+1.

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We show (b) occurs by proving that ri+1 is a w-slope. We show that ri+1 and ri+2 are both paths of length 1, starting with the latter.
We note that Px is equal to si+2 ri+2. Moreover, ri+2 has the face of n[G] bounded by the H-green cycle Cg containing B on one side and the face bounded by A6w on the other. Thus, ri+2 is just a single edge.
Claim 5 shows that the Ce-bridge B' containing ri+1 is just an edge and its ends. Thus, ri+1 is B' and so has length 1, as required, completing the proof in Case 1.
CD	Case 2: no edge of P2 is spanned by a global H-bridge'..
In this case, P2' = P2. We start by showing that every edge of P2 is H-green.
Because ew is w-consecutive for e, Definition 13.1 implies no edge of P2 is 00	red. By Theorem 11.3, we need only show that none is H-yellow. Suppose to the
contrary that there is an H-yellow edge f in P2, as witnessed by the H-yellow cycle Cy and the H-green cycle Cg. Lemma 11.2 implies there is a global H-bridge B contained in Cg.
The face of n[G] bounded by Cy (Lemma 11.2 (3)) is in M. Now the faces of n[G] bounded by Ae and A6w separate M into two parts, one of which contains f, and therefore P2. It follows that P1 is also in this part and Cy has at least a vertex in P1. We conclude that B spans an edge of P1. Claim 1 implies B = P{,
we = vi+5, ew G si+1, and ex G si+2. Because P2 = P2, and ew G si+1, we deduce that w = vi+1.
Since ew is not in an H-yellow cycle, we conclude that Qi+1 is not an H-yellow cycle. The other attachment of B, namely x, which is in [vi+7, ri+7, vi+8), must therefore be vi+7.
If the H-yellow edge f is in ri+1, then Cy n P1 is contained in the interior of the span of B. This implies that si+1 is in an H-yellow cycle and, therefore, ew is in an H-yellow cycle, contrary to the hypothesis.
We have noted that x = vi+7 is an end of B. Consequently, no edge of [vi+2, ri+2, x6w] can be H-yellow. That is, every edge of P2 is H-green.
We now complete the proof in Case 2. Let C be the H-green cycle containing
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the edge of P2 that is incident with w. Because ew is not in any H-green cycle, w is incident with an edge e' in C that is not in Ce. Let B be the Ce-bridge containing
Claim 5 implies B is just an edge with the two ends w and a second vertex z. The path C n P2 is in the boundary of the face of n[G] bounded by C (Lemma 6.6 (8)). Also, there is no global H-bridge spanning an edge of P2 (we are in Case 2). These two facts imply C n P2 is just an edge.
Suppose first that z G Px — x6w. Because C is H-green, it is disjoint from P1. Thus, Claim 3 implies that the cycle C' that is the union of the wz-subpath of P2Px and B bounds a face of n[G]. This face is contained in M, as is the face bounded by C. Both are incident with the edge of P2 incident with w and so they are the same face. We conclude that C = C'.
Now C n Px is in the boundary of a face inside the disc bounded by Ae' on one side and the face bounded by C on the other. Because G is 3-connected, this subpath has length 1. In this case, we have (b).
The other possibility is that z is in P2. We have already shown that w and z are the ends of a digon. If z = x6w, then we have (a). Therefore, we may suppose z = x6w.
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00	Since G is 3-connected, z has a neighbour y distinct from its neighbours in P2.
Let B' be the Ce-bridge containing zy. Claim 5 implies B' is just an edge joining z and y.
The choice of y shows y = w. Claim 4 (1) and (2) imply, respectively, that y G (PwP1} and that y = x. If y G P2, then (just as for w and z) z and y are the ends of a digon, so y is a neighbour of z in P2 , contradicting the choice of y. Therefore, y G (Px}.
Let C' be the cycle consisting of zy and the zy-subpath of P2Px. Claim 3 implies C' bounds a face of n[G].
To see that (c) holds, notice that C'nPx is in the boundary of the faces bounded by C' and Aew. Again, the 3-connection of G shows C' n Px is a path of length 1. Likewise, C' n P2 is in the boundary of the face bounded by C'. There is no global H-bridge spanning any edge of P2, so C' n P2 is also a path of length 1, completing the proof that (c) occurs and the proof of Claim 6.	□
It remains to consider the possibilities that ew is in either an H-yellow or an H-green cycle. We do the latter first.
Claim 7. If ew is in an H-green cycle C, then either
(d)	Ce U C contains Ce plus a back-slope or
(e)	Ce U C is Ce plus a crossbar.
Proof. Let F be the face bounded by C (Lemma 6.6 (8)). Obviously F is not inside the face bounded by Ae, and, since F is contained in M, F is Ce-interior. Let y be the end of ew nearer w in P2; then yGr¡. From the definition of H-green cycle (Definition 6.2), the edge of the yxew-subpath of P2 incident with y is in C.
If w is an attachment of a global H-bridge, then every edge of C n R is in two H-green cycles, which is impossible by Theorem 6.7. Therefore, P2 = P2', y = w, and C is the union of the wz-path CnP2 (this defines z) and an R-avoiding wz-path
P.
The path P contains a subpath P' joining a vertex of the zx-subpath of P2Px to a vertex of the component of Pw — ew containing we; we may assume P' is Ce-CO	avoiding. Claim 3 implies that the cycle contained in P' U PwP2Px bounds a face
of n[G]. As z is in this cycle, it must be that z is an end of P' and, moreover, this cycle is C. In particular, P is just P' plus a subpath of Pw. We know that C n P2 is just an edge. Since the path C n Pw is in the boundary of the faces bounded by both C and Ae, it is also just the edge ew.
If z = xew, then P' = P and the zwe-path contained in P U Pw contradicts Claim 4 (1). Therefore, z = xew.
Let B be the Ce-bridge containing P'. Claim 5 implies B has precisely two attachments w' G Pw and x' G Px: therefore, B is just the edge w'x' (this is also P'). If x' is xew, then B is a w-backslope.
Finally, suppose x' is in Px — xew. Then C bounds a face incident with C n Px. Since C n Px is also in the boundary of the face bounded by Aew, it has length 1.
On the P1 side, B together with the w'x'-subpath of (PwP{Px) is a cycle C' disjoint from P2'. By Claim 3, C' bounds a face of n[G]. As above, each of C' nPw, C' n Px, and P' all have length 1. Therefore, B is a crossbar.	□
Our final case is that ew is in an H-yellow cycle. Claim 8. If ew is in an H-yellow cycle C, then either
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(d)	Ce U C contains Ce plus a back-slope or
(e)	Ce U C is Ce plus a crossbar.
Proof. Let C' be the H-green cycle witnessing that the cycle C containing ew is H-yellow. Then C' contains an H-jump J and either both ends of J are in Pi or both ends of J are in P2. In either case, Claim 1 implies the span of J is all of Pi or P2. We treat these two possibilities separately.
CD
Subclaim 1. If both ends of J are in P2, then (e) occurs.
Proof. In this case, Claim 1 implies J has ends w and xew, xew = Vj+3, ew G sj+i, and ex G sj+2.
Because ew is both incident with vj+1 and in an H-yellow cycle as witnessed by the H-green cycle C' containing J, vj+1 is in the interior of the span of J; consequently, w G (rj). Therefore, the edge of r incident with vj+1 is H-green.
We observe that J witnesses that Qj+1 is an H-yellow cycle. It follows from Lemma 11.2 (3) that C = Qj+1. The same part of the same lemma combines with the fact that e is not H-green to show that Pw consists of [w, rj, vj+1, sj+1, Vj+6] and that Pw has length precisely two. Symmetrically, Px consists of sj+2 rj+2 and has length 2. Therefore, we have (e), as required.	□
It remains to consider the possibility that both ends of J are in Pi. Claim 1 implies J = wex, we = Vj+5, ew G sj+1, and ex G sj+2. Also, rj+5 is in the H-green cycle C' containing J, and so Pw contains rj+5 sj+1. Since Theorem 12.1 (3) implies Pw has at most one H-rim edge, we conclude that w = vj+1. Recall that Pw is
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in the boundary of the face of n[G] bounded by Ae. The path rj+5 is also in the boundary of the face bounded by C' and so is just an edge. The path sj+1 is also in the boundary of the face bounded by C, so it too is just an edge.
If J is not incident with Vj+7, then the situation is precisely that Subclaim 1 with the roles of (e, w) and (ew, x) interchanged. Therefore, Ce U C is (e), as required.
Therefore, we may assume J is incident with Vj+7. At this point, we know that sj+1 rj+5, J and at least the edge ex of sj+2 are contained in Ce. There is a Ce-bridge containing r.j+6; Claim 5 implies this Ce-bridge is precisely r.j+6 and this is just an edge.
The cycle C has a second edge e' incident with Vj+6. There is a Ce-bridge B containing e'. Claim 5 implies B has precisely two attachments, namely Vj+6 and some other vertex y.
If y G Px — xew, then B together with the yv.j+6-subpath of Ce — P2 contains a cycle disjoint from P2 and yet does not bound a face (it contains rj+6). We know that rj+5 rj+6 J bounds a face of n[G], so y is not in Jrj+5. Claim 4 implies y G P2' — xew . Thus, y = xew .
To finish the proof that (d) occurs, note first that sj+1 and B are both edges;
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thus, it suffices to prove that P2' is just an edge. In fact, Claim 3 implies P2' B sj+1 bounds a face of n[G]. In particular, P2 is not inside this face; therefore, P2' = P2. Consequently, P2' = P2 is just an edge.	□
In order to determine the 13 pictures, we remark that, from the perspective of both e and ew, any of (1)-(5) in Definition 13.7 can occur. However, if (5) occurs for either, then Claim 3 implies Ce and this crossbar is all that is in De. In the cases (2)(4) and (3)(4), there are two possibilities, as the slope and the back-slope
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00	can have either distinct or common ends in the spoke; the latter is denoted by a
+ in the listing below. There is no third possibility, since the slope and back-slope do not cross in De. Thus, there are the 13 pictures (1)(1), (1)(2), (1)(3), (1)(4), (2)(2), (2)(3), (2)(4), (2)(4)+, (3)(3), (3)(4), (3)(4)+, (4)(4), and (5)(5).
Label the red edges in G as e0, e1,..., ek_1 so that, for i = 0,1,..., k — 1, ei has ends ui and vi and so that, reading indices modulo k, ei+1 is the vi-consecutive red edge for ei. This implies that ei is the ui+1-consecutive red edge for ei+1.
Since there are no red edges between ei_1 and ei+1 on the "peak of Aei" portion of R, defining adjacency to mean "consecutive" shows the set of red edges make a cycle. Furthermore, vi and ui+1 are both in the cycle Cei that determines the picture pi between ei and ei+1. Taking any viui+1-path Pi in pi, we see that Pi together with either of the viui+1-subpaths of R makes a non-contractible cycle in RP2. In this sense, ei and ei+1 are on opposite sides of R.
If we think of e0 as being on "top" and e1 on the "bottom", then e2, e4, ... are all on top and e3, e5 ..., are on the bottom. When we get back to e0 from ek_1, we have gone once around the Mobius strip, so e0 is now on the bottom. It follows that ek_1 is on top and, therefore, k — 1 is even, showing k is odd.
It follows that G contains a subgraph H that is in T(S). (There may be edges in the interior of Ce "between" the structures we identified "near" P{ and P2'.) However, Theorem 5.5 implies H G M], so we conclude G = H. That is, GgT(S).	■
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CHAPTER 14
Graphs that are not 3-connected
The rest of this work is devoted to: describing all the 2-crossing-critical graphs that are not 3-connected, discussed in this section; finding all 3-connected 2-cros-sing-critical graphs that do not contain a subdivision of V8, treated in Section 15; and showing that the number of 3-connected 2-crossing-critical graphs that do not contain a subdivision of V2n is finite, which is Section 16. These last two combine 1	with the preceding work to show that there are only finitely many 3-connected 2-
crossing-critical graphs to be determined, namely those that have a subdivision of V8 but no subdivision of V1o.
In this section we show that every 2-crossing-critical graph that is not 3-connected is either one of a few known examples or is obtained from a graph in Mf by replacing 2-parallel edges with a "digonal" path (that is, a path in which every edge is duplicated). We remark that we continue assuming that the minimum degree is at least 3, as subdividing edges does not affect crossing number. We first determine all the 2-crossing-critical graphs that are not 2-connected.
14.1. 2-critical graphs that are not 2-connected
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Since the crossing number is additive over components, any 2-crossing-critical graph can have at most two components, each of them equal to either K3 3 or K5. Thus, there are only three different such graphs: two disjoint copies of K5, two disjoint copies of K3 3, and disjoint copies of each.
Similarly, the crossing number is easily seen to be additive over blocks. Thus, the blocks of a connected, but not 2-connected, 2-crossing-critical graph must be 1-critical graphs, and therefore all such graphs can be obtained from the aforementioned disconnected 2-crossing-critical graphs by identifying two vertices from distinct components. The identified vertex may be a new vertex that subdivides some edge. For example, there are three possibilities in which both blocks are K5: the identified vertex is a node in both, or only in one, or in neither. Likewise for K3 3. There are four 2-crossing-critical graphs in which one block is a subdivision of K5 and the other is a subdivision of K3 3.
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Proposition 14.1. The thirteen graphs in Figure 14-1 are precisely those 2-crossing-critical graphs that are not 2-connected.
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14.2. 2-connected 2-critical graphs that are not 3-connected
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In this subsection, we treat 2-crossing-critical graphs that are 2-connected, but not 3-connected. With 36 exceptions, these all arise from 3-connected 2-crossing-critical graphs that have digons (i.e., two edges with the same two ends). The digons may be replaced with arbitrarily long "digonal paths" — these are simply paths in which every edge is converted into a digon.
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Figure 14.1. The 2-crossing-critical graphs that are not 2-connected.
Tutte [34, 35] developed a decomposition theory of a 2-connected graph into its cleavage units, which are either 3-connected graphs, cycles of length at least 4, or for k > 4, k-bonds (a k-bond is a graph with k edges, all having the same two ends). We provide here a brief review of this theory. A 2-separation of a 2-connected graph G is a pair (H, K) of edge-disjoint subgraphs of G, each having at least two edges, so that H U K = G and H n K = ||{u, v}|| (recall ||{u, v}|| is the graph with just the vertices u and v and no edges.). Notice that a 3-cycle and a 3-bond have no 2-separations and, therefore, are to be understood in this context to be 3-connected graphs.
The 2-separation (H, K) with H n K = ||{u, v}|| is a hinge-separation if at least one of H and K is a ||{u, v}||-bridge and at least one of them is 2-connected. Another way to say the same thing, but in terms of H n K: ||{u, v}|| is a hinge if either there are at least three ||{u, v}||-bridges, not all just edges, or there are exactly two ||{u, v}||-bridges, at least one of which is 2-connected.
The theory of cleavage units develops as follows. Let G be a 2-connected graph.
(1)	If ||{u, v}|| is a hinge and (H, K) is a hinge-separation (possibly of another hinge), then there is some ||{u, v}||-bridge containing either H or K.
(2)	G has no hinge if and only if G is 3-connected, a cycle of length at least 4, or a k-bond, for some k > 4. (Recall that a 3-cycle and a 3-bond are 3-connected.) In each of these cases, G is its own cleavage unit.
(3)	If (H, K) is a hinge-separation and H n K = ||{u, v}||, then the cleavage units of G are the cleavage units of the two graphs H + uv and K + uv obtained from H and K by adding a virtual edge between u and v, respectively. This inductively determines the cleavage units.
(4)	There is a decomposition tree T whose vertices are the cleavage units of G and whose edges are the virtual edges. A virtual edge joins in T the two cleavage units of G containing it.
(5)	G contains a subdivision of each of its cleavage units.
(6)	If G contains a subdivision of some 3-connected graph H, then some cleavage unit of G contains a subdivision of H.
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units, each time we combine two graphs along a virtual edge, there are two possibilities for how to identify the vertices of the corresponding hinge. This ambiguity will play a small role in constructing the 2-crossing-critical graphs that are 2- but not 3-connected.
It is easy to see that G is planar if and only if every cleavage unit is planar. (We could apply Kuratowski's Theorem and Item 6 or prove it more directly.) Since we are interested in non-planar graphs, there are two relevant possibilities: one or more than one of the cleavage units of G is not planar. We start by treating the latter case. We remark that the following discussion makes clear that the crossing number is not additive over cleavage units. Related discussions can be found in Siran [32], Chimani, Gutwenger, and Mutzel [10] (but see [5] for significant comments about the latter), Beaudou and Bokal [5], and Leanos and Salazar [21].
LEMMA 14.2. Let G be a 2-connected graph. If two cleavage units of G are not planar, then cr(G) > 2.
It is an important consequence that, if G is 2-crossing-critical, 2-connected, and has 2 non-planar cleavage units, then G is simple, i.e., has no digons.
Proof of Lemma 14.2. Among all 2-separations (H, K) of G, we choose the one that has K minimal so that both H + uv and K + uv are not planar, where H n K = ||{u, v}||. If the crossing number of G is not at least 2, then cr(G) < 1, so, by way of contradiction, suppose D is a 1-drawing of G.
Let PK and PH be uv-paths in K and H respectively. Since G contains the subdivision H U PK of H + uv, G is not planar. Therefore, D has a crossing. Evidently, D(H U Pk) and D(K U Ph) both contain the crossing. We conclude 00	that the crossing in D is of an edge of PH with an edge of PK. It follows that
there are not edge-disjoint uv-paths in either H or K and that the crossed edges 2	are cut-edges in their respective subgraphs.
Let w and x be the ends of the edge in K that is crossed, labelled so that w is nearer to u in PK than x is. Let Ku and Kv be the two components of K — wx, with
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the former containing u. Since K + uv is not planar, either Ku + uw or Kv + vx is CO	not planar. We may assume it is the former. Notice that (H U Kv) + xu contains
a subdivision of H + uv and, therefore, is not planar. But then ((H U Kv), Ku) is a 2-separation contradicting the minimality of K.	■
We are now in a position to determine the 2-connected, 2-crossing-critical graphs having two non-planar cleavage units.
Theorem 14.3. Let G be a 2-connected, 2-crossing-critical graph having two non-planar cleavage units. Then G is one of the 36 graphs in Figures 14.2 and 14.3.
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Claim 1. G has at most three cleavage units: C1, C2 and possibly a 3- or 4-cycle; if there are three, then the 3- or 4-cycle is the internal vertex in the decomposition tree.
Proof. Let C1 and C2 be non-planar cleavage units of G.
Proof. For i = 1,2, let {ui, vi} be the hinge of G contained in Ci such that C1 and C2 are contained in different ||{ui, vi}|-bridges. For any other virtual edge xy
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Figure 14.2. 2-connected, not 3-connected, 2-crossing-critical graphs, 2 non-planar cleavage units
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Figure 14.3. 2-connected, not 3-connected, 2-crossing-critical graphs, 3 cleavage units, 2 of which are non-planar.
in Cj, there is a path Pxy in G that is C1 U C2-avoiding. Let C be Cj n G (i.e., Cj with none of its virtual edges) together with all these Pxy. Let H be the subgraph of G consisting of CJ1 U C2 U Q, where Q consists of two disjoint {u1, v1}{u2, v2}-paths in G. Evidently, H is 2-connected and C1 and C2 are cleavage units of H.
Lemma 14.2 implies cr(H) > 2. Since H C G and G is 2-crossing-critical, H = G. Since G has no vertices of degree 2, G consists of either two or three cleavage units, namely C1, C2 , and possibly a 3- or 4-cycle between them.	□
00	We next determine the possibilities for C1 and C2 .
Claim 2. For each i = 1, 2, one of the following occurs:
(1)	Cj is K5;
(2)	Cj is ^3,3;
(3)	C» — «»v» is a subdivision of K33.
Proof. Hall proved that every 3-connected non-planar graph is either K5 or contains a subdivision of K33 [16]. Since G is simple and C» is 3-connected, we deduce that C» is either K5 or contains a subdivision of K33. So suppose C» contains a subdivision K of K33.
Suppose C» — «»v» has an edge e for which C» — e is not planar. Since C» — e is 2-connected, G —e is 2-connected and has at least two non-planar cleavage units (C3-j and another contained in C» — e). By Lemma 14.2, cr(G — e) > 2, contradicting 2-criticality of G. So C» — «»v» C K. Thus, either C» = K or C» — «»v» = K, as claimed.	□
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Claim 3. There are five possibilities for Cj, namely:
(1)	Cj is K5;
(2)	Cj is ^3,3;
(3)	Cj — ujvj is K3 3 and ujvj joins two non-adjacent nodes of K3 3;
(4)	Cj — ujvj is K3 3 with one edge subdivided once and ujvj joins the degree 2 vertex to a node of K3 3 that is not incident with the subdivided edge; and
(5)	Cj — ujvj is K3 3 with two non-adjacent edges both subdivided once and ujvj joins the two degree 2 vertices.
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Proof. If C» is neither K5 nor K3 3, then it must be a subdivision K of K3 3 with the additional edge «»v». Clearly K has at most two vertices of degree 2. If K has no vertices of degree 2, then, since C» is simple, we have (3). Likewise, if K has only one vertex of degree 2, that vertex (one of and v») cannot be in a branch incident with the other one of and v», which is (4). Finally, suppose and v» are both of degree 2 in K. Then their containing branches cannot be incident with a common vertex w, as otherwise, we could delete the edge «»w and still have two non-planar cleavage units, contradicting 2-criticality. This proves (5).	□
Note that in all five cases of Claim 3, there is only one possibility for C», up to isomorphism. Only (4) has non-isomorphic labellings of and v».
Claim 4. If G has just two cleavage units, then G is one of the 16 graphs in Figure 14.2.
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Proof. If neither C1 nor C2 is (4) from Claim 3, then, with repetition allowed, there are 10 possible unordered pairs for C1 and C2. Each of the pairs uniquely produces the graph G. There are four graphs having C1 but not C2 satisfying Claim 3 (4), and there are two graphs having both C1 and C2 satisfying Claim 3 (4). □
CD	Claim 5. If G has three cleavage units, then at least one of C1 and C2 is either
K5 or K3 3.
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Proof. Let e be an edge of G in the third cleavage unit of G; recall that this cleavage unit is either a 3- or a 4-cycle. The blocks of G — e include C1 — «1v1 and C2 — «2v2; if these were both non-planar, then cr(G — e) > 2, contradicting
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Claim 6. If G has three cleavage units, then G is one of the 20 graphs in Figure 14.3.
00	2-criticality of G. Hence, at least one of Cl — w1v1 and C2 — u2v2 is planar. By
Claim 3, such a one must be either K5 or K3 3.	□
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Proof. There are three pairs in which both Cl and C2 are one of K5 and K3 3 and two possibilities for the third cleavage unit, yielding six graphs. Now suppose Ci is one of K5 and K3 3 and C2 is not. There are three possibilities for C2 and two possibilities for the third bridge. However, when the third bridge is a 3-cycle, there are two ways to attach C2 when it is of Type (4) from Claim 3. Thus, there are 6 graphs with the third cleavage unit a 4-cycle and 8 when it is a 3-cycle. □
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From the claims, we see that the 36 graphs shown in Figures 14.2 and 14.3 are all the cases in which G is 2-connected, but not 3-connected, and has two non-planar cleavage units.	■
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In the remaining cases of 2-connected, but not 3-connected, 2-crossing-critical graphs, there is only one non-planar cleavage unit C. The graph C is simple. The following result shows how to obtain G from a 3-connected 2-crossing-critical graph. It requires the following definition.
Definition 14.4. A digonal path is a graph obtained from a path P by adding, for every edge e of P, an edge parallel to e.
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Theorem 14.5. Let G be a 2-crossing-critical graph with minimum degree at least 3. Suppose that G is 2-connected but not 3-connected and has only exactly one non-planar cleavage unit, C. The graph C obtained from C by replacing each of its virtual edges with a digon is 2-crossing-critical and 3-connected. The graph G is recovered from C by replacing these virtual edge pairs by digonal paths.
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Proof. That C is 3-connected is a trivial consequence of the fact that C is 3-connected.
As for the rest, let uv be a virtual edge in C. Then ||{u, v}|| is a hinge of G. We consider the ||{u, v}||-bridges in G; let Buv be the one that contains C n G, and let B#y be the union of the remaining ||{u, v}||-bridges. We have two objectives: to show that C is 2-crossing-critical and that, for each uv, B#y is a digonal uv-path.
For the former, we first show cr(C') > 2. Otherwise C has a 1-drawing D. Obviously no edge in a digon of C is crossed in D. For each virtual edge uv of C, B# + uv is planar, so it may be inserted into D in place of the uv-digon in D to obtain a 1-drawing of G, which is a contradiction. Therefore, cr(C') > 2.
We next claim that each B#, consists of digonal uv-paths. Assume first that B#V has a cut-edge e separating u and v. Since G has no vertices of degree 2 and B#y is not just a single edge, B# contains some edge e' so that B#v — e' still contains a uv-path.
If no edge of B#y is crossed in a 1-drawing De/ of G — e', then, since B#y — e' contains a uv-path, B#, may be substituted for B#y — e' in De' to obtain a 1-drawing of G, which is impossible. So some edge of B#, is crossed in De/. Deleting edges from B#y — e' to leave only a uv path shows that De/ restricts to a 1-drawing of Buv + uv in which there is at most one crossing; if there is a crossing, then uv is crossed. Since every planar embedding of B#y + uv has uv and e on the same face,
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produce a 1-drawing of G in which e is crossed. This contradiction that shows B#y contains edge-disjoint uv-paths.
Let e be an edge of B#y. Then a 1-drawing De of G — e must have a crossing of some edge e' of B#,. If B#, — {e, e'} has a uv-path P, then De restricts to a planar embedding of C by using P to represent u-y. But C is non-planar, so every edge of B# — uv is in an edge-cut of size at most 2 separating u and v. Combining this with the preceding paragraph shows that every edge of B#, is in an edge-cut of size exactly 2. It is an easy exercise to see that this implies B#, is a pair of digonal paths.
We conclude by showing that, for every edge e of CJ, cr(CJ — e) < 1. Suppose first that e is not in a digon. Each B#, has a uv-path Puv that is clean in De. Thus, De [G — e] contains a subdivision of C — e in which no virtual edge (represented in the subdivision by Puv) is crossed. Therefore, the virtual edges may be replaced with digons to give a 1-drawing of CJ — e, as claimed.
Now suppose e is in the uv-digon. Let e' be any edge of B#y. Then DPj contains a 1-drawing of C, in which every other virtual edge wx is represented by a wx-path Pwx in B#x that is clean in De/. All these other virtual edges may be replaced with digons to give a 1-drawing of C — e, as required.	■
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CHAPTER 15
On 3-connected graphs that are not peripherally-4-connected
In this chapter, we reduce the problem of finding all 3-connected 2-crossing-critical graphs to the consideration of non-planar, peripherally-4-connected graphs. Our motivation for doing this is to use a known characterization of internally-4-connected graphs (a concept intimately related to peripherally-4-connected graphs) with no subdivision of V8 to find all the 3-connected, 2-crossing-critical graphs with no subdivision of V8.
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Definition 15.1. A graph G is peripherally-4-connected if G is 3-connected and, for any 3-cut S of G and any partition of the components of G — S into two non-null subgraphs H and K, at least one of H and K has just one vertex.
We begin this section by finding the four 3-connected, not peripherally-4-connected, 2-crossing-critical graphs that are not obtained from planar substitutions into a peripherally-4-connected graph. The bulk of the section is devoted to explaining in detail how to obtain the remaining 3-connected 2-crossing-critical graphs from peripherally-4-connected graphs. Finally, this theory is used to explain how to find all the 3-connected 2-crossing-critical graphs that do not contain
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a subdivision of V8.
15.1. A 3-cut with two non-planar sides
In this section we find the four 3-connected, not peripherally-4-connected, 2-S	crossing-critical graphs that are not obtained by substituting planar pieces into
degree-3 vertices in a peripherally-4-connected graph (this substitution process being the remainder of the section). We start by describing the four graphs and showing that they are 2-crossing-critical.
Definition 15.2. The graph K| 4 is obtained from disjoint copies of K2 , 3 by joining the parts of the bipartition having three vertices in each of the copies by a perfect matching M.
Observe that K3 4 is obtained from K3 4 by contracting all the edges of the matching M. The following generalizes the well-known fact that K3 4 is 2-crossing-critical.
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Lemma 15.3. If H is obtained from K3 4 by contracting some subset of M, then
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Proof. Suppose e is an edge of K3 4 not in M. Then there is a 1-drawing of K3 4 — e in which no edge of M is crossed. Thus, cr(H — e) < 1. If e G M, then H — e is planar. It remains to show cr(H) > 2.
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K2 , 3 subgraphs of H contained in K| 4 — M. For each vertex v of degree 3 in H2, there are three disjoint vH1-paths in H; adding v and these paths to H1 yields a subdivision Hv of K3 , 3 in H. Thus, D[Hv] has a crossing, and, since there are two choices for v, this crossing involves only edges of H1 and M.
Interchanging the roles of H1 and H2 shows the crossing in D involves only edges of M. But then D[Hv ] has its only crossing on branches incident with v, which is impossible.	■
We remark that there are splits of K34 that have crossing number 1 — split two of the degree 4 vertices so that the two partitions of the four neighbors are different. Fortunately, they do not occur in our context.
00	In order to show that these are the only four graphs with "non-planar 3-cuts",
we need to understand just what "non-planar 3-cuts" are.
Definition 15.4. Let S be a 3-cut in a 2-connected graph, so there are subgraphs H and K of G such that G = H U K and H n K = ||S||. For L G {H, K}, L+ denotes the graph obtained from L by the addition of a new vertex adjacent to precisely the vertices in S.
Lemma 15.5. Let G be a 3-connected non-planar graph different from K5 and let v be a vertex of G. Then G has a subdivision H of K3 3 in which v is an H-node.
We will see that, in the case G is 2-crossing-critical, with the exception of K3 , 4, there are at most three non-trivial S-bridges, and so at least one of H and K is an S-bridge. Our next goal is to show that the four graphs in Lemma 15.3 are the only four that have both H+ and K+ non-planar. We start with the following, which is likely well-known; however, we could not find a reference. It extends Hall's Theorem [16] that there is a subdivision of K3 3.
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Proof. Here is an outline of the easy, but tedious, proof. As a first step, we show that there is a subdivision of K3 , 3 containing v. By Hall's Theorem [16], G contains a subdivision L of K3 3. If v G L, then there are three disjoint vL-paths.
CO	There are three possibilities for the ends of these paths in L: two are in the same
closed L-branch; two are in L-branches incident with a common L-node; and the L-branches containing the ends of the paths are pairwise disjoint. In the first case, v is incorporated into the interior of a branch of a new subdivision of K3 , 3, while in the other cases, v is incorporated as a node of the new subidivision of K3 , 3.
So now assume that v is in L, but not as a node. Then v is interior to some L-branch b with ends u and w. Let L' = L — (b) — this is a subdivision of K3 3 less an edge. Because there are, in G, disjoint L'-avoiding v{u, w}-paths, standard proofs of Menger's Theorem imply that there are three disjoint L'-avoiding vL'-paths, having u and w among their three L'-ends. Therefore, we may assume not only is v in the interior of b, but there is an L-avoiding vx-path from v to some other vertex x of L'.
Up to symmetry, there are three possibilities for x: it is a node of L other than u and w; it is interior to an L-branch incident with u but not with w; and it is interior to an L-branch not incident with either u or w. Let y and z be nodes of L' (note that u and w are not actually nodes of L'). We can assume x is either y, or in the L-branch [w,y], or in the L-branch [y, z]. Let Y be a {u,w, x}-claw with centre v, so that Y n L' is just u, w, and x.
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v as a node. If x is in (y, z), then (L' U Y) — (w, y) is a subdivision of K33 having v as a node.	■
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We are now ready for the classification of the 3-connected 2-crossing-critical graphs with two non-planar sides to a 3-cut.
THEOREM 15.6. Let G G M2 have subgraphs H and K of G and a set S of three vertices of G such that:
(1)	G = H U K;
(2)	H n K = ||S||;
(3)	H and K both have an ^S^-bridge having all of S as attachments; and the two graphs H + and K + are both non-planar.
Then G is one of the four graphs obtained from K| 4 by contracting some subset of M.
Proof. Let u, v, and w be the vertices in S. For L G {H, K}, let v+ denote the vertex in L+, but not in L. The graph L+ is a subdivision of a 3-connected graph (the only possible vertices of degree 2 are u, v, and w). Since L+ is not planar and has a vertex of degree 3, it is not a subdivision of K5 and, therefore, by Lemma 15.5 contains a subdivision L' of K3 3 in which v+ is a node. Now G' = (H' — v+) U (K' — v+)
contracted. By Lemma 15.3, cr(G') = 2, so G' = G, as required.	■
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15.2. 3-reducing to peripherally-4-connected graphs
In this section, we discuss the general details of reducing a 3-connected graph to a peripherally-4-connected graph. These results apply in some generality and not just in the context of 2-crossing-critical graphs. These are the first of several steps toward finding all the 3-connected 2-crossing-critical graphs that do not contain a subdivision of V8.
These results are fairly technical but essential to this part of the theory.
Definition 15.7. (1) A 3-cut S in a 3-connected graph is reducible if G — S has at most 3 components and they partition into two subgraphs each having at least two vertices.
(2) The set K consists of those 3-connected graphs that do not contain a subdivision of K3 4.
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The following result is obvious from the definitions and begins to explain the appearance of K3 4 in Definition 15.7 (2).
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Lemma 15.8. Let G be a 3-connected graph that is not peripherally-4-connected. Then either G has a reducible 3-cut or G has K34 as a subgraph.
The next result sets up the basic scenario that we will use throughout our reduction to peripherally-4-connected graphs.
Lemma 15.9. Let G G K. Then there is a sequence G0,G1,...,Gk of 3-connected graphs in K so that: Go — G; G^ is peripherally-4-connected; and, for each i = 1, 2, .. ., k, there is a 3-cut Sj in Gj_1 and a non-trivial, planar Si-bridge Bj so that Nuc(Bi) has at least two vertices and Gi is obtained from Gi-1 by contracting the nucleus of Bi.
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so that Nuc(Bj) has at least two vertices, choose Bj to be inclusion-wise maximal. We claim that the graph Gj obtained from Gi-1 by contracting Nuc(Bj) to a vertex is 3-connected.
Otherwise, there is some pair {u, v} of vertices so that Gj — {u, v} is not connected. If the vertex of contraction of Nuc(Bj) is neither u nor v, then {u,v} is a 2-cut in Gj_1, a contradiction. Therefore, we can assume u is the contraction of Nuc(Bj).
e	Let H and K be components of Gj — {u, v}, with the labelling chosen so that
|Sj n V (H)| > |Sj n V (K )|; in particular, |Sj n V (K)| < 1. Let he V (H); if there is a vertex k e V(K) \ Sj, then {v} U (Sj n V(K)) separates k from h in Gj_1, which contradicts the assumption that Gj-1 is 3-connected.
Therefore V(K) C Sj, so there is a single vertex s in K, and s e Sj. It follows that s is adjacent to only vertices in Nuc(Bj) and possibly to v. But this contradicts the maximality of Bj: let S' = (S \ {s}) U {v}. Observe that Bj + s is a planar S'-bridge, contradicting maximality of Bj.
Lastly, we show that if Gj_1 does not have a subdivision of K3 4, then neither does Gj. Any subdivision of K3 4 in Gj must contain the vertex vj of contraction. Since vj has degree 3 in Gj and Bj_1 is an S-bridge, we can reroute the subdivision
of K3 4 in Gj into Bj_1 to obtain a subdivision of K3 4 in Gj_1.	■
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Definition 15.10. Let G e K. (1) Then G reduces to G' by 3-reductions if there is a sequence G0, G1,..., Gk
of 3-connected graphs so that G0 = G; Gk = G'; and, for each i =
1, 2,..., k, there is a 3-cut Sj in Gj_1 and an Sj-bridge Bj, whose nucleus at least two vertices, so that Gj is obtained from Gj-1 by contracting the nucleus of Bj.
(2)	For each vertex v of G' and each i = 0,1, 2 ..., k, KV denotes the connected subgraph of Gj that contracts to v. We also set Kv = K°.
(3)	If v has just three neighbours x, y, and z in G', then Gv is the graph obtained from Kv by adding x, y, and z, and, for each t e {x, y, z} and each edge v't' of G with v' e Kv and t' e Kt, adding the edge v't.
We now commence a lengthy series of technical lemmas that all play vital roles in usefully reducing the 3-connected graph 2-crossing-critical graph G to a smaller 3-connected 2-crossing-critical graph Grep(v). The culmination of this part of the work is Theorem 15.25 in the next section, showing that Grep(v) is 2-crossing-critical. This will lead to a program for determining all the 3-connected 2-crossing-critical graphs that reduce to a particular peripherally-4-connected graph.
LEMMA 15.11. Let G e K and suppose G reduces by 3-red,uctions to the peri-pherally-4-connected graph Gp4c. For any two vertices u, v of Gp4c, there is a single vertex in G incident with all edges having one end in Ku and one end in Kv.
Proof. Let G = G0, G1,..., Gk = Gp4c be a sequence of 3-reductions. Choose i to be largest so that there are disjoint KU_1KV_1-edges ab and cd with a, c e KU_1 and b, d e KV_1. In Gj, either a and c have been identified or b and d have; by symmetry, we may assume the former.
The vertices b and d are obviously attachments of Bj and so these are in Sj. Let zj be the third vertex in Sj. Since K^_1 is connected and since, by Definition
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00	15.10, u has three neighbours in Gp4c, zj G Ku. Continue using the label a for the
vertex obtained by contracting Nuc(Bj).
At some point in the later 3-reductions, a and zj are identified and at another point b and d are identified. We show that neither can be done before the other, which is impossible.
Suppose zj and a are identified first. When this identification occurs, a 3-cut Sj and an Sj-bridge Bj so that zj and a are in Nuc(Bj). The vertices b and d are again attachments of Bj and so are in Sj; let zj be the third vertex in Sj.
Because i is largest so there are disjoint K^-1KV-1-edges, all edges between KjU and Kj at this moment are incident with a. It follows that {a, zj} is a 2-cut in the current graph, separating zj from b. But this contradicts the fact that Gj-1 is 3-connected. Therefore, zj and a are not identified before b and d.
On the other hand, suppose b and d are identified first, by the contraction of Nuc(Bj). When b and d are identified, the only neighbours of a are b, d, and zj. Following the identification of b and d, the only neighbours of a are zj and the vertex of identification, again contradicting 3-connection of Gj.	■
We need a slight variation on a standard definition.
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Definition 15.12. Let G be a connected graph.
(1)	An isthmus is a set I of parallel edges so that G — I is not connected.
(2)	A cut-edge is an edge e so that G — e is not connected.
Obviously, e is a cut-edge of G if and only if {e} is an isthmus, but an isthmus may have more than one edge. The distinction comes into play because at various points we will consider edge-disjoint paths in certain subgraphs of our 2-crossing-
CM
critical graph; if there are not two edge-disjoint uv-paths, then there is a cut-edge separating u and v. On the other hand, the 3-connection of G does not preclude the possibility of parallel edges; at several points we will be able to identify that two vertices u and v have the property that they must be adjacent, but be unable to distinguish whether they are joined by 1 or 2 edges. A common scenario will have the set of edges between them making an isthmus in some subgraph.
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In particular, the case that Kv has an isthmus is a central one in reducing 2-crossing-critical graphs.
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LEMMA 15.13. Let G G K reduce to the peripherally-4-connected graph Gp4c by a sequence of 3-reductions. Suppose there is a vertex v of Gp4c so that the graph Kv has an isthmus I. Then, for each component K of Kv — I, there are at least two neighbours x and y of v in Gp4c so that there are KKx- and KKy-edges in G.
Proof. At some moment in the reduction of G, Gj-1 has a 3-cut Sj and Bj is the planar Sj-bridge in Gj-1 that contains I. Then Bj — I is not connected; the ends u and w of the edge or edges in I are in different components K and L, respectively, of Bj — I.
Let x, y, and z be the neighbours of v in Gp4c and let t be any vertex of Gj-1 not in KV U KX U Ky U KZj. (Since Gp4c is not planar, it has at least five vertices.) In Gj_1 there are three pairwise internally-disjoint ut-paths. These three paths leave Bj through distinct attachments of Bj; these are the vertices in Sj. The same argument applies for wt-paths.
In particular, two of the ut-paths leave K on edges incident with vertices in Sj. Likewise for L. Therefore, K and L are both joined by edges to the same
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00	attachment s G Sj. It follows that s is not in KV, so s is in Kx, say. Moreover,
since the KV-ends of these two edges are not the same, Lemma 15.11 implies all the edges between KV and Kx are incident with s.
Since Gj_1 is 3-connected, Gj-1 — ({s} U I) is connected. Therefore, there are edges of Gj_1 leaving each of K and L; each of these edges is also leaving KV and, therefore, has its other end in one of Kx, Ky, and KZj. However, this other end cannot be s and, consequently, cannot be in Kx, as required.	■
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The connectivity of G has further implications about the structure of the KV.
Lemma 15.14. Let G G K reduce by 3-reductions to a peripherally-4-connected graph Gp4c. Let v be a vertex of Gp4c with just the three neighbours x, y, and z and suppose KV has at least two vertices. For each t G {x, y, z}, let t' be any vertex incident with all the KVKt-edges. Then x', y', and z' are all distinct.
Proof. Suppose x' = y'. Then x' is in KV. Observe that no vertex of KV — {x', z'} is adjacent to any vertex of of G — {x',z'} not in KV. Since G is 3-connected, it follows that KV consists of just x' and z'. In particular, z' = x'. Also, recall that KV contracts to a single vertex in the sequence of planar 3-reductions.
At the moment of contraction of KV, Gj_1 is 3-connected and x'z' is an isthmus. Therefore, Lemma 15.13 implies that z' is joined to at least one of Kx and Ky; this
contradicts the fact that all edges from KV to Kx U Ky are incident with x'. ■
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The vertices x', y', and z' are not uniquely determined. It is possible that there is only one vertex in each of KV and Kx incident with all KVKx-edges; one obvious instance is if there is only one KVKx-edge. We will follow up on this a little later. Here is a very simple and very useful observation.
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Lemma 15.15. Let H be a simple, non-planar, peripherally-4-connected graph. There is no 3-cycle of H having two vertices with just 3 neighbours.
Proof. Suppose to the contrary there are three vertices x, y, z making a 3-cycle, with x and y having only three neighbours each. Let v and w be the other neighbours of x and y. Then x and y are the vertices of one component of H — {v, w, z}.
Observe that H is non-planar, 3-connected, and has a vertex of degree 3. Therefore H is not K5 and so contains a subdivision of K3 , 3. It follows that H has at least six vertices. Thus, there is another component of H — {v, w, z}.
Since H is peripherally-4-connected, the only possibility is that there is exactly one other component and it consists of a single vertex u, adjacent to all of v, w, and z. The only other possible edges in H are between v, w, and z. However, the resulting graph is planar, a contradiction.	■
The following result assures us that useful (and expected) paths exist in each
Kv .
Lemma 15.16. Let:
<D	4
(1)	G G K reduce by 3-reductions to the peripherally-4-connected graph Gp4c;
(2)	Gp4c have at least five vertices;
(3)	v be a vertex of Gp4c so that KV has at least two vertices; and
(4)	x, y, and z be the neighbours of v in Gp4c, with corresponding vertices x', y', and z' in G as in Lemma 15.14.
a	y,	4
00	Then:
a)	for any vertex w in Kv — {x', y', z'}, there are three w{x', y', z'}-paths in Gv that are pairwise disjoint except for w; and
b)	if x' G Kv, then there are x'y' - and x'z' -paths in Gv — x that are disjoint except for x'.
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Proof. For a), let u be any vertex of G not in Kv U Kx U Ky U Kt3. Since G is 3-connected, there are three pairwise internally-disjoint wu-paths in G. The result follows from the observation that w and u are in different components of
G — {x',y',z'}.
If b) fails, then there is a vertex w of Gv — x that separates x' from {y',z'}. Since Kv is an ||{x, y, z}||-bridge in Gv, w is in Kv (possibly w = y' or w = z'). 00	Since {x', w} is not a 2-cut in G, x' and w are adjacent in Kv. But now they are
joined by an isthmus I in Kv. Since x' is a component of Kv — I joined only to
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15.3. Planar 3-reductions
In this section we now turn our attention to the particular case G G M2. We want to show that the 3-reductions can be taken to be contractions of planar bridges. So suppose S is a non-peripheral 3-cut in G.
If there are four or more non-trivial S-bridges (that is, having a nucleus), then G has a subdivision of K3j4 and so is K3 4- In the remaining cases, there are at most three non-trivial S-bridges. If there are three and B is one of them so that B+ is not planar (as in Subsection 15.1), then the union K of the remaining S-bridges has K+ not planar. Theorem 15.6 implies that G is one of four 2-crossing-critical 00	graphs. Thus, if there are three non-trivial S-bridges, we may assume that, for
each one B, B+ is planar. Finally, consider the case that there are precisely two non-trivial S-bridges Bi and B2. Since S is not peripheral, both Bj have at least two vertices. If both B+ are non-planar, then we are in the case dealt with in Theorem 15.6, so we may assume that one of them is planar. In summary, in every case, we may assume that Gp4c is obtained from 3-reductions in G in which the
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contracting Sj-bridge Bj is always planar.
Definition 15.17. Let G be a 3-connected graph and let Gp4c be a peripherally-4-connected graph. Then G reduces to Gp4c by planar 3-reductions if there is a sequence G = G0, G1, G2,..., Gk = Gp4c of 3-reductions so that, for each i = 1, 2,..., k, Gj is obtained from Gj-1 by contracting Nuc(Bj-1) and Bj+1 is planar.
We need two results about Kv in the context of planar 3-reductions. This requires further definitions.
Definition 15.18. Let G be a 3-connected graph that reduces by 3-reductions to the peripherally-4-connected graph Gp4c. Suppose v is a vertex of Gp4c having only the neighbours x, y, and z. For each t G {x, y, z}, let mt denote the number of vertices in Kv adjacent to vertices in Kt and let nt denote the number of vertices in Kt adjacent to vertices in Kv. (Lemma 15.11 implies that at least one of mt and nt is 1.)
(1) The subgraph KV"ax induced by Kv together with, for each t G {x,y, z} with nt = 1, the vertex of Kt adjacent to vertices in Kv.
Kx, we have a contradiction of Lemma 15.13.
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u
CD
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00	(2) The subgraph KV"ln induced by Kv together with, for each t G {x, y, z}
with mt > 1, the vertex of Kt adjacent to vertices in Kv.
We remark that Kv C KV"ln C KV"ax, and, for t G {x, y, z}, KV"ax has a vertex t' G Kt that is not in KV"ln precisely when nt = mt = 1.
Lemma 15.19. Let G G K reduce by 3-reductions to a peripherally-4-connected graph Gp4c. Let v be a vertex of Gp4c with just the three neighbours x, y, and z and suppose Kv has at least two vertices. Then there is a cycle C in KV"ln containing all of x', y' and z'.
Proof. Suppose w is a cut-vertex of KV"ln, so there are subgraphs X and Y of K™n with X U Y = K™n, X n Y = ||w||, and both X — w and Y — w are not 00	empty. We may choose the labelling so that X has at least the two vertices x' and
z' from {x', y', z'}, while Y — w has at most one; we may further assume x' = w. If y' G Y — w, then w is a cut-vertex of G, contradicting the fact that G is 3-connected.
1	Therefore, y' Y — w.
However, if y' G Kv, then we have a contradiction to Lemma 15.16 (b). Therefore, y' G Kv. If there is a vertex in Y other than w and y', then we contradict 3-connection of G, so y' is adjacent only to w in Gv. But then y' G KV"ln.
It follows that there is no cut-vertex in KV"m. Thus, there is a cycle C in KV"ln containing x' and y'. Obviously, we are done if z' G C, so we assume z' G C.
Since there is no cut-vertex in KV"ln, there are two z'C-paths P1 and P2 that are disjoint except for z'. If the C-ends of P1 and P2 are not both on the same x'y'-subpath of C, then G+ contains a subdivision of K3 3. This contradicts the fact that we are doing planar 3-reductions. Therefore, the C-ends of P1 and P2 are on the same x'y'-subpath of C and it is easy to find the desired cycle through all of x', y', and z'.	■
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The following is the last lemma we need to get the main result of this section.
Lemma 15.20. Let G G M2 and suppose G reduces by planar 3-reductions to the peripherally-4-connected graph Gp4c. Let v and x be adjacent vertices in Gp4c. Then there are at most two vertices in Kv adjacent to vertices in Kx.
Proof. This is obvious if Kv has at most one vertex. In the remaining case, v has degree 3 in Gp4c; let y and z be its other neighbours.
Suppose by way of contradiction that s, t, and « are distinct vertices in Kv all adjacent to vertices in Kx. By Lemma 15.11, there is a vertex x' incident will all the Kv Kx-edges and, evidently, x' G Kx.
In the planar embedding D+ of G+, letting w denote the new vertex adjacent to each of x, y, and z, we may choose the labelling so that the edges xw, xs, xt, x« occur in this cyclic order around x.
Claim 1. There is an sw-path in Kv containing t.
Proof. As Kv is connected, there is an s«-path P in Kv. We are obviously done if t G P, so we assume t G P. Let C be the cycle obtained by adding x' to P and joining it to s and
The rotation at x implies that t is on one side of D+[C], while w, y, and, consequently, z, are on the other. Therefore, every t{y, z}-path in G+ goes through either x or P.
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00 CM CM
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00	If there is a cut-vertex r in Kv separating t from P, then {r, x'} is a 2-cut in
the 3-connected graph G, which is impossible. Therefore, there are tP-paths Q and R in Kv that are disjoint except for t. We can now reroute P through t to obtain the desired path.	□
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Since G is 2-crossing-critical, there is a 1-drawing D of G — x't. From Claim 1, there is an su-path P in Kv containing t. Let C be the cycle obtained from P by adding x', x's and x'u.
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CD	Claim 2. All the vertices of G — (Kv U Kx) are in the same face of D[C].
Proof. Suppose by way of contradiction that there are vertices in G — (Kv U Kx) that are in different faces of D[C].
Case 1: there is a vertex p in Gp4c so that Kp contains vertices that are in different faces of D [C].
In this case there is an edge f of Kp that crosses D[C]. As D has at most one crossing, f is a cut-edge of Kp. Lemma 15.13 implies each component of Kp — f is adjacent to at least two different Kn's. If one of them is adjacent to both Kx and Kv, then we have a 3-cycle pxv in Gp4c in which both p and v have degree 3, contradicting Lemma 15.15.
Therefore, we may assume each is adjacent to one, say Kq and Kr, that is neither Kx nor Kv. However, now {v, x,p} is a 3-cut in Gp4c separating q and r in Gp4c. Therefore one of them — say q — is adjacent to precisely these three vertices in Gp4c, producing the 3-cycle {q, v, x} in Gp4c that contradicts Lemma 15.15.
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Case 2: any two vertices of G — (Kv U Kx) in different faces of D[C] are in CM	different Kp's.
Since G — (Kv U Kx) is connected, there is a path in G — (Kv U Kx) joining vertices in different faces of D[C]. Therefore, there is, for some vertices q and r of Gp4c, a KqKr-edge f that crosses D[C]. It follows that D[C] has no self-crossings, so D[C] has only two faces.
Clearly Gp4c — {x, v, f} has Kq and Kr in different components. Since Gp4c has at least six vertices, it has a vertex m different from all of v, x, q and r. We may choose the labelling so that D[Kq] is in one face of D[C], while D[Kr U Km] is contained in the other. It follows that { v, x, r} is a 3-cut in Gp4c separating q from
Since Gp4c is peripherally-4-connected, one of q and m — say q — is adjacent precisely to v, x, and r, yielding the 3-cycle {v,x, q} in Gp4c that has two vertices with only three neighbours, contradicting Lemma 15.15.	□
We note that the crossing in D cannot involve two edges, each incident with a vertex in Kv, as otherwise Gp4c is planar. In particular, D[C] is not self-crossing.
Claim 3. ODG+ (C) is isomorphic to ODG(C). In particular, ODG(C) is bipartite.
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Proof. The main point is that there is a single C-bridge in G containing G — (Kv + x'). To prove this, we show that any two vertices in G — (Kv + x') are connected by a C-avoiding path. For vertices not in Kv U Kx, this is easy: for any two vertices p and q in Gp4c — {v,x}, there is a pq-path in Gp4c — {v,x}, showing that any two vertices in Kp U Kq are joined by a path in G — (Kv U Kx).
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w
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00	If p G Kx — x', then Lemma 15.14 implies that the three vertices separating Kx
from its neighbours are distinct. For one of these vertices w' that is not x', Lemma 15.16 implies there is a pw'-path in Kx — x', completing the proof that there is a single C-bridge B in G containing G — (Kv + x').
Every other C-bridge in G is contained in Kv + x'. These are all C-bridges in G+; the only other C-bridge in G+ is the one containing the vertex joined to x, y, and z. This C-bridge has precisely the same attachments as B. This shows that ODG(C) and ODG+ (C) are isomorphic.
Since G+(C) is planar, ODG+ (C) is bipartite, yielding the fact that ODG(C) is bipartite.	□
Suppose first that C is clean in D. Since B is the unique non-planar C-bridge 00	in G, D yields a 1-drawing of C U B with C clean. Therefore, Corollary 4.7 implies
cr(G) < 1, a contradiction.
If, on the other hand, C is not clean in D, then C is crossed by an edge f. By Claim 2, f is incident with a vertex in Kv U Kx. If f is incident with a vertex in Kv, then contract Kv (with a vertex inserted at the crossing point, if necessary, to get a 1-drawing of Gp4c so that both edges incident with the crossing are incident with v. This implies the contradiction that Gp4c is planar.
If f is not incident with x', then Kx — x' has vertices on both sides of D[C]. One of these is in a component Kx of Kx — f that is on the side of D[C] that does not contain any vertex of G — (Kv U Kx). Lemma 15.13 implies Kx — x is joined to a vertex in some other Kw, w = v, which cannot happen without crossing D[C] a second time, a contradiction. It follows that f is incident with x'. Furthermore, Lemmas 15.19 and 15.16 (a) imply that f is in a cycle Cf in G — Kv. The ends of the edge ev of Kv crossed in D are separated by D[Cf ], so ev is a cut-edge of Kv. Moreover, ev is in C.
We now see that the C-bridges are B, those contained in one component of Kv — ev, and those contained in the other component of Kv — ev. Notice that B is a cut-vertex of ODG(C), and so it overlaps C-bridges of both the other types.
Since ODG(C) is connected and bipartite, it follows that the C-bridges in either of the components of Kv — ev occur on the same side of D[C] that they do in D+. In particular, x't may be reintroduced to D to obtain a 1-drawing of G, which is impossible.	■
Strategy. The strategy now is to show that if we replace any Kv with a smallest possible representative subject to the preceding observations, then we produce a 2-crossing-critical graph. This is the last part of this section. This implies that Gp4c turns into a 2-crossing-critical graph by choosing these smallest possible representatives. From this, it is then possible to determine (although not in a theoretical sense, but rather in a definite, finite — really manageable — way that we shall describe) all the 3-connected 2-crossing-critical graphs that have these configurations and reduce to Gp4c by planar 3-reductions.
There will remain the issue of determining all the possible Gp4c. Of course, one can list them all, but it is not clear at what point to stop. Fortunately, Theorem 2.14 shows that we do not need to do this when G contains a subdivision of V10, as we already know what G looks like. When G does not contain a subdivision of Vg, a theorem of Robertson plus some analysis implies that Gp4c has at most 9 vertices. We are left with the open question of finding the graphs in M3 that
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00	contain a subdivision of V8 but do not contain a subdivision of V1o. In Section 16,
we show that any such graph has at most about 4 million vertices.
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We next characterize certain properties of the graphs Gv; our goal is to show that these (more or less) determine the crossing number of G.
Definition 15.21. Let x, y, and z be vertices in a graph H so that H is an ||{x, y, z}||-bridge. Then:
CD	• T is the set of vertices w G {x, y, z} so that there are edge-disjoint
w({x, y, z} \ {w})-paths in H; and
•	U is the set of vertices w G {x, y, z} for which there are edge-disjoint paths in H — w joining the two vertices in {x, y, z} \ {w}.
•	(H, {x,y, z}) is a (T, U)-configuration if the graph H + obtained from H by adding a new vertex adjacent just to x, y, and z is planar.
Our entire argument depends on the fact, to be proved in the next section, that the pairs (T, U) effectively characterize 2-criticality. Theorem 15.24, the main point of this section, shows that substituting one (T, U)-configuration for another retains the fact that the crossing number is at least 2.
For a (T, U)-configuration, obviously there are only four possibilities for |T|. It is a routine analysis of cut-edges to see that, if |T| < 1, then U is empty, while if, for example, T = {x, y}, then U = {z}. Thus, for |T| < 2, U is determined by T. This is not the case for |T| = 3. In this instance, if z G U, then there is a cut-edge in Gv — z separating x and y. From here and the fact that T = {x, y, z}, one easily sees that x, y G U. Thus, if T = {x, y, z}, then |U| can be either 2 or 3. Therefore, there are in total five possibilities for the pair (|T|, |U|).
We first show that replacing a (T, U)-configuration with another (T, U)-con-figuration does not lower the crossing number below 2. First the definition of
CM	substitution.
Definition 15.22. Let G reduce by planar 3-reductions to the peripherally-4-CO	connected graph Gp4c. Suppose v is a vertex of Gp4c with neighbours x, y, and z so
that (Gv, {x, y, z}) is, for some subsets T and U of {x, y, z}, a (T, U)-configuration. Let N be the set of vertices t in {x, y, z} for which KV"ax n Kt is null. (See Definition 15.18 for Kvmax.) Let Nv denote the attachments of KV"ax: these are the vertices that are of the form t', t G {x, y, z}, chosen to be in Kt whenever possible.
(1)	A (T, U)-configuration (H, {x, y, z}) is (G, Kv)-compatible if:
(a)	for each t N, then there is only one neighbour of t in H;
(b)	the degrees of each t G {x, y, z} are the same in both Gv and H; and
(c)	setting Nh to consist of the union of the set of vertices of H in {x, y, z} \ N together with the neighbours in H of the vertices in N, H — N either has a single vertex or contains a cycle through all the vertices in NH.
(2)	The substitution of the Kv-compatible (T, U)-configuration (H, {x, y, z}) for Kv in G is the graph G^ obtained from G by adding H — N by identifying the vertices in Nv with those in NH in the natural way, and then deleting all vertices in KV"ax — Nv.
We are almost ready for a major plank in the theory.
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00	Our plan is to show that we can replace a "large" (T, U)-configuration by a
"small" (T, U)-configuration and still be 3-connected and 2-crossing-critical. There is one special case that requires particular attention.
Definition 15.23. A (T, U)-configuration (H, {x,y, z}) is doglike with nose n if |T| = 3 and |U| = 2 and n is the vertex in T \ U.
Theorem 15.24. Let G reduce by planar 3-reductions to the peripherally-4-connected graph Gp4c. Suppose v is a vertex of Gp4c with precisely the neighbours x, y, and z so that has at least two vertices so that (Gv, {x, y, z}) is, for some subsets T and U of {x,y,z}, a (T, U)-configuration. Let (H, {x,y,z}) be a (G, )-compatible (T, U)-configuration. If cr(G) > 2, then cr(GH) > 2.
Proof. We remark that the non-planarity of G and the fact that we are doing planar 3-reductions implies Gp4c is not planar. This fact will be used throughout the proof.
Let H' = H — {x, y, z} and let N be the set of vertices t in {x, y, z} so that Kvmax n is null. By way of contradiction, we suppose GH has a 1-drawing D. We start with two simple observations.
Claim 1. Some edge of H' is crossed in D.
Proof. If no edge of H' is crossed in D, then Definition 15.22 (1b) implies we may resubstitute for H' to obtain a 1-drawing of G, a contradiction.	□
Claim 2. There is no drawing D' of GH in which each crossed edge is incident with a vertex in H'.
Proof. Otherwise, insert a vertex at each crossing point, and add this vertex to H'. Then contract every edge in the new graph that has both ends in H', and also contract all the to single vertices. The result is a planar embedding of Gp4c, a contradiction.	□
Therefore, we may assume the crossing edges are ev G H' with some other edge f not incident with any vertex in H'. Observe that H' cannot be a single vertex.
Claim 3. f is not a cut-edge of G^ — H'.
Proof. Suppose f is a cut-edge of GH — H'. Since D[GH — H'] has no crossing, it is planar. Therefore, the faces on each side of f in D[GH — H'] are the same. Thus, the ends of ev are in the same face of D[GH — H'].
Consider now the planar embedding D[GH — ev ]. The two ends of ev are in the same face of the subembedding D[GH — H'] and so may be joined by an arc that is disjoint from D[GH — H']. This produces a drawing of GH in which all the crossings involve ev and edges incident with at least one vertex in H'. This contradicts Claim 2.	□
Since f is not a cut-edge of GH — H', there is a cycle Cf of GH — H' containing f. Moreover, D[Cf ] separates the two ends of ev, so ev is an cut-edge of H'. Let CD	H1 and H2 be the two components of H' — ev.
The next claim is central to the remainder of the argument.
Claim 4. Let t e {x,y, z} be a common neighbour of H1 and H2. Then f is incident with t' e Kt and one of the faces of H' +1' incident with both t' and ev is empty except for the segment of f from t' to the crossing with ev.
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CM
I
CM 00 CM CM
00	Proof. Let C be any cycle in H' +1' containing ev. Since ev is a cut-edge of
H', t' G C. Since Gp4c — {v,t} is connected, G — (Kv U Kt) is connected.
Suppose by way of contradiction that there are vertices u and w of G — (Kv UKt) on both sides of D[C]. By the preceding paragraph, there is a uw-path P in G — (Kv U Kt). Since P is graph-theoretically disjoint from C, but D[u] and D[w] are on different sides of D[C], D[P] crosses D[C]; this must be at the unique crossing of D, so f G P and the crossing of D[P] with D[C] is the crossing of f with ev.
Moreover, D[Cf] crosses D[C] at the crossing of D and so they must cross somewhere else. As Cf and H' are disjoint, the second crossing is at the vertex t'. Since this is true of any cycle Cf in G — Kv, f is a cut-edge of (G — Kv) — t'.
We now consider two cases.
,	4
Case 1: there are distinct vertices t1 and t2 of Gp — {t, v} so that D[Ktl] and D[Kt2 ] are on different sides of D[C].
In this case, either (i) for some vertex s of Gp4c, f G Ks, in which case t1 and t2 are in different components of Gp4c — {t, v, s}, or (ii) since Gp4c is non-planar and so has at least five vertices, for some vertex s of Gp4c that is an end of f, we may choose t1 and t2 to again be in different components of Gp4c — {t, v, s}.
In either case, the internal 4-connection of Gp4c implies that there is an i G {1, 2} so that tj is the only vertex in its component of Gp4c — {t, v, s}. But then tvtj is a 3-cycle in Gp4c having v and tj as degree 3 vertices, contradicting Lemma 15.15.
Case 2: there are not distinct vertices t1 and t2 of Gp4c — {t, v} so that D[Ktl ] and D[Kt2 ] are on different sides of D[C].
In this case, there is a vertex s of Gp4c — {t,v} so that f G Ks and all the vertices of G — (Kv U Kx) on one side of D[C] are in one component K^ of Ks — f, while all the other vertices of G — (Kv U Kx), including the other component K^ of Ks — f, are on the other side of D[C].
Lemma 15.13 implies that K^ has neighbours in two Kr's. According to D, these can only be Kv and Kt. But now the 3-cycle tvs has the two degree 3 vertices v and s, contradicting Lemma 15.15.
Since f is on both sides of D[C], but one side has no vertex, it must be that the end of f on that side is in C. But f is disjoint from H', and so this end can only be t'.	□
Our proof proceeds by considering how many common neighbours among Kx, Ky, and Kz there are for H1 and H2. We start by noting that there cannot be three, since then the graph H + is not planar, contradicting Definition 15.21.
Claim 5. H1 and H2 have exactly one common neighbour.
Proof. We have already ruled out the possibility that H1 and H2 have three common neighbours.
To rule out two common neighbours, suppose by way of contradiction that H1 and H2 have the two common neighbours Kx and Ky. By the preceding remark, at least one of H1 and H2 does not have a neighbour in Kz. Since H' does have a neighbour in Kz, we may choose the labelling so that H1 has a neighbour in Kz and H2 does not.
Claim 4 implies f is incident with both x' and y'. But now D[f ] can be rerouted along the other side of the x'H2-edges, around H2, and on to y' so that G^ has
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00	no crossings. This implies the contradiction that Gp4c is planar. We conclude that
H1 and H2 have at most one common neighbour.
If they have no common neighbours, then H1 has neighbours just in Kx, while H2 has neighbours in and , but not in Kx. In this case, e^ is a cut-edge in H separating x from {y, z}. It follows that x G T. Since Gv is also a (T, U)-configuration, there is an cut-edge eV of Gv separating x from {y, z}. Now we can replace H' in T with in such a way that eV (in fact the only edge of Gv incident with x) is crossed by f to yield a 1-drawing of G. This contradiction completes the proof of the claim.	□
We conclude from Claim 5 that H1 and H2 have precisely one common neighbour x'. Claim 4 implies that f is incident with x'.
If, for some i G {1, 2}, H® has no other neighbour, then we may reroute f to go around D[H®], yielding a planar embedding of GH and, therefore, of the non-planar graph Gp4c, a contradiction.
1	Thus, we may choose the labelling so that H1 has at least one neighbour in
, while H2 has at least one neighbour in . If, say, H1 is joined to by only one edge, then y G T ; therefore, y is incident with a unique edge in Gv and we can replace D[H] with the planar embedding of so that it is the yKV-edge that is crossed by f. This yields that contradiction that G has a 1-drawing.
Thus, we may assume that T = {x, y, z}. However, there are not edge-disjoint yz-paths in H — x (ev is a cut-edge separating y and z). Therefore, U = {y, z}, showing Gv is doglike. It follows that Gv — x has a cut-edge eV separating y and z. We may substitute the planar embedding of for D[H] so that eV crosses f, yielding the final contradiction that G has a 1-drawing.	■
00
15.4. Reducing to a basic 2-crossing-critical example
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In this section, we show that if G is a 3-connected 2-crossing-critical graph that reduces by planar 3-reductions to a peripherally-4-connected graph, then there is a "basic" 3-connected 2-crossing-critical graph from which G is obtained by the regrowth mechanism of the preceding section.
Theorem 15.25. Let G gM2
reduce by planar 3-reductions to a peripherally-4-connected graph Gp4c. Let v be a vertex of Gp4c with just the three neighbours x, y, and z, so that (Gv, {x, y, z}) is a (T, U)-configuration and has at least two vertices. Let Grep(v) be the graph obtained from G by contracting as indicated in the following cases.
(1)	If (Gv, {x, y, z}) is doglike, then let e be the cut-edge of and contract each component of — e to a vertex.
(2)	If (Gv, {x, y, z}) is not doglike, then we have the following subcases.
(a)	If none of Gx, Gy, and Gz is doglike, then contract to a vertex.
(b)	If (|T|, |U|) = (3, 3), then contract to a vertex.
(c)	If Gx is doglike and y G T, then let C be a cycle
in G+ containing
x', y', and z', delete everything in — E(C) and contract the edges of C to the 3-cycle x'y'z'. Then Grep(v) G Mf.
There is one clarification that is required to understand one fine detail of Grep(v). If, for example, the vertex x' is in , then we proceed precisely as described in the
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00	statement. If, however, x' is in Kx and x G T, then in Grep(v) we retain only two
edges between x' and the contracted vertex in Krep(v) to which it is joined. This especially applies in the case 2c: if z' G Kz, then we keep only the two edges of C incident with z', while if z' G Kv, then we keep all the z'Kz-edges.
There is also an important remark to be made. We had long thought that it was possible to reduce each Kv to a single vertex and retain 2-criticality. This might be true in the particular cases of 3-connected 2-crossing-critical graphs with no subdivision of Vg, but it is certainly not true of all 3-connected 2-crossing-critical graphs.
In Definition 2.10 we described the set S of all graphs that can be obtained from the 13 tiles and the two frames. These graphs are all 3-connected and 2-cros-sing-critical. Consider any one of these that uses the right-hand frame in Figure 2.1 and uses the second picture in the third row of Figure 2.2. With appropriate choices of the neighbouring pictures, the 3-cycle in the upper half of the picture is part of a doglike Gv that contains the parallel edges in the picture and the parallel edges in the frame: the horizontal edge in the 3-cycle is Kv. The vertical edge in the other 3-cycle in the picture is a Kx. When we do the planar 3-reductions in this case, the contractions of Kx and Kv produce a pair of parallel edges not in the rim. The conclusion is that the resulting peripherally-4-connected graph plus parallel edges is not 2-crossing-critical. Thus, the technicalities we must endure in the statement of Theorem 15.25 seem to be unavoidable.
Proof. We use the notation Krep(v) for the contraction of Kv in Grep(v).
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Phase 1: showing Grep(v) is 3-connected.
Let t and n be vertices of Grep(v). We show Grep(v) — {t,«} is connected.
Let wt and wu be the vertices of Gp4c so that t G Kwt and « G KWu (taking, for example, Kwt to be Krep(v) if t G Krep(v)). It follows from Lemma 15.16 that every vertex of every Ks has a path in G — {t,«} to at least one neighbour of Ks that is not one of Kwt or KWu. This is also true of Krep(v), as may be seen by checking the analogues for Krep(v) of Lemma 15.16 in the three cases for which Krep(v) has at least two vertices. (Note there are two possible outcomes for Krep(v) in Case 2c, depending on whether z' G Kv, in which case Krep(v) is a 3-cycle, or z' G Kz, in which case Kv is an edge.)
Since each Ks is connected, Grep(v) — {t,«} is connected.
Phase 2: showing cr(Grep(v)) > 2.
f-r.
The graph Krep(v) obtained from Krep(v) by adding x, y, and z is a (G, Kv )-compatible (T, U)-configuration. Therefore, Phase 2 follows immediately from Theorem 15.24.
Phase 3: showing that Grep(v) is 2-crossing-critical.
Let e be any edge of Grep(v). Then there is an edge eG in G naturally corre-
sponding to e (in the sense that precisely the same contractions and deletions of G
and G — eG can be used to obtain both Grep(v) and Grep(v) — e).
Special situation. There is one case where the choice of must be made with special care. Suppose Kv contracts down to the single vertex v and e is one of two parallel edges vx. In the case Kv has a cut-edge e', Lemma 15.13 implies each component of Kv — e' is joined to two of the neighbours of v. Suppose that Kx
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00	is the only common neighbour of these two components. Since Gv is not doglike,
some component L of Kv — e' is joined by exactly one edge to its other neighbour; choose eg to be an xL-edge.
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Definition 15.26. For each vertex w of Krep(v), Lw denotes the subgraph of Kv that contracts to w.
Since G is 2-crossing-critical, there is a 1-drawing D of G — eG. If no edge of any Lw C Kv is crossed in D, then these may each be contracted to obtain a
CD	1-drawing of Grep(v) — e, and we are done.
Claim 1. If there is a drawing of G — eG in which all the crossings are between edges incident with vertices in Lw, then Grep(v) — e is planar.
Proof. Insert vertices at each crossing point and contract every edge in the new graph that has both ends in some Lu. The result is a planar embedding of
Grep(v) e.
Therefore, we may assume the crossing edges are ev G Lw C Kv with some other edge f not incident with any vertex in Lw.
Case 1: f is a cut-edge of (G — eG) — Lw.
In this case, D[(G — eG) — Lw] has no crossing, so it is planar. Therefore, the faces on each side of f in D[(G — eG) — Lw] are the same. Thus, the ends of ev are in the same face of D[(G — eG) — Lw].
Consider now the planar embedding D[(G — eG) — ev]. The two ends of ev are in the same face of the subembedding D[(G — eG) — Lw] and so may be joined by an arc that is disjoint from D[(G — eG) — Lw]. This produces a drawing of G — eG in which all the crossings involve ev and edges incident with at least one vertex in Lw. Claim 1 implies Grep(v) — e is planar, as required.
Case 2: f is not a cut-edge of (G — eG) — Lw.
In this case, f is in a cycle Cf of (G — eG) — Lw. Moreover, D[Cf] separates the two ends of ev, so ev is a cut-edge of Lw. Let Lw and Lw be the components of Lw — ev.
We consider separately two cases for Gv.
Subcase 2.1: Gv is doglike.
In this subcase, Krep(v) is two vertices w and w joined by a cut-edge e' of Gv — x, each joined by an edge to x', w is joined by at least two edges to Ky and w is joined by at least two edges to Kz. Lemma 15.20 implies that Kx has at most two neighbours in Kv. We already know there is one in each of Lw and Lw. Lemma 15.11 now implies there is a vertex x' G Kx incident with all the KvKx-edges in G. Thus, we may choose the labelling of Lw and Lw so that the neighbour of x' in Lw is in Lw.
We see that x' and the end of ev in Lw are neighbours of vertices in Lw, and neither of these vertices is in Lw. The only other possibilities for neighbours of Lw outside of Lw are in Ky and Lw, the latter being the end of e'. A similar remark holds for Lw: it has the neighbour (via ev) in Lw, and can have at most neighbours in Ky and Lw (via e').
Since G is 3-connected, for each i = 1, 2, Lw has at least two neighbours outside of Lw other than x'. From the neighbour analysis of the preceding paragraph, there
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00	are at most three in total: two to Ky and one to LW. There are two ways this can
happen.
In the first way, both edges from LW to Ky have their ends in LW, while e' has an end in LW. But then ev is a cut-edge of Kv that violates Lemma 15.13: the edge eG cannot connect LW to either x' (Lemma 15.20 or Kz (because e' is a cut-edge of Gv — x), so the component LW of Kv — ev is joined only to Ky.
Therefore, e' has one end in LW and the two KvKy edges have ends in different ones of LW and LW. It follows that y' is incident with these edges, so Lemma 15.20 implies y' has precisely these neighbours in Kv.
Contract D[ev ] so that LW is pulled across f and, if necessary, shrink D[LW ] so that we obtain a new drawing D1 of G — eG in which f crosses the edges from x' and y' to LW,.
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Claim 2. f e LW.
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Proof. If f e Lw, then exactly the same analysis as for Lw implies that Lw — f has two components LW, from which there is an edge to x' and an edge to z', and LW, from which there is an edge to z' and LW. But now the graph-theoretically disjoint cycles in Lw + y' containing ev and Lw + z' containing f cross exactly once in D, which is impossible.	□
It follows from Claim 2 that f e Lw. We contract the uncrossed Dl[Lw ] and Dl[Lw] to obtain a drawing D2 of Grep(v) — e, in which the only crossings are of f with the edges from x' and y' to LW. In D2, there are parallel edges y'w; the one from y' to LW is not crossed in D2, so we may make all the others go alongside the uncrossed one. This yields a drawing D3 of Grep(v) — e in which the only crossing is x'w with f, so D3 is a 1-drawing of Grep(v) — e, as required.
Subcase 2: Gv is not doglike.
Subcubcase 2.1: there is a neighbour x of v in Gp4c so that Gx is doglike and x' e Kv is the nose of Gx.
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Let C be the cycle in Gv that be the subgraph of G obtained by deleting all edges between the various Lu except the one or three edges in C. Choose the labelling so that y is a neighbour of v in Gp4c so that there is exactly one Kv Ky-edge in G; thus y' e Kv.
Let r be that element of {x, y, z} so that r' e Lw. There are precisely two edges eL and e2 in GC coming out of Lw in Gv — r.
Let LW be the component of Lw — ev containing r' and let LW be the other. Since C goes through r', at least one of eL and e2 is incident with a vertex in LW. Therefore, at most one of eL and e2 has an end in LW.
We claim that LW is not joined to any other vertex in GC. The only possibility is that there is an edge from LW to Kx U Ky U Kz. Since all the Kv Kx- and Kv Ky-
•T—I	"	- -n
edges in G are incident with x' and y', respectively and x' and y' are not in L2
there are no edges in G from LW to Kx U Ky.
As for the possibility of an LWKz-edge, this can only exist if z' e Kz. But z' already has two known neighbours in Kv, namely the Kv-ends of the edges of C incident with z'. Lemma 15.20 implies these are the only vertices of Kv adjacent to vertices in Kz. Therefore these known z'-neighbours are the only ones; in particular, z' has no neighbour in LW, as claimed.
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necessary, scaling D[L^] down so that L^ and L^ are now drawn on the same side of f. The only crossing in this new drawing is of the edge of D[GC], if it exists, that is not ev and joins L^ to the rest of GC. Now we may contract all the Lu to single vertices to obtain the required 1-drawing of Grep(v)
— e.
Subsubcase 2: there is no neighbour x of v in Gp4c so that Gx is doglike and x' G Kv is the nose of Gx..
At this stage, Kv contracts to a single vertex of Grep(v). In this case, Kv — ev has two components Kj and Kj. Lemma 15.13 implies each of Kj and Kj are connected in G to at least two of Kx, Ky and Kz. Because G+ is planar, at most two of Kx, Ky, and Kz can be adjacent to both Kj and Kj.
If both Kx and Ky have neighbours in both Kj and Kj, then there is an i G {1, 2} so that Kj has adjacencies only in those two. Now pull D[KV] across f and, scaling D[KV] if necessary, to obtain a planar embedding of G — eG. This contracts to a planar embedding of Grep(v) — e, as required.
Thus, we may assume Kj and K^ have precisely one common neighbour in G. Each has its own neighbour. Since Gv is not doglike, one of these, say Kj, is joined by a single edge to that unique neighbour and now we can drag Kj across f. This works unless e goes to Kj and Kj is joined to its unique neighbour by two edges. But this is the special situation, and e is joined to Kj, not Kj.	■
15.5. Growing back from a given peripherally-4-connected graph
CM
The important corollary of Theorem 15.25 is that, if we replace each Kv with its Krep(^), then we get a 2-crossing-critical model of Gp4c with very simple replacements for the vertices of Gp4c. In this section, we explain how to obtain all the 3-connected 2-crossing-critical graphs that reduce by planar 3-reductions to a particular peripherally-4-connected graph.
CO	Let L be a non-planar peripherally-4-connected graph. For each vertex v of
L having only three neighbours x, y, and z, we decide on the type of v; that is, we choose Tv C {x,y, z} and, in the case |Tv | = 3, we decide on : either = {x,y, z}, or consists of two of {x,y, z}. For each edge of L joining two vertices of degree at least 4, we decide whether the edge will be a single edge or a parallel pair.
The choices must be made so that x G Tv if and only if v G Tx. If, for some v, (|T!y |, |Uv |) = (3, 2) (v is chosen to be doglike), then some other implications (as in Theorem 15.25) must be maintained. Choose the labelling so that x G U. Then x is the nose of the dog, v is replaced with Kv , so that Kv is an edge y'z', so that y' incident with two edges going to Ky, and likewise for z' to Kz. Each of y' and z' is also incident with an edge to x' G Kx. Furthermore, Kx can be either a vertex, or, if |Tx| = 3, an edge, or a 3-cycle.
Once all these choices have been made, the resulting graph is tested for 2-criticality. Thus, for a given peripherally-4-connected graph L, there will be many graphs that require testing. If one of the resulting graphs L' is found to be 2-cros-sing-critical, then there may be many other 3-connected 2-crossing-critical graphs that arise from L'. Recall that, for each vertex of L that has only three neighbours,
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explains what may replace the vertex of each type.
LEMMA 15.27. Suppose the peripherally-4-connected graph L has choices as explained in the preceding paragraphs to produce a 3-connected 2-crossing-critical graph L'. Suppose G is a 3-connected 2-crossing-critical graph that reduces by planar 3-reductions to L so that L' is the graph obtained from G by the replacements described in Theorem 15.25. Then, for each in L', is replaced by one of the o	possibilities shown in Figures 15.1, depending on (Tv, ).
Proof. We only illustrate the tedious proof in a couple of cases. Case 1: (T„, U) = ({x,y,z}, {y,z}).
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Let e be a cut-edge in Gv — x separating y and z. Let Kv — e have the two components Ky, containing the neighbour(s) of y, and KV, containing the neighbour(s) of z. If Ky, for examples, is not just either a single vertex or an edge joining the two neighbours of y, then it contains a subdivision of one of these (either pick a path in Ky joining the neighbour of y to the Ky-end of e or pick a path joining the two neighbours of y). It is easy to see that the subdivision (making a similar choice on the z-side) is also a (Tv, Uv )-configuration. By Theorem 15.24, the subgraph has crossing number 2, and so is all of G. Thus, Kv can be at most one of the three figures in Figure 15.1 corresponding to (|T|, |U|) = (3, 2).
Case 2: Tv = {x, y, z} = Uv.
In this case, Gv — x contains edge-disjoint yz-paths. Therefore, it contains two such paths P and Q that make a digonal pair. If P and Q are internally disjoint, then there is a (P — {y, z})(Q — {y, z})-path R. If P and Q are not internally disjoint, then set R = 0. In either case, set M = P U Q U R. There are two x(M — {y, z})-paths R1 and R2 in Gv.
If the ends of P and Q are in the same digon of P U Q, then planarity of G+ implies R1 and R2 have their ends in the same one of P and Q. It follows that M U R1 U R2 is a (Tv, Uv )-configuration, and so is Gv by 2-criticality and Theorem 15.24.
The fact that G is 3-connected implies that there cannot be more than four common internal vertices to P and Q, as if there were six digons, then some two consecutive ones would not contain an end of either R1 or R2. This would readily yield a 2-cut in G, which is impossible. This is why the number of possibilities for
In some of the larger (T, U)-configurations, there are edges that are not required to produce the relevant paths between s, t, and u, but, rather, are there to maintain the connectedness of the configuration. These edges might be deletable without reducing the crossing number below 2. Thus, each candidate 3-connected graph produced by the method described needs to have its criticality checked.
15.6. Further reducing to internally-4-connected graphs
In order to find the 2-crossing-critical graphs that do not contain V8, we wish to use the characterization by Robertson of V8-free graphs. This characterization, described in the next section, is in terms of internally-4-connected graphs. These
Gv in this case is finite.
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Figure 15.1. The possible (T, U)-configurations.
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graphs are very closely related to peripherally-4-connected graphs and it is the purpose of this section to describe the reduction of a peripherally-4-connected graph to an internally-4-connected graph, and back again.
Definition 15.28. A hug in a graph G is an edge e in a triangle T whose vertex v not incident with e has degree 3. The triangle T is the e-triangle, v is the head of the hug and the two edges of T other than e are the arms of the hug.
Definition 15.29. A G is internally-4-connected if it is peripherally-4-connec-ted and has no hugs.
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Figure 15.2. The thick edge is a bear hug. The dotted edges tw and vz might be subdivided, and the dashed edge uw need not be present. If uw is not present, then {ux, uy} is a simultaneously deletable pair of bear hugs.
It is not correct that simply deleting (successively) the hugs from a peripherally-4-connected graph produces an internally-4-connected graph. There is a particular situation that arises that needs special care.
Definition 15.30. (1) A hug e with head v is a bear hug if there is an end u of e, incident with a second hug uy whose head t is different from v, and so that, with w the other end of e, the neighbours of u are contained in the union of {t, v, w} and the set of neighbours of t. (See Figure 15.2.)
(2)	A hug is deletable if it is not a bear hug.
(3)	A pair of bear hugs having a common end is simultaneously deletable.
We are now in a position to reduce a peripherally-4-connected graph to an internally-4-connected graph.
Theorem 15.31. Let G be a non-planar peripherally-4-connected graph and let G = Go, Gi, ..., Gfc be a sequence of graphs so that, for each i = 1, 2, ..., k, there is either a hug h or a simultaneously deletable pair h of bear hugs in Gj_i so that
Gi
Gj_i — hj. Then, for i = 0,1, 2,
, k:
(1)	Gj is a subdivision of a non-planar peripherally-4-connected graph;
(2)	if v has degree 2 in Gj but not in Gj-i, then hj is a simultaneously deletable pair of bear hugs in Gj-i, both incident with v; and
(3)	every degree 2 vertex in Gj has two degree 3 neighbours in Gj. Furthermore, if the sequence Go, Gi,. .., Gk is maximal, then Gk is a subdivision of an internally-4-connected graph.
We emphasize that, in the reduction process described in the statement, Gi is obtained from Gi_i by the deletion of either one or two edges.
Proof. Suppose by way of contradiction that i is least so that Gi is planar. Since G0 is not planar, i > 0, so Gi = Gi_i — hi. Each edge in hi joins two neighbours of a degree 3 vertex in Gi and so may be added to the planar embedding of Gi to produce a planar embedding of Gi together with that edge of hi. In the case
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00	|hj| = 2, the heads of the hugs are not adjacent. Thus, both hugs may be added
simultaneously, while preserving planarity. Thus, Gj_1 is planar, contradicting the choice of i.
By way of contradiction, we may let i be least so that Gj is not a subdivision of a peripherally-4-connected graph. Thus, i > 1. Throughout the proof, when we refer to the vertices t, u, v, w, x, y, z, we are always referring to the labelling in Figure 15.2. In each of the three cases, there are two possibilities for h to be considered.
It will be helpful to notice that, in the case h consists of a simultaneously deletable pair of bear hugs, the vertex u is not a node of Gj and is incident with both deleted edges.
Claim 1. Gj is a subdivision of a 3-connected graph.
Proof. Let a and b be distinct nodes of Gj. Then a and b are distinct nodes of Gj_1, so there are three internally disjoint ab-paths P^P2,P3 in Gi-1.
If e G hj, then the head c of the e-triangle has degree 3. If e is in some Pj and T is the triangle containing e and its head, then we may replace Pj n T with the path in T complementary to Pj n T. The at most two modifications result in three internally disjoint paths that are also paths in Gj.	□
Claim 2. If a has degree at least three in Gj_1 and degree 2 in Gj, then:
(1) |hj| = 2;
(2)	a is incident with both edges in hj; and
(3)	both neighbours of a have degree 3 in Gj.
Proof. Let e g hj. The head b of the e-triangle has degree 3 in Gj_1 and, since Gj_1 is a subdivision of a peripherally-4-connected graph, no other vertex of the e-triangle has degree 3, so Lemma 15.15 shows they both have degree at least 4. It follows that if e is the only edge in hj, then the ends of e have degree at least 3 in Gj and no new vertex of degree 2 is introduced in Gj.
Therefore hj is a deletable pair. The only new vertex of degree 2 in Gj is u, so a = u. Also, the only neighbours of u in Gj have degree 3 in Gj.	□
The remaining possibility is that there is a set { a, b, c} of nodes of Gj and a 3-separation (H, J) of Gj so that H n J = ||a, b, c|| and both H — {a, b, c} and J — {a, b, c} have at least two nodes of Gj.
Because Gj_1 is a subdivision of a peripherally-4-connected graph, there is an edge e G hj having one end rH in H — {a, b, c} and one end rj in J — {a, b, c}.
Suppose for the moment that hj has a second edge. Since Gj_1 is a subdivision of a peripherally-4-connected graph, not all the neighbours of u in Gj_1 can be in the same one of H and J. We may choose the labelling so that x = rJ. As t is a common neighbour of u = rH and x = rJ, we conclude that t G {a, b, c}, say t = a.
It follows that at least one of v and y (the other two neighbours of u) is in H — {t, b, c}. Since v and y are adjacent, it follows that both are in H and, furthermore, uy is also in H. In particular, there is a unique edge in hj that has one end in H — {a, b, c} and one end in J — {a, b, c}.
Now the two possibilities for hj are merged: e is the unique edge in hj having one end rH in H — {a, b, c} and one end rJ in J — {a, b, c}. The head q of the e-triangle must be in {a, b, c}, say q = a.
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Gi-1 is a subgraph of the illustrated
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Since q has degree 3, we may choose the labelling so that rH is the only neighbour of q in H — {q, b, c}. The neighbour rj of q is in J — {q, b, c}. Note that rH and rJ are both nodes of Gi-1.
The third neighbour s of q is in J, so {rH, b, c} is a 3-cut in Gi-1. Since Gi-1 is peripherally-4-connected, there is a unique node p in H — rH, which is joined by branches in Gi-1 to all of rH, b, and c.
If s € {b, c}, then the discussion in the preceding paragraph applies with rJ and J in place of rH and H, respectively. The nodes of Gi-1 are now all known (there are only 7), and the edges are almost completely determined. In particular, Gi-1 is a subgraph of the planar graph shown in Figure 15.3, contradicting the fact that Gi-1 is non-planar. Therefore, s is in J — {q, b, c}.
The vertex rH is the only candidate for the second branch vertex (after p) of Gj in H — {q, b, c}, so it must be joined by a Gj-branch to at least one of b and c; choose the labelling so that b is an end of such a Gj-branch.
If b has only one neighbour in J — { q, b, c} , then p and b are both degree 3 vertices in a triangle in Gj-1; since Gj-1 is a subdivision of a peripherally-4-con-nected graph, this contradicts Lemma 15.15. The same reasoning implies that both rH and b have degree at least 4 in Gj-1. These imply that rHp, rHb, and pb are all edges of Gj-1.
Because rHrJ is in hj and q is the head of the rHrJ-triangle, we know that rHrJ, qrH, and qrJ are all edges of Gj-1. Furthermore, rHs is not a Gj-1-branch (it would yield a second edge with one end in each of H — {a, b, c} and J —{a, b, c}).
The triangles prHb and qrHrJ show that rHrJ is a bear hug. Since it was deleted, it must be in a simultaneously deletable pair of bear hugs. This implies that rHb is the other edge in that pair. Thus, H — {a, b, c} has only one node in Gj, a contradiction that completes the proof that each Gj is a subdivision of a peripherally-4-connected graph.
We move on to showing that a maximal sequence ends in a subdivision of an internally-4-connected graph. So suppose Gj is not a subdivision of an internally-4-connected graph. Since it is a subdivision of a peripherally-4-connected graph
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00	H, there is a 3-cut {a, b, c} in H so that ab is an edge of Gj. Since H is periphe-
rally-4-connected, there is a vertex p adjacent in H to all of a, b, c and with no other neighbours in H. Lemma 15.15 shows that the triangle p, a, b has at most one vertex of degree 3; since p is such a vertex, a and b have degree at least 4 in H. It follows that pa and pb are edges of Gj and, therefore, ab is a hug in Gj.
It is evident from the definitions that, as soon as Gj has a hug, then either Gj has a hug that is not a bear hug or Gj has a pair of simultaneously deletable bear hugs. In either case, Gj is not the last in a maximal sequence.	■
We conclude this section with a brief discussion of the reverse process: how to generate all the peripherally-4-connected graphs that reduce to a given non-planar internally-4-connected graph G. Every graph created through iterating the 00	following procedure is peripherally-4-connected and non-planar. We choose either
two non-adjacent neighbours of a degree 3 vertex and add the edge between them, or we choose an edge e joining degree 3 vertices and a neighbour of each vertex incident with e, subdivide e once, and join both the chosen neighbours to the vertex of subdivision.
Every internally-4-connected graph produces only finitely many peripherally-4-connected graphs through this process, as the number of possible additions is initially finite and strictly decreasing.
15.7. The case of Vg-free 2-crossing-critical graphs
In this section, we complete our analysis of peripherally-4-connected 2-crossing-
CM
critical graphs by considering the case of 3-connected 2-crossing-critical graphs that do not contain a subdivision of Vg. This is the whole reason for studying periphe-
rally-4-connected graphs, since there is a characterization of the closely related internally-4-connected graphs that do not contain a subdivision of Vg. Two important classes of graphs in this context are the following.
Definition 15.32. (1) A bicycle wheel is a graph consisting of a rim, which is a cycle C, and an axle, which is consists of two adjacent vertices CO	x and y not in the rim, together with spokes, which are edges from {x, y}
to C.
(2) A 4-covered graph is a graph G containing a set W of four vertices so that G — W has no edges.
Maharry and Robertson [22] prove Robertson's Theorem that an internally-4-connected graph with no subdivision of Vg is one of the following:
(1)	a planar graph;
(2)	a non-planar graph with at most seven vertices;
(3)	C3 □ C3;
(4)	a bicycle wheel; and
(5)	a 4-covered graph. Suppose G is a 3-connected graph that does not contain a subdivision of Vg
and G reduces by planar 3-reductions to the peripherally-4-connected graph Gp4c. It follows that Gp4c has no Vg. Eliminating hugs as described in Theorem 15.31 produces an internally-4-connected graph Gl4c. Deleting hugs does not affect the planarity of the graph; since Gp4c is not planar, so is Gl4c. By Robertson's Theorem, one of the following happens:
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00	(1)	Gl4c is not planar and has at most seven vertices;
(2)	Gl4c is C3 □ C3;
(3)	Gl4c is a bicycle wheel; and
(4)	Gl4c is a 4-covered graph.
Our ambition in the remainder of this section is to explain how to determine all the peripherally-4-connected graphs Gp4c that can be the outcome of a sequence of planar 3-reductions starting from a 3-connected, 2-crossing-critical graph G that has no subdivision of V8. Any peripherally-4-connected graph with no subdivision of V8 that either has crossing number exactly 1 or is itself 2-crossing-critical needs to be tested. Those with crossing number 1 might extend to a 2-crossing-critical example by duplication of edges and/or replacing vertices of degree 3 by one of the basic (T, U)-configurations, as explained in the preceding section.
The first two items arising from Robertson's Theorem are easily dealt with. A computer program can easily find all internally-4-connected graphs with at most 7 vertices and determine which ones either have crossing number 1 or are 2-crossing-critical. The graph C3 □ C3 is itself 2-crossing-critical, so this is one of the 3-connected, 2-crossing-critical graphs that do not contain a subdivision of Vg.
Definition 15.33. Let Gp4c be a peripherally-4-connected graph and let Gl4c be the internally 4-connected graph obtained from Gp4c by simplifying (that is, leaving only one edge in each parallel class) and eliminating hugs. Then Gp4c is a peripherally-4-connected extension of Gl4c.
We conclude this section by showing how to which bicycle wheels and 4-covered graphs Gl4c can have such a 2-crossing-critical Gp4c as an extension. In particular, Gl4c must either have crossing number 1 or itself be 2-crossing-critical; in the latter case Gp4c = Gl4c.
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CASE 1: the bicycle wheels.
Let x and y be the adjacent vertices making the axle of the bicycle wheel Gl4c, and let C be the cycle that is the rim. Our goal is to provide sufficient limitations on C to show that the computation is feasible. Here is our first limitation, which can very likely be improved.
Lemma 15.34. Suppose G G M3 reduces by planar 3-reductions to the graph Gp4c that is a peripherally-4-connected extension of Gl4c. If Gl4c is a bicycle wheel with axle xy and rim C, then x is not adjacent in Gl4c to six consecutive vertices on C, none of which is adjacent to y.
Proof. Suppose by way of contradiction that x1, x2, x3, x4, x5, x6 are six consecutive (in this order) vertices of C adjacent to x but not y. Lemma 15.15 implies no two consecutive ones of these vertices have only three neighbours in Gp4c. By symmetry, we may assume x3 has a neighbour u that is not adjacent to x3 in Gl4c.
Because Gp4c is a peripherally-4-connected extension of Gl4c, there are vertices w and z so that x3, u, and w are the neighbours (in both graphs) of z and no other vertex has just these three neighbours. Since y is not adjacent to x3 and x has more than 3 neighbours, z G C .If follows that x3 and u are the C -neighbours of z and w is the neighbour of z that is in {x, y}. In particular, z, being a neighbour of x3 is either x2 or x4, so w = x. In either case, three consecutive vertices from x1, x2,..., x6 are
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Claim 1. K is clean in D.
00	such that the outer two are adjacent by a chord in Gp4c; if necessary, we relabel so
these are x1,x2,x3. In particular, x2 has just three neighbours in Gp4c.
Let D be a 1-drawing of Gp4c — xx2 and let K be the subgraph of Gp4c — xx2 induced by x, x1, x2, and x3.
& g
Proof. In Gp4c — xx2, x2 has only two neighbours, so the edge x1x3 and the path (x1,x2,x3) make a pair of parallel edges. Therefore, we may assume neither e	of these is crossed in D.
It suffices to prove that xx1 is not crossed in D, as the proof for xx3 is symmetric. Suppose by way of contradiction that xx1 is crossed in D and consider the planar embedding of Gp4c — {xx1, xx2} induced by D. Since Gl4c — {xx1, xx2} is a subgraph, it is also planar, embedded in the plane by D.
Since x3 has only three neighbours in Gl4c — {xx1,xx2}, we can add the edge xx2 alongside the path (x, x3, x2) to obtain a planar embedding of Gl4c — xx1. Then we may add the edge xx1 alongside the path (x,x2,x1) to get a planar embedding of Gl4c. However, this contradicts the fact that Gl4c is not planar.	□
Now let K be the subgraph of Gp4c — xx2 induced by x, x1, x2, and x3. Because x1, x2, and x3 are consecutive along C, there is a unique K-bridge B in Gp4c — xx2. The claim shows K is clean in D, so D[B] is contained in one face F of D[K].
Adjusting which of D[x1x3] and D[(x1, x2, x3)] is which, if necessary, we may arrange D so that both x and x2 are incident with a face of D[K] that is not F. This permits us to add xx2 to D without additional crossings, to obtain a 1-drawing of G. This final contradiction yields the result.	■
Along the same lines, we have the following limitation.
Lemma 15.35. Suppose G G M3 reduces by planar 3-reductions to the graph Gp4c that is a peripherally-j-connected extension of Gl4c. If Gl4c is a bicycle wheel
with axle xy and rim C, and there are four distinct vertices of C adjacent to both
x and y, then these are the only six vertices of Gl4c.
Proof. Suppose to the contrary that u1, u2, u3, and u4 are distinct vertices of C adjacent to both x and y in Gp4c and there is another vertex u5. We may choose the labelling of x and y so that xu5 G Gl4c. Let D be a 1-drawing of Gp4c — xu5.
Let K be the subgraph of Gp4c — xu5 consisting of C and all edges between x and vertices of C. (We do not include any chords of C that might exist in Gp4c.) If x and y are both in the same face of D[C], then y is in some face F of D[C] and at least two of u1, u2, u3, and u4 are not incident with F. This implies the contradiction that D has at least two crossings.
We conclude that y is not in the same face of D[C] with x. It follows that xy crosses C in D and this is the only crossing. We claim we can add the edge xu5 to D to obtain a 1-drawing of Gp4c.
Let F be the unique face of D[K] incident with both u5 and u and let C' be the cycle bounding F. If we cannot add xu5 in F, then there is an edge e of Gp4c that has an end in each of the two components of C' — {x, u5}. Since C' — x C C, it follows that both ends w1 and w2 of e are in C.
Since e is not an edge of Gl4c, there are vertices w3 and z of Gp4c so that z has just the neighbours w1, w2, and w3. Since both x and y have at least four
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00	neighbours, z G {x, y}. Since one of x and y is a neighbour of z, w3 G {x,y}.
Finally, z has at least two neighbours in C, so these are w1 and w2. We conclude that z = «5.
We note that xy cannot cross the 3-cycle u5w1w2 in D. Therefore, we can move w1w2 to the face of D[C] that contains y; in this new 1-drawing of Gp4c — xu5, x and u5 are incident with the same face, giving the contradiction that Gp4c has a 1-drawing.	■
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e	The final limitation is the following.
LEMMA 15.36. Suppose G G M2 reduces by planar 3-reductions to the graph Gp4c that is a peripherally-4-connected extension of Gl4c. Suppose Gl4c is a bicycle wheel with axle xy and rim C, and there are six distinct vertices x1, y1, x2, y2, x3, y3 in this cyclic order on C, so that, for i = 1, 2, 3,, xj is adjacent to x and yj is adjacent to y. Then these are the only six vertices of C.
We remark that we allow for the possibility that some (or all) of the xj are also
6	adjacent to y a„d, likewise some of the y ca„ be ^»t to ..
Proof. Suppose to the contrary that there is another vertex u in C. If possible, choose the xj, yj and u so that u is adjacent to only one of x and y. We may assume that u occurs between x1 and y1 in the cyclic order on C. By the choice of the xj, yj, and u, if u is adjacent to both x and y, then so are x1 and y1 and all vertices between them on C.
Let D be a 1-drawing of Gp4c — xu. Let K be the subgraph of Gp4c — xu consisting of C and all edges between x and vertices of C. (We do not include any CM	chords of C that might exist in Gp4c.) If x and y are on the same side of D[C],
00	then at most one of the yj is incident with the face of D[K] containing y, showing
CM	D has at least two crossings, a contradiction. Therefore, the crossing of D is of xy
CM	with an edge of C.
There is a face of D[K] incident with both x and u; let C' be its bounding cycle. If we cannot add xu to D, it is because there is an edge e of Gp4c — xu with an end in each of the components of C' — {x, u}. Since C' — x C C, it follows that CO	the ends w1 and w2 of e are both in C. Because Gp4c is a peripherally-4-connected
extension of a bicycle wheel, there are vertices z and w3 so that z has only the neighbours w1, w2, and w3.
Both x and y have at least four neighbours in Gl4c, so z G {x,y}; thus, z G C. Since z has two neighbours in C and at least one in {x,y}, it follows that w3 G {x, y}, while w1 and w2 are the two C-neighbours of z. Therefore, z = u. As u is adjacent to x, we conclude that u is not also adjacent to y. But now we can move the edge w1w3 to the other side of C so that the resulting 1-drawing of Gp4c — xu extends to a 1-drawing of G, a contradiction.	■
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Lemmas 15.34, 15.35, and 15.36 effectively limit the possibilities for Gl4c. Each of these must be checked for either having crossing number 1 or being 2-crossing-critical. Those with crossing number 1 must have their peripherally-4-connected extensions tested for 2-criticality. No matter what improvement is made to Lemma 15.34, this will require computer work to complete.
CASE 2: the 4-covered graphs.
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00	We begin our analysis by describing three particular internally-4-connected 2-
crossing-critical graphs that are 4-covered.
Definition 15.37. (1) The 3-cube Q3 is the 3-regular, 3-connected, planar, bipartite graph with 8 vertices.
S	(2) The graph Q3 is the bipartite graph obtained from Q3 by adding one new
vertex joined to all four vertices on one side of the bipartition of Q3.
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(3)	The graph Q3e is the bipartite graph obtained from Q3 by adding two of the four missing (bipartite-preserving) edges.
(4)	The graph Q3 is the graph obtained from Q3 by adding a 3-cycle abc on one side of the bipartition of Q3 together with one edge joining the fourth vertex d of the same part to the non-adjacent vertex in the other part of
00
the bipartition.
Lemma 15.38. The graphs QV, Q3e, and Q3 are all '¿-crossing-critical. Proof. We start with the following observation.
Claim 1. If D is a 1-drawing of Q3, then D is the unique planar embedding of Q3.
Proof. If e and f are two non-adjacent edges of Q3, then it is easy to see that they are in disjoint cycles. Therefore, no two edges of Q3 cross in D.	□
We use Claim 1 to show that cr(Q3) > 2, cr(Q;;e) > 2, and cr(Q3) > 2. Adding the one vertex to the planar embedding of the 3-cube yields 2 crossings, since each face of the 3-cube is incident with only 2 of the four vertices joined to the new vertex. This shows cr(QV) > 2.
For Q3e, each of the two new edges joins vertices not on the same face of Q3 and so each has a crossing with Q3. Thus, cr(Q3e) > 2.
For Q3, the new edge e incident with d must cross Q3 in any drawing D of Q3 for which D[Q3] has no crossings. If the 3-cycle D[abc] also has a crossing with Q3, then D has two crossings. Otherwise, D[abc] separates the two ends of D[e], so D[e] crosses D[abc]. Thus, cr(Q3) > 2.
We now consider 2-criticality in each case.
For QV, deleting any edge of the 3-cube makes a face incident with 3 of the four vertices and so yields a 1-drawing. Likewise deleting one of the edges incident with the new vertex yields a 1-drawing.
For Q3e, obviously deleting either of the edges not in Q3 yields a 1-drawing. On the other hand, if e is an edge of Q3 incident with at most one of the vertices of Q3e of degree 4, then deleting e makes one of the newly adjacent pairs now lie on the same face, yielding the required 1-drawing. If e is one the remaining two edges of Q3, there is a 1-drawing of Q3 — e with one crossing that extends to a 1-drawing
of Q2e - e.
3
For Q3, criticality of all the edges not incident with d is obvious, as it is the new edge e incident with d. The remaining three edges are symmetric. Deleting any one of these results in a subgraph that has crossing number 1 (we may move the other end of e to the other side of abc to get a 1-drawing).	■
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00	Lemma 15.39. Suppose G G M3 reduces by planar 3-reductions to a peripherally-
4-connected Gp4c with at least 8 vertices that is an extension of the internally-4-connected 4-covered graph Gl4c. Then either G is one of the graphs Q3, Q3e, or Gp4c has exactly 8 vertices.
Proof. Let a, b, c, d be the four vertices so that Gl4c — {a, b, c, d} is an independent set I. For each x G {a, b, c, d}, let X be the set of vertices in I adjacent to everything in {a, b, c, d} \ {x}, and let R be the remaining vertices in I; a vertex in R is joined o	to all of {a, b, c, d}.
Note that a vertex in R has degree 4 in Gl4c, so it is also a vertex of G; it cannot be the outcome of any 3-reductions. If |R| > 3, then G contains K34 and so G = K3 4, a contradiction. Thus, |R| < 2.
If, for some x G {a, b, c, d}, |X| > 2, then {a, b, c, d} \ {x} is a 3-cut in Gl4c that separates any two vertices v, w in X from all the other vertices in I \ {v, w}, of which there are at least two. This contradicts the fact that Gl4c is internally 4-connected. Thus, |X | < 1.
This implies that Gp4c has at most 10 vertices, but we can proceed a little further.
If R = 0, then Gl4c is planar (adding the K4 on {a, b, c, d} does not affect planarity), which is a contradiction. Thus, |R| > 0.
If, for each x G {a, b, c, d}, |X| = 1, then the bipartite subgraph of Gl4c consisting of {a, b, c, d} and the four vertices in A U B U C U D is the 3-dimensional cube Q3. Adding one of the vertices in R to Q3 produces Q3. That is, if all of A, B, C, and D are not empty, |R| = 1 and G = Q3. Thus, we may assume R = 0 and D = 0.
If |R| = 2, then for Gl4c to have at least 8 vertices, at least two of A, B, and C are not empty. Thus, Q2e C Gp4c, so Gp4c = Q2e.
In the final situation, we have |R| = 1 and, because Gp4c has at least 8 vertices, all of A, B, and C are not empty. In particular, Gp4c has exactly 8 vertices, as required.	■
A computer search can find all the peripherally-4-connected graphs having 8 vertices. These will include all the examples that are peripherally-4-connected extensions of internally-4-connected, 4-covered graphs having 8 vertices. This completes our analysis of 3-connected, 2-crossing-critical graphs with no subdivision of Vs.
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CHAPTER 16
Finiteness of 3-connected 2-crossing-critical graphs
with no V2n
This section is devoted to showing that, for each n > 3, there are only finitely many 3-connected 2-crossing-critical graphs that do not contain a subdivision of V2n. In particular, Theorem 16.14 asserts that if G has a subdivision of V2n but no subdivision of V2n+2, then |V(G)| = O(n3).
The finiteness has been proved previously by completely different methods in [13]. In our particular context, this shows that there are only finitely many 3-connected 2-crossing-critical graphs that have a subdivision of V8 but do not have a subdivision of V1o; these are the only ones missing from a complete determination of the 2-crossing-critical graphs.
The first subsection shows that, if G is a 3-connected 2-crossing-critical graph that does not contain a subdivision of V2n+2, then, for any V2n = H C G, each H-bridge in G has at most 88 vertices. The second subsection shows that, for a particular subdivision H of V2n, there are only O(n3) H-bridges having a vertex that is not an H-node. These easily combine to give the O(n3) bound of Theorem 16.14.
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16.1. V2n-bridges are small
The main result of this subsection is to show that if G G and V2n = H C G, then any H-bridge B is a tree with a bounded number of leaves, so that | V(B) | < 88. In the next subsection, we show that there are only O(n3) non-trivial H-bridges.
S	The next lemma will have as a corollary the first main result of this subsection.
„
Lemma 16.1. Let G G M2, V2n = H C G, n > 3, and B an H-bridge. Then |att(B)| < 11n +12.
Proof. Let e be an edge of B incident with x G att(B) and y G Nuc(B). Then De[B — e] is contained in a face F of De[H]. Because we know the 1-drawings of V2n, we know that each face of De[H] is incident with at most n +1 H-branches. Moreover, B — e is an H-bridge in G — e and attG_e(B — e) is either attG(B) or attG(B) \ {x}.
If B has at least 11(n + 1) + 2 attachments, then some H-branch b contains at least 12 attachments of B — e. Let a1... a12 be any 12 distinct attachments of B — e occurring in this order in b. Let T C B be a minimal tree that meets att(B) at a1, a3, a4, a6, a7, ag, a1o, and a12, so that these aj are the leaves of T, and let Q = [a1,b,a12]. Set Y = T U Q.
For i = 1,4, 7,10, there is a unique cycle Cj C Y that meets b precisely in ajQaj+2. Let I C {1,4, 7,10} be the subset such that, for i G I, x G C; clearly
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For each i G I, let Mj be the Cj-bridge in G — e with H C Mj U Cj. As x G Cj, x G Nuc(Mj). Let Bj be the Cj-bridge in G — e containing y or Bj = y if y G Cj. Let Pj be a minimal subpath of Cj containing Bj n Cj, so that ajQaj+2 C Pj.
Claim 1. Let i, j, k g I be distinct. If y G Mj U M,, then:
Bj = ;
•	P = Pj C Cj n Cj; and
•	y G Mk.
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Proof. If « and v are vertices in Cj n Cj, then « and v are not in b and there is a unique «v-path P in T. We note that P C Cj n Cj. Thus, Cj n Cj is a path.
If there were a yCj-path disjoint from Cj, then y G Mj, a contradiction. Therefore, every yCj-path meets Cj and, symmetrically, every yCj-path meets Cj. Thus, every y(Cj U Cj)-path has one end in Cj n Cj. It follows that if y G Cj U Cj, then y G Cj n Cj, so in this case Bj = Bj = ||y||.
In the case y G Cj U Cj, let B be the (Cj U Cj)-bridge containing y. The preceding paragraphs show that att(B) C Cj n Cj, so that in fact B is also both a Cj- and a Cj-bridge. In particular, Bj = Bj = B.
For the last part, we assume y G Mk and note that B = Bj = Bj = Bk and Cj nCj nCk is a non-null path P'. If P' has length at least one, then P' UCj UCj UCk contains a subdivision of K2 3 and yet has all three of the vertices on one side incident with a common face, which is impossible. Therefore, P' consists of a single vertex z.
If z is not y, B has only z as an attachment in G — e. It follows that either z or {z,x} is a cut-set of G, contradicting the fact that G is 3-connected. Thus, z = y, and so, for some t G {i,j, k}, y is an attachment of Mt; in particular, y G Mt, a CM	contradiction.	□
By Claim 1, there is an i G I such that y G Mj. For such an i, set C = Cj and note that x G Mj — att(Mj), so that M = Mj + e is a C-bridge in G. Furthermore, attG(M) = attG-e(M — e).
Notice that De[C] is clean, since the crossing of De is between disjoint H-branches. Thus, C has BOD in G — e. Also, any C-bridge B' = M has C U B' planar. As attG(M) = attG-e(M — e), C has BOD in G.
Recall that the H-bridge B has aj, aj+1, and aj+2 as attachments. For any vertex « of B not in b, there is an H-avoiding «a^-path, whose edge e' incident with « is in some C-bridge B'. Since x and y are on the same side of De[C], M is contained on that side of De[C] and e' is on the other side. Therefore, B' = M.
In Dei, the crossing is in H and De> [C] is clean. That is, De> [C U M] is a 1-drawing with C clean. Corollary 4.7 shows cr(G) < 1, the final contradiction. ■
The following corollary is the first main result of this section.
Corollary 16.2. Let G e M2, V2n = H C G, n > 3, B an H-bridge. Then |att(B)| < 45.
Proof. If n = 3, then the result is an immediate consequence of Lemma 16.1. Thus, we may assume n > 4. If B has attachments in the interiors of non-consecutive spokes, then G is the Petersen graph and the result clearly holds.
Otherwise, B has attachments in at most two consecutive spokes. Thus, there is a subdivision H' of V6 contained in H that contains all the attachments of B. Applying Lemma 16.1 to H', we again see that |att(B)| < 45.	■
o
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00	We now turn to the other half of the argument that bounds the number of
vertices in an H-bridge, namely, that the bridge is a tree. We need a new notion.
Definition 16.3. Let T* be a graph consisting of subdivision of a K23 together with three pendant edges, one incident with each of the three degree 2 vertices in the K23. A tripod is any graph T obtained from T* by contracting any subset of the pendant edges; if all three pendant edges are contracted, then an edge is added between the two copies of K13, but not having a vertex of contraction as an end — this may be done in any of three essentially different ways. The attachments of CD	the tripod are the degree 1 and 2 vertices in T.
We are now ready for the second half of the main result of this section.
LEMMA 16.4. Suppose G € M2, V2n = H C G, n > 3, G has no subdivision of V2(n+1), and B is an H-bridge. Then either B is a tree or B has a tripod, n = 3 and |V(G)| < 10.
Proof. By way of contradiction, suppose B has a cycle C. If att(B) n C = 0, let e be an edge of C incident with u € att(B). If C n att(B) = 0, then let e be any edge of C. The choice of e shows that B — e is an H-bridge in G — e and that attG-e(B — e) = attG(B). Since De[H] contains the crossing in De[G — e], De[B — e] is contained in a face F of De[H].
Let C' = dFx, so C' is a cycle in G' = (G — e)x. Since G' is planar, C' has BOD in G' and C' U B' is planar for each C'-bridge B' in G'. If C' U B were planar, then G' + e would be planar, in which case cr(G) < 1, a contradiction. Therefore, C' U B is not planar.
We now introduce a convenient notion.
CM
Claim 1. (C' U B)4 is 3-connected.
Proof. Let L = (C' U B)4. If |V(Nuc(B))| = 1, then L is a wheel and the claim follows. So assume |V(Nuc(B))| > 2. We show that any two vertices of L
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are joined by three internally disjoint paths. For u, w € Nuc(B), this is true in G,
CO	so let P1, P2, P3 be such paths in G. If at least one Pj is contained in B — C', then
we can easily modify the others to use C' rather than G — B to get three paths in L. If all three intersect Ce, then B n (P1 U P2 U P3) is two claws YU and YW. There is a YUYW-path in Nuc(B), which returns us to the previous case.
If u € Nuc(B) and w € C', then w is an attachment of B. Let Y be a claw in B with centre u and talons on C'. Using a C'-avoiding wY-path in B, if necessary, we can assume w is a talon of Y. It is then easy to use C' to extend the other two paths in Y to w.
Finally, if u, w € C', then both u and w are attachments of B, so there is a C'-avoiding path joining them. This path and the two uw-paths in C' yield the required three paths.	□
Definition 16.6. Let C be a cycle in a graph G and let P1 and P2 be disjoint C-avoiding paths in G. Then P1 and P2 are C-skew paths if the two C-bridges in C U P1 U P2 overlap.
Definition 16.5. Let G be a graph. The graph G4 is the graph whose vertices are the G-nodes and whose edges are the G-branches.
t
As C' U B has no planar embedding, [25] implies B has either a tripod whose attachments are in C' or two C'-skew paths.
Claim 2. If B has a tripod T, then n = 3, G = H U T and |V(G)| < 14.
Proof. Let S be the attachments of T. As H U T is 2-connected and, relative to the cut S, both H' + (taking H' to be any V6 containing S) and T+ are non-planar. By Theorem 15.6, cr(H' U T) > 2. Thus, G = H' U T, so n = 3 and, again by Theorem 15.6, |V(G)| < 10.	□
u
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Thus, we can assume B has no tripod. Then B has C'-skew paths, say P1 and P2. Since these do not exist in B — e, e is in one of them. If C n att(B) = 0, choose e' any edge of C not in Pi U P2. If C n att(B) = 0, choose e' to be the other edge of C incident with the same attachment as e.
Repeat with G — e'. This yields C'' so that B has C''-skew paths ulU and wiw2 (e' incident with ui). Since m1m2 U wiw2 C B — e', they are not C''-skew. In C', we have the cyclic order u1, w1, u2, w2, say. In C'' we have u1 u2w1w2. Likewise
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in C' we have u1 u2w1 w2, while in C" we have u1w1 u2w2.
Let D and D' be 1-drawings of H having all attachments of B on faces F, F', respectively, so that the cyclic orders of att(B) are different in dF and dF'.
Claim 3. n > 4.
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Proof. Let H be a subdivision of V6 in G. We remark that if f and f' are any disjoint H-branches having internal vertices that are ends of an H-avoiding path P in G, then H U P is a subdivision of V8 in G.
We consider first the case that att(B) is not contained in any 4-cycle of H. Because we know the 1-drawings of H and att(B) is contained in the boundary dF
CM
of a face F of such a 1-drawing, dF is xv1v2v3x. If B has attachments in both (xv1) and (v3x), then G has a subdivision of V8, as required. Thus, we may assume that att(B) is contained in a 4-cycle Q of H, which we may take to be [v1v2v3v4v1].
In at least one of D and D', Q is self-crossed (otherwise the cyclic orders of att(B) are the same) and B is drawn in the face xv1v6v3x. However, in this case att(B) C (x,v1] U [v3, x) and at least two attachments of B are in each. In this case, we again have a subdivision of V8 in G, as required.	□
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Claim 4. B has no (interior) spoke attachment.
Proof. From Claim 3, we know that n > 4. By way of contradiction, we assume B has an attachment in (s0). From the listing of the faces of 1-drawings of V2n, the only possibilities for each of dF and dF' are:
:	(1) [vo,ro,V1,S1,v„+1,r„,v„,so,vo];
:	(1') [vo,r_1,v_1,s_1,vn_1,r„_1, v„, so, vo];
:	(2) (v1,ro,vo,so,v„,r„,vn+1,rn+1,vn+2);
:	(2') (v_1,r_1,vo,so,v„,r„_1,v„_1,r„_2,v„_2);
:	(3) (v„_1,r„_1, v„, so,vo,r_1,v_1,r_2,v_2);
: (3') (v„+1,r„,v„, so,vo,ro,v1,r1,v2];
: (4) (v_1 ,r_1, vo, so,v„,r„,v„+1); : (4') (v„_1,r„_1, v„, so,vo,ro,V1); : (5) [vo,v1,v2,...,v„,so,vo]; : (5') [vo,so,v„,v„+1,v„+2,...,v_1 ,vo].
We now consider these possibilities in pairs. In every case, the ends of the skew paths will occur in the same cyclic order on the boundaries of the two faces, which is impossible.
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1,1') att(B) Ç s0 2,2') att(ß) Ç s0 3,3') att(ß) Ç s0 4,4') att(ß) Ç s0 5,5') att(ß) Ç s0
same cyclic order, a contradiction. same cyclic order, a contradiction. same cyclic order, a contradiction. same cyclic order, a contradiction. same cyclic order, a contradiction.
1.2)	att(B) Ç (v1,r0,v0, s0,vn,rn, vn + 1]; same cyclic order, a contradiction. 1,2') att(B) Ç [v0,s0,vn]; same cyclic order, a contradiction.
1.3)	att(B) Ç s0; same cyclic order, a contradiction.
1,3') att(B) Ç (vn+1, rn, vn, s0, v0, r0, v1]; same cyclic order, a contradiction.
1.4)	att(B) Ç (v1,r0,v0, s0,vn]; same cyclic order, a contradiction. 1,4') att(B) Ç [v1,r0,v0,s0,vn]; same cyclic order, a contradiction.
1.5)	att(B) Ç [v1 ,r0, v0, s0,vn]; same cyclic order, a contradiction. 1,5') att(B) Ç [vn+1, rn, vn, s0,v0]; same cyclic order, a contradiction.
2.3)	att(B) Ç s0; same cyclic order, a contradiction.
2,3') att(B) Ç (vn+1, rn, vn, s0, v0, r0, v1]; same cyclic order, a contradiction.
2.4)	att(B) Ç (vn+1,rn,vn, s0,v0]; same cyclic order, a contradiction. 2,4') att(B) Ç (v1, r0, v0, s0, rn]; same cyclic order, a contradiction.
2.5)	att(B) Ç (v1,r0,v0, s0,vn]; same cyclic order, a contradiction. 2,5') att(B) Ç [v0, s0, vn, rn, vn+1, rn+1, vn+2); same cyclic order, a contradiction.
3.4)	att(B) Ç (v_1, r_1, v0, s0,vn]; same cyclic order, a contradiction. 3,4') att(B) Ç (vn_1, rn_1, vn,s0,v0]; same cyclic order, a contradiction.
3.5)	att(B) Ç (vn_1, rn_1, vn, S0,v0]; same cyclic order, a contradiction. 3,5') att(B) Ç (v_2, r_2, v_1, r_1, V0, s0,vn]; same cyclic order, a contradiction. 4,5) att(B) Ç [v0,s0,vn]; same cyclic order, a contradiction. 4,5') att(B) Ç (v_1,r_1,v0, s0, vn, rn, vn+1); same cyclic order, a contradiction.
As any pair gives the same cyclic order, we always get a contradiction. □
Claim 5. B is not a local H-bridge.
Proof. Suppose B is local, with att(B) Ç Q0. From Claims 3 and 4, we may assume n > 4 and B has no spoke attachment. Thus, att(B) Ç r0 U rn. Moreover, B cannot have attachments in both (r0) and (rn) because G has no subdivision of V2(n+1). On the other hand, B has at least two attachments in both r0 and rn or else the cyclic order of the ends of the skew paths is always the same. So we may assume att(B) n r0 = {v0, v1}. We need two attachments in rn. From the listing of faces in 1-drawings of V2n, the only possibilities for dF and dF occur when Q0 is not self-crossed and so the cyclic orders of the attachments of B are the same in both cases, a contradiction.	□
Claim 6. For some ¿, att(B) Ç r U ri+n+1.
Proof. By Claims 3, 4, and 5, n > 4, B has no spoke attachments, and B is not local.
We consider in turn the possibilities for the face of De[H] that contains B — e. We know B is not local, so it can only be contained in a face whose boundary has one of the following forms:
(1)	[x,rj,vj, Si,vi+„,ri+„_1, x];
(2)	[x,ri,vi,ri_1,vi_1,si_1,v„+i_1,r„+i_1, x];
(3)	[x,ri, vi+1,ri+2, .. ., vi+„_1, ri+„_1, x];
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CM
00
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m
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(4)	[vi,si,v„+i,rn+i,v„+i+i,...,ri_i,vi]; or
(5)	[x,rj, vi+i,ri+i, .. ., r„+i_i, v„+i, r„+i, x].
As in the proof of Claim 4, the faces of De [H] and De/ [H] containing B — e CD	and B — e', respectively, cannot both be of one of the types (3, 4, 5): the vertices
of att(B) will occur in the same order in both cases.
If one of the drawings has B — e or B — e' in a face of type (1), then we are done: att(B) C r U ri+n_1. The remaining case is that one of the drawings has B — e or B — e' drawn in a face of type (2).
All other possibilities having been eliminated, we may assume (taking i = n+1)
att(B) C [x,ri, vi,ro, vo, so,v„,r„, x] .
Because B is not local, att(B) n (rL) = 0. Because att(B) occurs in different orders in dF and dF', att(B) n rn = 0. By way of contradiction, we suppose B also has an attachment in [v0,r0,vL). The only other face which could allow these three attachments is [x, r0, vL, rL,..., vi_1, ri_1, vn, rn, x]. Notice v0 is not in this second boundary, so one attachment is in (r0). Because V2(n+i) C G, no attachment is in (rn). Thus att(B) n rn = {vn}. But then, once again, the attachments of B occur in the same cyclic orders in dF and dF', a contradiction.	□
As we have seen above, the alternative to "B is neither a tree nor contains a tripod" is that B has the C'-skew paths Pl and P2, as well as the C''-skew paths Pl' and P2'. Claim 6 shows the four ends of Pl and P2 are in r0 U rn+1. If three of them are in r0, say, then they occur in the same cyclic order in dF and dF', a contradiction. So two are in r0 and two in rn+1. If Pl has both ends in r0, say, then the ends of Pi and P2 can never interlace, a contradiction as they interlace in dF. So each has one end in each of r0 and rn+1. Likewise for Pl, P2'.
Adding at most 3 paths in B — att(B) to Pl U P2 U Pl U P2', we obtain B' C B containing Pl U P2 U Pl U P2' so that B' is an H-bridge in H U B'.
Recall that n > 4 by Claim 3. All the attachments of B' are in H — (s3). Suppose D'' is a 1-drawing of (H U B') — (s3). Then D''[B'] is in a face F'' of D''[H — (s3)]. Since r0 and rn+1 both have at least two attachments of B', they are both incident with F''. Thus one of the pairs Pl,P2 and Pl,P2' is a dF''-skew pair. Therefore, cr((H U B') — (s3)) > 2, contradicting the fact that G is 2-crossing-critical.	■
Combining Corollary 16.2 and Lemma 16.4, we immediately have the main result of this section.
Theorem 16.7. Let G e M\,
V2n = H C G, n > 3, and suppose G has no subdivision of V2(n+i). If B is an H-bridge, then |V(B)| < 88.
This completes the first main step of our effort to show that 3-connected, 2-crossing-critical graphs with no subdivision of V2n have bounded size.
16.2. The number of bridges is bounded
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This subsection, the final leg of this work, is devoted to showing that there is a particular subdivision H of V2n in G so that there are at most O(n3) H-bridges in G that have a vertex that is not an H-node. Theorem 16.7 shows that, for any V2n = H C G, all H-bridges have at most 88 vertices (when there is no subdivision of V2(n+1)). The combination easily implies G has at most O(n3) vertices.
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00	Definition 16.8. Let G be a graph and let n be an integer, n > 3. A subdivi-
sion H of V2n in G is smooth if, whenever B is an H-bridge with all its attachments in the same H-branch, B is just an edge that is in a digon with an edge of H.
We begin by showing that every G G M3 with a subdivision V2n has a smooth subdivision H of V2n. For such an H, every vertex of G either is an H-node or is in an H-bridge that does not have all its attachments in the same H-branch. So it will be enough to show that the number of these H-bridges is O(n3). o	This analysis is completed in three parts. We start with the result that there
e	are not many H-bridges having an attachment in a particular vertex of H and an
attachment in the interior of some H-branch. This is useful for H-bridges having both node and branch attachments, but is also used in the second part, which is to bound the number of H-bridges having attachments in the interiors of the same two H-branches. The final part puts these together with those H-bridges having attachments in three or more H-nodes.
We start by showing that every G G M3 with a subdivision of V2n has a smooth subdivision of V2n.
LEMMA 16.9. Let G G Mf and suppose G contains a subdivision of V2n, with n > 3. Then G has a smooth subdivision of V2n.
Proof. Choose H to be a subdivision of V2n in G that minimizes the number of edges of G that are in H. We claim H is smooth.
To this end, let B be an H-bridge with all attachments in the same H-branch b and let P be a minimal subpath of b containing att(B). Set K = B U P and notice that K is both H-close and 2-connected. By Lemma 5.13, K is a cycle, so B is just a path and, since G is 3-connected, just an edge. It remains to prove that P is just an edge as well.
Let H' = (H U B) — (P}. Evidently H' is a subdivision of V2n in G and |E(H')| = |E(H)|-|E(P)l + 1. Since |E(H)| < |E(H')| by the choice of H, we see that |E(P)| < 1, and, therefore, P is just an edge, as required.	■
We now turn our attention to the H-bridges of a smooth subdivision H of V2n. There are three main steps.
Step 1: Bridges attaching to a particular vertex and branch.
The first step in bounding the number of H-bridges is to bound the number of them that can have an attachment at a particular vertex of H and in the interior of a particular H-branch. This is the content of this step.
Lemma 16.10. Let G g Mf,
V2n = H C G, n > 3 and suppose H is smooth. For a vertex (not necesssarily a node) u of H and an H-branch b, there are at most 41 H-bridges with an attachment at u and an attachment in (b) — u.
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Proof. Suppose there are 42 such H-bridges. Let B0 be one of them, let e G E(Bo) and let D be a 1-drawing of G — e. If u G (b), then at most 4 faces of D[H] are incident with (b), and therefore at least 11 of these H-bridges (other than B0) are in the same face F of D[H]. If u G (b), then precisely two faces of D[H] are incident with u, so at least 21 of these bridges are in the same face F of D[H] and of these at least 11 have an attachment in the same component of D [b — u] n (dF)x. In both cases, let B be the set of 11 bridges, contained in F, having u as an
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c
00	attachment and an attachment in the same component b' of D[b — u] n (dF)x. As
D[(dF)x U (UBeBB)] is planar with (dF)x bounding a face, no two (dF)x-bridges in B overlap.
Let P = b' and Q = (dF)x — (P). Lemma 4.8 applies to (dF)x, P, Q, B. As there are no digons disjoint from H, there is a unique (up to inversion) ordering Bi,... ,Bii of B so that P = PBl .. .PBl1 and Q = QSl .. . QSll.
Because u G QBl nQB2 n — nQBl1 and the QBi are internally disjoint subpaths of Q, all of QB2,..., QBl0 are just u. For i = 1,..., 11, let aj and aj be the ends of Pj, so that P =(..., a2,. .., a2, .. ., a3, .. ., a3, .. ., aio,.. ., aio,. ..).
Claim 1. For i g {2,..., 9}, aj = aj+i.
Proof. Otherwise, aj = aj = aj+i = aj+i, implying that Bj and Bj+i constitute a digon disjoint from H, which is impossible.	□
For i, j G {2, 3,..., 10} with i < j, set Kjj = (Uk,=jBk) U ajPaj.
Claim 2. For i, j G {2,..., 10} with i < j, Kj is 2-connected.
Proof. Let Rj be an H-avoiding uaj-path in Bj, and Rj an H-avoiding uaj-path in Bj. Then Cjj := Rj U Rj U ajPaj C Kjj is a cycle containing u and
aj Paj.
For x G Bk, i < k < j, x G H, for any H-node w = u, G has 3 internally disjoint xw-paths; at least two of these leave Bk in akPa'k, and so no cut vertex of Kjj separates x from Cjj.	□
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Since b' is not crossed in D, D[Kj j+2] is clean and is contained in F U dF. There is a unique face Fj of D[Kj j+2] so that Fj C F; since Kj j+2 is 2-connected, Fj is bounded by a cycle Cj. As D[KM+2] C F U dF, dF C Fj U dFj. As D[u] G dF n D[KM+2], D[u] G dFj. Likewise D[ajPaj+2] C dFj.
Thus, u G Cj and ajPaj+2 C Cj. Therefore, Cj n H is u and ajPaj+2, from which we deduce that there is a Cj-bridge Mj so that H C Cj U Mj. Observe that Bj+i is a Cj-bridge different from Mj.
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For i = 2, 5, 8, let ej be an edge of Bj+i incident with u, and let Dj be a 1-drawing of G — ej.
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Claim 3. For i g {2,5, 8}, Cj has BOD in G and Dj[Cj] is not clean.
Proof. At most one of D2[Cj], i G {2,5, 8} is crossed, so for at least one i G {5, 8}, De[Cj] is clean. It follows that Cj has BOD in G — e.
By Claim 1, a3 = aj, whence B3 C Mj, and B3 — e C Mj — e. Furthermore, u G H, so u G att(Mj — e). Thus attG_e(Mj — e) = attG(Mj) and Mj — e is a Cj-bridge in G — e. We conclude that the overlap diagrams for Cj in G — e and G are isomorphic and, therefore, Cj has BOD in G.
We now show that all three Cj, j G {2, 5, 8}, have BOD in G. If Dj[Cj] is clean, then Dj[CjUMj] is a 1-drawing of CjUMj, implying via Corollary 4.7 that cr(G) < 1, a contradiction. So Dj[Cj] is not clean, and, therefore, for j G {2, 5, 8} \ {i}, Dj[Cj] is clean. Thus, Cj has BOD in G — ej, and, following the argument above for Cj, we deduce that Cj has BOD in G.	□
Claim 4. For i G {2, 5, 8}, one face of Dj[Cj] contains all H-nodes, other than (possibly) u.
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00
u
a
CD U
00	Proof. Let e' be the edge of H so that Dj[e'] crosses Dj[ajba-+2] and let b'
be the H-branch containing ej. If n = 3, let R be a hexagon in H containing b and b'. For n > 4, both b and b' are in the rim R of H.
Since b and b' are disjoint, for n > 3, R — ((b) U (b'}) has two components, each with at least two nodes of H. Either of these with < n nodes has all its nodes adjacent by spokes to the other component. Obviously, there is at least one such.
Observe that if A is any path in R — ((b) U (b'}) such that [A] has a vertex in each face of [C], then u G V(A) and the two paths P, P' in A having u as an end are such that [P] and Dj [P'] are in different faces of [Cj].
Let K be a component of R — ((b) U (b'}) not containing u and let L be the other. Then Dj [K] is in the closure of a face F, of Dj [C]. We claim that Dj [L] C Fj U{u}.
Any H-node w in L that is joined by a spoke to an H-node w- in K has D [w] C F; U D [u], since otherwise [ww'] crosses [Cj].
If there is an H-node w in L that is not adjacent by a spoke to any vertex in K, then w is adjacent by a spoke to another H-node w' in L and, moreover, w and w' are the first and last nodes of L. As [ww'] is disjoint from [Cj], we deduce that there is a face F of [Cj] so that [w] and [w'] are both in F U [u]. Therefore, [L] is contained in that face. As at least one H-node in L is adjacent by a spoke to an H-node in K, we conclude that [L] C F; U [u].	□
Let F, be the face of [C] containing all the H-nodes and let Fj be the other face of Dj [Cj].
Claim 5. For i g {2,5, 8}, the crossing in D.j is not in (a.j +1, b, aj+1).
CM
Proof. Suppose by way of contradiction that ej is an edge of G — ej so that D.j [ej] crosses (a.j+1,b, a'+1). Clearly, aj+1 = a'+1. Since H — (b) is 2-connected, there is a cycle C' C H containing e'. Let P be an H-avoiding aj+1aj+1-path in B; +1 and let C be the cycle PU[aj+1, b, a'+1 ]. Then C and C' are graph-theoretically disjoint and [C] n [C'] contains the crossing of . But then [C] and [C'] must cross a second time, a contradiction.	□
Claim 6. The only C;-bridge that overlaps B; +1 is .
CO
Proof. Let B be a C-bridge different from overlapping B;+1. Then att(B) C [ajba'+2] U {u}. As H is smooth, u G att(B). We claim both Bj+1 and B overlap .
By Claim 1, aj = a'+1, so B;+1 either has an attachment in (aj,a'+^ or it has both aj and a'+2 as attachments. In either case, B;+1 overlaps (which has attachments at u, aj, a'+2).
Likewise B either has two attachments in [aj, a'+2] or at least one attachment in (a.j +1,a'+^ C (a.j,a'+2), so B overlaps Mj. But now B.j+1, B.j, and Mj make a triangle in OD(C;), contradicting Claim 3.	□
CD
Let b' be the H-branch that crosses C; in and let x be the H-node so that the crossing is in [x, b', u].
Claim 7. Let L be the graph [D;[G —ej]n(cl(F'))]x UB;+1. Then the C-bridge containing [x,b', u] overlaps B;+1 in L.
Proof. If L embeds in the plane with C; bounding a face, then this embedding combines with restricted to the closure of F to yield a 1-drawing of G, which
o
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00	is impossible. As each individual Cj-bridge B in L has Cj U B planar, there are
overlapping Cj-bridges in L.
By definition, L is planar with all Cj-bridges other than Bj+1 on the same side of Cj. Therefore Bj+1 overlaps some other Cj-bridge in L. By Claim 6, this is not any Cj-bridge other than Dj[Mj]x n Dj[L], that is, the one containing [x, b', u]. □
Step 2: H-bridges joining interiors of disjoint H-branches.
By Claim 4, [aj, b, aj+2] — x has a component A containing att(Bj+1) — u. Let CD	z be the one of aj and aj+2 that is an end of A and let Q be the minimal subpath
of A containing all of z, aj+1 , aj+1. By Claim 7, Mj has an attachment wj € [zQ) and an H-avoiding path Qj from wj to a vertex xj € (x,b', u). Notice that, if j € {2, 5, 8} \ {i}, then Qj n C = 0.
There are at most two H-branches (or subpaths thereof) incident with u that can cross b. Thus for some i,j € {2, 5, 8}, bj = bj. Choose the labelling so that xj is no further in bj from u than Xj is. Since xbju contains xj, Dj[xj] C Fj but Dj [wj] C Fj. Since Qj n Cj = 0, Dj [Qj] crosses Cj, the final contradiction. ■
i—l
The other steps in the argument are to show that a smooth subdivision H of V2n in G has few bridges with attachments in the interiors of distinct H-branches. There are two parts to this: either the branches do or do not have a node in common. We first deal with the latter case.
o
Lemma 16.11. Let G € M\, V2n = H C G, n > 3, H smooth and suppose G has no subdivision of V2(n+1). If b1,b2 are disjoint H-branches, then there are at
CM	most 164n + 9 H-bridges having attachments in both (b1) and (b2).
00
Proof. Suppose there is a set B of 164n + 10 H-bridges having attachments in both (b1) and (b2). Let B0 € B and let e € B0. In De, at most 4 faces are incident with (b1), so there is a set B' consisting of 41n + 3 elements of B\{B0} in the same face of De[H]. By Lemma 4.8, there is a unique ordering (B1,..., B41n+3) of the elements of B' so they appear in this order in both (b1) and (b2). It follows that B2,..., B41n+2 have all attachments in (b1) U (b2). By Lemmas 4.8 and 16.10, Bj and Bj+41 are totally disjoint. So there are n + 1 totally disjoint (b1) (b2)-paths with their ends having the same relative orders on both.
We aim to use these disjoint paths to find a subdivision of V2(n+1) in G. We need the following new notion.
Definition 16.12. Let e = uw and f = xy be edges in a graph G. Two cycles C and C' in G are ef-twisting if C = (u, e, w,..., x, f, y,...) and C' = (u, e, w,..., y, f, x,...), i.e., C and C' traverse the edges e and f in opposite ways.
We note that V6 has edge-twisting cycles: if e = uw and f = xy are disjoint edges in V6, with u, x not adjacent, then the 4-cycle (u,w,x,y, u) and the 6-cycle (u, w, z, y, x, z', u) are ef-twisting.
Next suppose n > 4. There are three possibilities for b1 and b2. : Case 1: Both b1 and b2 are in R. We may assume without loss of generality (recall that b1 and b2 are not adjacent) that b1 = r0, b2 = rj,2 < i < n. Set H' = R U s0 U s1 U s2, so H' = V6. Then b1 and b2 are in disjoint H'-branches and so H', and therefore H, contains b1b2-twisting cycles.
o
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00	: Case 2: One is in R, the other is a spoke. We may assume without loss of
generality that b1 = r0, b2 = sj, i G {0,1}. Set H' = RU s0 U s1 U sj. Then b1 and b2 are in disjoint H'-branches, so H', and therefore H, contains b1 b2-twisting cycles.
: Case 3: Both b1 and b2 are spokes. We may assume without any loss of generality that b1 = s0, b2 = sj. Then there exists j G {0,... ,n — 1} \ {0, i}. Set H' = R U s0 U sj U Sj. Then b1 and b2 are in disjoint H'-branches and so H', and therefore H, contains b1 b2-twisting cycles. Choose the cycle C in the twisting pair in H for b1 and b2 so that C traverses b1 and b2 in order so that the ends wj of the n +1 disjoint paths occur in C as «1, «2,..., «„+1,..., w1,..., wn+1. Then C and these paths are a subdivision of V2(n+1) in G, contradicting the assumption that G has no subdivision of V2(n+1).
C^	■
Next is the third and final consideration.
Step 3: H-bridges joining interiors of H-branches having a common node.
Lemma 16.13. Let G g M\, V2n = H C G, n > 3, and let b1,b2 be adjacent H-branches. Then at most 2 H-bridges have attachments in both (b^ and (b2).
Proof. By way of contradiction, suppose there is a set {B1,B2,B3} of 3 such H-bridges. For each i G {1, 2, 3}, let ej G B.. There is precisely one face P., of a 1-drawing D. of G — e., that is incident with both (b1) and (b2). Thus, for each Bj, j = i, Dj[Bj] C P.. Clearly for {j,k} = {1, 2, 3} \ {i}, B^ and Bk do not overlap on P.. In particular, their attachments in b1 and b2 are in the same order as we
CM
traverse them from their common end Thus we may assume B1, B2, B3 appear 00	in this order from « on both b1 and b2 .
Notice that att(B3) = att(B2). Therefore, there is a cycle C C B2 U b1 U b2 consisting of a (b1) (b2)-path in B2 and a subpath of b1 U b2 containing such that C does not contain some attachment w of B3. Reselect e3 G B3 to be incident with w. Let MC be the C-bridge so that H C C U MC.
CO
Then w G Nuc(Mc), so B3 C Mc. Furthermore, if e3 is incident with an CO	attachment x of MC, then x is contained in R. In particular, it is incident with
another edge of MC. Thus, MC — e3 is a C-bridge in G — e3 having the same attachments as MC has in G. Because C is H-close, D1[C] is clean; furthermore, D1[C U MC] is a 1-drawing of C U MC. Since D3[C] is also clean, C has BOD in G — e3 and hence in G. Corollary 4.7 implies the contradiction that cr(G) < 1. ■
We end this section with the asserted finiteness of 3-connected 2-crossing-cri-tical graphs with no subdivision of V2n+2.
Theorem 16.14. Suppose G G M3 and there is an n > 3 so that G has a subdivision of V^n, but no subdivision of V2(n+1). Then |V(G)| = O(n3).
Proof. By Lemma 16.9, G has a smooth subdivision H of V2n. We may assume no H-bridge contains a tripod, as otherwise |V(G)| < 14 by Lemma 16.4.
We first claim that a vertex u of H that is not an H-node is an attachment of some H-bridge B not having all its attachments in the same H-branch. Since u has degree 2 in H and degree greater than 2 in G, u is an attachment of some H-bridge. Because H is smooth, an H-bridge that has all its attachments in the
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CO	same H-branch is an edge in a digon. If all the H-bridges attaching at u are such
edges, then u has only two neighbours and G is not 3-connected, a contradiction.
Thus, every vertex of G is either an H-node or is in some H-bridge that does not have all its attachments in the same H-branch. We bound the number of these H-bridges as follows.
We claim that, for any three H-nodes u,v,w, at most two H-bridges have all three of u,v,w as attachments. To see this, suppose three nontrivial H-bridges Bj,i = 1, 2, 3, all have all of u, v, w as attachments. Each Bj contains a claw Yj having u, v, w as talons. Then Y1 U Y2 U Y3 U H contains a subdivision of K3,4, in which case 2-criticality implies G is K3,4. Thus, at most two H-bridges have attachments in any three nodes. So there are at most 2(nontrivial H-bridges with only node attachments.
Every other H-bridge of concern has an attachment in the interior of some H-branch and at some vertex of H not in that H-branch. Lemma 16.10 implies that there are at most (2n)(3n)41 H-bridges with an attachment in an H-node and in an open H-branch.
Lemma 16.11 implies there are at most ((32") — 6n)(164n + 9) H-bridges having attachments in the interiors of disjoint H-branches.
Lemma 16.13 implies there are at most 2 H-bridges with attachments on two given adjacent H-branches and so there are at most 6n(2) H-bridges with attachments on two adjacent H-branches.
Every H-bridge has at most 88 vertices, and every vertex of G is either an H-node or in one of these enumerated H-bridges. Therefore,
|V(G)| < 88 < 2 +2n • 3n • 41 + 6n(2) +
2n
3
3n 6n 2 — 6n
164n + 9
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CHAPTER 17
Summary
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This short section provides a single theorem and some remarks summarizing the current state of knowledge about 2-crossing-critical graphs.
Theorem 17.1 (Classification of 2-crossing-critical graphs). Let G be a 2-crossing-critical graph.
(1)	Then G has minimum degree at least two and is a subdivision of a 2-cros-sing-critical graph with minimum degree at least three.
Thus, we henceforth assume G has minimum degree at least three.
(2)	If G is 3-connected and contains a subdivision of V10, then G e T(S) (Definition 2.12). That is, G is a twisted circular sequence of tiles, each tile being one of the 42 elements of S (Definition 2.10).
(3)	If G is 3-connected and does not have a subdivision of V10, then G has at most three million vertices (so there are only finitely many such examples).
CM
Each of these examples either
•	has a subdivision of Vg or
•	is either one of the four graphs described in Theorem 15.6 or obtained from a 2-crossing-critical peripherally-4-connected graph with at most ten vertices by replacing each vertex v having precisely three neighbors with one of at most twenty patches, each patch having at most six vertices (so G has at most sixty vertices).
(4)	If G is not 3-connected, then either
•	G is one of 13 examples that are not 2-connected, or
•	G is 2-connected, has two nonplanar cleavage units, and is one of 36 graphs, or
•	G is 2-connected, has one nonplanar cleavage unit, and is obtained from a 3-connected 2-crossing-critical graph by replacing digons with digonal paths.
We conclude with some remarks on what remains to be done to find all 2-crossing-critical graphs.
Remark 17.2. In Section 15.7, we provided a method for finding all 3-connected, 2-crossing-critical graphs not containing a subdivision of V8. It would be desirable for this program to be completed.
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Remark 17.3. The remaining unclassified 3-connected, 2-crossing-critical graphs have a subdivision of Vg but not of V10. The works of Urrutia [36] and Austin [3] have found many of these, but more work is needed to find a complete set. It may be helpful to note that we have found all such examples that do not have a representativity 2 embedding in the projective plane. The known instances are all
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or so.
ACKNOWLEDGEMENTS
Initial impetus to this project came through Shengjun Pan, who described mechanisms for proving a version of Theorem 2.14 (for G containing a subdivision of V2n, with 2n likely somewhat larger than 10).
We are grateful to CIMAT for hosting us on multiple occasions for work on this project. In particular, we appreciate the support of Jose Carlos Gomez Larranaga, then director of CIMAT.
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Bibliography
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Index
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158
(T, U)-conflguration, 137 <uPv>, 17 <uPv], 17 AB-path, 14 Au, 84 Aw, 84 BOD, 14 C-exterior, 24 C-interior, 24 C-skew paths, Ce, 112 GH, 137 Gv, 158 Gp4c, 137 Gi4c, 151 Gv,130 H-avoiding, 14 H-bridge, 14 H-close, 17 H-face, 19 H-friendly, 31 H-green, 30 H-hyperquad, H-node, 11 H-quad, 17 H-rim, 12 H-yellow, 74 K-prebox, 17 K&, 79 K^ax, 133 K^in, 134 K„, 130 compatible, compatible substitution, Ki, 130
17
137
137
3,4
127
rep(v)
141
L - <g) AP B, 17 L = H, 11 L+, 128 Lw, 142 MAe, 84 NBOD, 14
OD(C), 14 i* 112
PQ, 17 Pu, 84 Pw, 84 Q3, 154 Q3, 154 Q3,154 Q33e, 154
Qi-local H-bridge, 61 R, 12
R-separated, 76 V2n, 1, 11
embeddings, 12 [uPv >, 17 [uPv], 17 Ae, 84 Au, 84 Aw, 84 Pu, 84 Pw, 84 Ue, 84 We, 84 Xe, 84 peak, 87 sharp, 87 Loc(H), 65 Nuc(B), 14 n-pretidy, 63 n-tidy, 64 a, 12 ß, 12 P i, 79 P i, 79 5, 7
cl(Q), 74 H#, 19 Y, 12 l|W ||, 14 [uPvQw], 17 □i, 79 M2, 12 Nh, 137
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N, 51 D, 12 M, 12 K, 129
Me), 62
<8>, 5 iC, 79 tcr, 5 T(S), 9
Pi, 79
Pi, 79
{x, y, z}-claw, 17
a,	12
att(B), 14
b,	12
e-triangle, 146 ef -twisting, 165 fc-bond, 121 fc-drawing, 5 u-consecutive, 99 ue, 84
w-backslope, 115 w-chord, 115 w-chord+w-slope, 115 w-consecutive, 99 w-slope, 115 we, 84 Xe, 84
(G, H, n,Y)), 50 internally-4-connected, 146 peripherally-4-connected, 127 extension, 151
1-drawing,	5
2-jump,	69
2-separation,	121 2.5-jump, 69
3-equivalent	bridges, 14 3-jump, 69 3-reductions, 130
planar, 133
3-rim	path, 74
4-covered	graph, 150
arm (of a hug), 146 attachment, 14 attachments
of a tripod, 158 avoiding, 14 axle, 150
backslope, 115 bearhug, 147 bicycle wheel, 150 axle, 150 rim, 150 spokes, 150 bipartite overlap diagram, 14 BOD, 14
bond, 121 box, 20 bridge, 14
attachment, 14
bipartite overlap diagram, 14
equivalent, 14
global, 61
local, 61
Mobius, 17
nucleus, 14
overlap, 14
overlap diagram, 14
planar C-bridge, 15
residual arc, 14
skew, 14
skew paths, 158
centre, 17 chord, 115 chord+slope, 115 chordless, 57 claw, 17 centre, 17 talon, 17 clean, 15
cleavage unit, 121 close, 17 closure, 74 compatible, 137
substitution, 137 complement, 79 configuration, 137 consecutive, 99 crossbar, 115 crossing-critical, 1 cut-edge, 131
deletable (hug), 147
simultaneously deletable, 147 digon, 120
digonal path, 120, 125 doglike, 138 nose, 138
equivalent, 14 exceptional, 30 exposed, 23 exterior, 24
face, 19 friendly, 31
friendly, standard quadruple, 50 fsq, 50
global H-bridge, 61
green, 30
head (of a hug), 146 hinge, 121
hinge-separation, 121
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hug, 146 arm, 146 bearhug, 147 deletable, 147 head, 146
simultaneously deletable, 147 hyperquad, 17
inside, 52 interior, 24 isthmus, 131
jump, 69
local H-bridge, 61
Mobius bridge Mobius ladder H-rim, 12 H-spoke, 11 rim, 11 rim branch, spoke, 11
17 11
11
NBOD, 14 node, 1, 11
non-planar C-bridge, 15 non-trivial Zi ¿C-path, 79 nose, 138 nucleus, 14
open H-claw, 17 outside, 52 overlap, 14 overlap diagram, 14 bipartite, 14
path, 14
AB-path, 14 peak, 87
planar C-bridge, 15 planar 3-reductions, 133 prebox, 17 pretidy, 63
quad, 17
red, 30
reduces (by 3-reductions), 130 reducible (3-cut), 129 representativity, 10 residual arc, 14 rim, 9, 11, 12
rim (of a bicycle wheel), 150 rim branch, 11 rim path, 74
scope, 79 separated, 76 separation, 121 sharp, 87
simultaneously deletable, 147
skew bridges, 14
skew paths, 158
slope, 115
smooth, 162
span, 69
spanned by, 69
spine, 79
spoke, 11
exposed, 23 spokes (of a bicycle wheel), 150 standard labelling, 23 substitution, 137
talon, 17 tidy, 64 tile, 5
fc-degenerate, 6 compatible, 5 crossing number, 5 cyclization, 6 join, 5 tile drawing, 5 triangle (e-), 146 tripod, 158
attachments, 158 trivial Zi ¿C-path, 79 twisting, 165
virtual edge, 121
yellow, 74