A bipartite graph $B$ is called a brace if it is connected and every matching of size at most two in $B$ is contained in some perfect matching of $B$ and A cycle $C$ in $B$ is called conformal if $B-\V{C}$ has a perfect matching.
We show that there do not exist two disjoint alternating paths that form a cross over a conformal cycle $C$ in a brace $B$ if and only if one can reduce $B$, by an application of a matching theoretic analogue of small clique sums, to a planar brace $H$ in which $C$ bounds a face.
We then utilise this result and provide a polynomial time algorithm which solves the $2$-linkage problem for alternating paths in bipartite graphs with perfect matchings.

Keywords: Perfect Matching, Disjoint Paths, Matching Minor, Bipartite Graphs, Planar Graphs

§ INTRODUCTION

Two of the key aspects in graph minor theory are those of embeddability into surfaces and linkedness, where the latter is the ability to link up a number of prescribed pairs of vertices by pairwise vertex disjoint paths.
In their series of papers on graph minors, Robertson and Seymour proved a general structure theorem that explains how graphs excluding a fixed minor $H$ can be constructed with a surprisingly comprehensive list of ingredients (see [12] for a summary).
Two of these ingredients are graphs embeddable in surfaces and the, so-called, clique sums.
For a non-negative integer $k$, a graph $G$ is a $k$-clique sum of two graphs $G_1$ and $G_2$, if both $G_i$ contain a clique $Q_i$ of size $k$ and $G$ can be obtained from $G_1$ and $G_2$ by identifying $Q_1$ and $Q_2$ into a single clique $Q$ and possibly
forgetting some of the edges of $Q$.
Especially the concept of clique sums has proven itself in the search for exact characterisations of $H$-minor free graphs, at least for relatively small graphs $H$.
By adding graphs on surfaces, or planar graphs in this case, as the next ingredient, one reaches the notion of flatness which plays a key role in a weaker version of the structure theorem for graphs excluding large complete graphs as minors known as the Flat Wall Theorem [21].
In essence, a graph $H$ can be said to be flat in some graph $G$, if $G$ can be constructed via (relatively) small clique sums from smaller graphs $G_1,\dots,G_n$, $n\geq 1$, such that the graph $G_i$ with $H\subseteq G_i$ is planar.
Please note that this is not the actual definition but rather a simplification to illustrate the flavour of the idea behind flatness.

An exceptional case occurs if we restrict $H$ to be some cycle $C$.
We say that $C$ is $C$-flat in a graph $G$ if $G$ can be constructed via $k$-clique sums, where $k\leq 3$, from graphs $G_1,\dots,G_n$, where the graph $G_i$ that contains $C$ is planar, and $C$ bounds a face of $G_i$.
In the case of flat cycles, one immediately finds a flavour of linkedness.
A cycle $C$ is said to have a $C$-cross in $G$, if there exist distinct vertices $s_1,s_2,t_1,t_2$ that occur on $C$ in the order listed, and paths $P_1$ and $P_2$ such that:
Both $P_i$ are internally vertex disjoint from $C$, $P_1$ and $P_2$ are vertex disjoint and each $P_i$ has $s_i$ and $t_i$ as its endpoints.
A classic theorem, to which we will refer as the Two Paths Theorem, links the notions of $C$-flatness and $C$-crosses.
The theorem has been obtained in many different forms with various techniques by a plethora of authors over time [10, 23, 24, 25, 20].

A cycle $C$ in a graph $G$ has no $C$-cross in $G$ if and only if it is $C$-flat in $G$.

By considering Menger's Theorem, one can get the impression that in terms of connectivity via paths, graphs and digraphs are not that different.
Indeed, if $D$ is an acyclic digraph, i.e. a digraph without directed cycles, then the directed version of the $k$-Linkage Problem can be solved in polynomial time [4].
In the case $k=2$ there even exists a directed analogue for the Two Paths Theorem [26].

However, once one allows for the existence of directed cycles, even the Directed $2$-Disjoint Paths Problem becomes NP-hard [4].
This hardness result has far-reaching consequences for structural digraph theory.
Deciding whether a given digraph $D$ contains a subdivision of a digraph $H$ can be NP-complete even for relatively small digraphs [3], while the complexity of deciding whether $D$ contains $H$ as a butterfly minor[A notion of minors suited for digraphs.] remains unknown for almost all digraphs $H$.
Similarly, attempts of generalising the Flat Wall Theorem to the world of digraphs [5] obtain weaker structural
insight due to the lack of a Two Paths Theorem.

So can we hope for a generalisation of the Two Paths Theorem, at least in some sense, for the directed case?
In this paper we present a possible answer to this question in terms of a matching theoretic result, how exactly this relates to digraphs will be discussed in the [sec:conclusion]conclusion.
For this let us introduce some definitions and the structural result that acts as the key tool for the results of this paper.

Let $G$ be a graph, a set $F\subseteq\E{G}$ of pairwise disjoint edges is called a matching, and a vertex $v\in\V{G}$ is said to be covered by $F$ if $F$ contains an edge that has $v$ as an endpoint.
The set of all vertices covered by $F$ is denoted by $\V{F}$.
A matching $M$ is perfect if $\V{M}=\V{G}$ and an edge $e\in\E{G}$ is admissible if there exists a perfect matching $M'$ of $G$ with $e\in M'$.
A graph $G$ is matching covered if it is connected and all of its edges are admissible.

A set $X\subseteq\V{G}$ is conformal if $G-X$ has a perfect matching.
If $M$ is a perfect matching and $\E{G-X}\cap M$ is a perfect matching of $G-X$, we call $X$ $M$-conformal.
A subgraph $H$ of $G$ for which $\V{H}$ is ($M$-)conformal is called ($M$-)conformal.
If $v\in\V{G}$ is a vertex of degree exactly two, we call the process of contracting both edges incident with $v$ and removing all loops and parallel edges afterwards the bicontraction of $v$.
A graph $H$ that can be obtained from a conformal subgraph of $G$ by repeatedly applying bicontractions is called a matching minor.
A path $P$ is $M$-alternating if there exists a set $S\subseteq\V{P}$ of endpoints of $P$ such that $P-S$ is a conformal subgraph of $G$ and we say that $P$ is alternating if there exists a perfect matching $M$ of $G$ such that $P$ is $M$-alternating.
$P$ is $M$-conformal if $S=\emptyset$ and $P$ is internally $M$-conformal if $S$ contains both endpoints of $P$.

Let $G$ be a graph and $X\subseteq\V{G}$.
We denote by $\Cut{X}$ the set of edges in $G$ with exactly one endpoint in $X$ and call $\Cut{X}$ the edge cut around $X$ in $G$.
Let us denote by $\Perf{G}$ the set of all perfect matchings in $G$.
An edge cut $\Cut{X}$ is tight if $\Abs{\Cut{X}\cap M}=1$ for all perfect matchings $M\in\Perf{G}$.
If $\Cut{X}$ is a tight cut and $\Abs{X}\geq 2$, it is non-trivial.
Identifying the shore $X$ of a non-trivial tight cut $\Cut{X}$ into a single vertex is called a tight cut contraction and the resulting graph $G'$ can easily be seen to be matching covered again.
A bipartite matching covered graph without non-trivial tight cuts is called a brace.
It follows from a famous result of Lovász [11] that the braces of a bipartite matching covered graph $B$ are uniquely determined.

Similar to how every $2$-connected minor of a graph $G$ must be a minor of one of its blocks, every brace that is a matching minor of some bipartite matching covered graph $B$ be a matching minor of some brace of $B$.
One brace in particular, namely $K_{3,3}$, has been at the centre of attention because of its connections to $K_{3,3}$-free Orientations (see [16] for an overview on the topic).

Notational Conventions
Let us also fix some conventions for this paper.
Given integers $i,j\in\Z$ we denote by $[i,j]$ the set $\CondSet{z\in\Z}{1\leq z\leq j}$.
This means in particular that $[i,j]=\emptyset$ in case $j<i$.
Given two sets $X$ and $Y$ we denote by $X\Delta Y$ their symmetric difference $\Brace{X\setminus Y}\cup Y\setminus X$.

Since the whole paper mainly deals with bipartite graphs it is convenient for us to always assume that any bipartite graph $B$ comes together with a bipartition of its vertex set into two stable sets which we call $V_1$ and $V_2$.
These two sets are called the colour classes of $B$ and in our figures $V_1$ is usually depicted as a family of black vertices, while the vertices of $V_2$ are depicted as white.
We sometimes write $\Vi{i}{B}$ in case we want to emphasise that we are speaking of the vertices of colour $i\in [1,2]$ in a specific bipartite graph $B$.

Given a bipartite graph $B$ with a perfect matching we say that $B$ contains $K_{3,3}$ if it has a matching minor isomorphic to $K_{3,3}$.
If $B$ does not contain $K_{3,3}$ we say that $B$ is $K_{3,3}$-free.

The following definition can be seen as a matching theoretic variant of clique sums.

For every $i\in\Set{1,2,3}$ let $B_i$ be a bipartite graph with a perfect matching and $C_i$ be a conformal cycle of length four in $B_i$.
A $4$-cycle-sum of $B_1$ and $B_2$ at $C_1$ and $C_2$ is a graph $B'$ obtained by identifying $C_1$ and $C_2$ into the cycle $C'$ and possibly forgetting some of its edges.
If a bipartite graph $B''$ is a $4$-cycle-sum of $B'$ and some bipartite and matching covered graph $B_3$ at $C'$ and $C_3$, then $B''$ is called a trisum of $B_1$, $B_2$ and $B_3$.

The Heawood graph is the bipartite graph associated with the incidence matrix of the Fano plane, see <ref> for an illustration.
Including one exception in the form of the Heawood graph, the structure theorem for $K_{3,3}$-free braces bears a striking resemblance to Wagner's characterisation of $K_5$ minor free graphs [27].

in 2,4,6,8,10,12,14

[v:main] () at (*25.71:15mm);

in 1,3,5,7,9,11,13

[v:mainempty] () at (*25.71:15mm);

in 2,4,6,8,10,12,14

[e:mainthin] (*25.71:15mm) to (128.55+*25.71:15mm);

in 1,...,14

[e:main] (*25.71:15mm) to (25.71+*25.71:15mm);

The Heawood graph $H_{14}$.

A brace is $K_{3,3}$-free if and only if it either is isomorphic to the Heawood graph, or it can be obtained from planar braces by repeated application of the [def:cyclesum]trisum operation.

There exists an algorithm that decides, given a brace $B$ as input, whether $B$ contains $K_{3,3}$ as a matching minor in time $\Fkt{\mathcal{O}}{\Abs{\V{B}}^3}$.

§.§ A Matching Theoretic Two Paths Theorem

The Two Paths Theorem describes the existence of crosses over cycles.
In the setting of matching theory, the cycles of a graph each fall into one of two categories: conformal cycles and non-conformal cycles.
The following lemma servers as an argument that, for the purpose of linking the existence of a cross to a topological property, it suffices to consider conformal cycles.

Let $B$ be a bipartite and planar matching covered graph, then every facial cycle of $B$ is conformal.

If we are interested in a statement like the one of <ref>, then we would expect the cycle which does not have a cross to bound a face in some kind of reduction of the original graph.
If this is the case, then <ref> implies that our cycle is conformal.
Indeed, if our cycle is conformal in the end, our reductions should not have changed this and thus it should have been conformal even before applying any sort of reduction.
Hence it makes sense to only consider 'crosses' over conformal cycles.

Next we need a notion of reduction that is appropriate for the setting of bipartite graphs with perfect matchings.
We have already seen the use of the $4$-cycle sum in the [thm:trisums]characterisation of bipartite graphs excluding $K_{3,3}$ as a matching minor.
The significance of the [def:cyclesum]trisum-operation is that, besides one small exception, it provides a way to combine braces into larger braces.

Let $T_{10}$ be the $4$-cycle-sum of three $K_{3,3}$ at a $4$-cycle $C$ such that no edge of $C$ is in $\E{T_{10}}$.

Let $k\geq 3$, and let $B,B_1,\dots,B_k$ be bipartite graphs such that $B$ is not isomorphic to $T_{10}$.
Suppose $B$ is a $4$-cycle-sum of $B_1,\dots,B_n$ at the $4$-cycle $C$, then $G$ is a brace if and only if $B_1,\dots,B_n$ are braces.

The last piece we need is the definition of a 'cross' itself.

Let $B$ be a bipartite graph with a perfect matching $M$ and let $C$ be a conformal cycle in $B$.
Let $P$ be an $M$-alternating path in $B$.
If $P$ is internally $M$-conformal we say that $P$ is of type 1, if $M$ is a perfect matching of $P$, we say $P$ is of type 2, and otherwise exactly one of the end-edges of $P$ must belong to $M$, in this case $P$ is of type 3.

Let $B$ be a bipartite graph with a perfect matching $M$ and let $C$ be a conformal cycle in $B$.
The cycle $C$ is said to have a matching cross if there exists a perfect matching $M$ and vertices $s_1$, $s_2$, $t_1$, $t_2$, called the pegs of the cross, that appear on $C$ in the order listed such that there exist paths $P_1$ and $P_2$ satisfying the following properties:

* for each $i\in[1,2]$, $P_i$ has endpoints $s_i$ and $t_i$ and is otherwise disjoint from $C$,

* $P_1$ and $P_2$ are $M$-alternating, and

* $P_1$ and $P_2$ are vertex disjoint.

A matching cross over a conformal cycle $C$ is said to be strong if it also meets the following requirements:

* $\Abs{V_1\cap\Set{s_1,s_2,t_1,t_2}}=\Abs{V_2\cap\Set{s_1,s_2,t_1,t_2}}$, and

* $P_1$ and $P_2$ are of the same type

In case $C$ has a matching cross with paths $P_1$ and $P_2$ such that $C+P_1+P_2$ is a conformal subgraph of $B$ we say that $C$ has a conformal cross.

Please note that any path of type 1 or 2 must be of odd length and every path of type 3 is of even length.
Hence the two paths of a conformal cross are either both of type 1 or 2, or both of type 3.
Moreover, if exactly one of the two paths is of type 2, then this path together with one of the two subpaths of $C$ connecting its endpoints forms an alternating cycle.
By switching the perfect matching along this cycle we arrive at a perfect matching for which both paths are of the same type.
Therefore, if we are faced with a conformal cross we may assume this cross to be strong.

Let $B$ be a brace and $C$ a conformal cycle in $B$.
A brace $H$ is called a first order $C$-reduction of $B$, if there exist braces $H$, $B_1,B_2$, and a $4$-cycle $K$ such that $B$ is a trisum of $H$, $B_1$, and $B_2$ at $K$, there exists $i\in[1,2]$ such that $\V{C}\cap\V{K}\subseteq V_i$, and $C\subseteq H$.
A brace $H_{\ell}$ is called a $C$-reduction of $B$ if there is a sequence of braces $H_1,\dots,H_{\ell}$ such that $B=H_1$ and $H_{i+1}$ is a first order $C$-reduction of $H_i$ for all $i\in[1,\ell-1]$.

We can now state the main result of this paper, a Two Paths Theorem for braces.

Let $B$ be a brace and $C$ a conformal cycle in $B$, then $C$ has no [def:matchingcross]matching cross in $B$ if and only if $B$ does not contain $K_{3,3}$ as a matching minor and there exists a planar $C$-reduction of $B$ in which $C$ bounds a face.

Some of the intermediate results that lead to <ref> can be used to solve a slightly altered version of $2$-DAPP.

Let $B$ be a bipartite graph with a perfect matching, $k\in\N$ a positive integer, and let $s_1,\dots,s_k\in V_1$ as well as $t_1,\dots,t_k\in V_2$ be $2k$ pairwise distinct vertices in $B$.
A matching linkage in $B$ for the terminals $s_1,\dots,s_k,t_1,\dots,t_k$ is a perfect matching $M$ and a collection $P_1,\dots,P_k$ of pairwise disjoint and internally $M$-conformal paths such that $P_i$ has endpoints $s_i$ and $t_i$ for each $i\in[1,k]$.

The (bipartite) $k$-Matching Linkage Problem ($k$-MLP) is the question whether, given tuples $\Brace{s_1,\dots,s_k}$ and $\Brace{t_1,\dots,t_k}$ of vertices as above, there exists a matching linkage for the terminals $s_1,\dots,s_k,t_1,\dots,t_k$ in $B$.

Please note that one can always turn an instance of $k$-DAPP into polynomially many instances of $k$-MLP by replacing vertices which appear several times as a terminal with a selection of their distance-$2$-neighbours[If $G$ is a graph and $v\in\V{G}$, then a distance $2$-neighbour of $v$ is a vertex from $\NeighboursG{G}{\NeighboursG{G}{v}}\setminus\Brace{\NeighboursG{G}{v}\cup\Set{v}}$.].
Hence $k$-DAPP and $k$-MLP are polynomial time equivalent, the difference is that $k$-MLP can be easier to work with since the possibility of several terminals being the same vertex does not have to be taken into account.

Let $B$ be a bipartite graph with a perfect matching, and let $s_1,s_2\in V_1$ as well as $t_1,t_2\in V_2$ be four distinct vertices.
There exists an algorithm that decides $2$-MLP for the terminals $s_1,s_2,t_1,t_2$ in time $\Fkt{\mathcal{O}}{\Abs{\V{B}}^5}$.

This result is somewhat surprising since, as we discuss in the [sec:conclusion]conclusion, the variant of the $2$-MLP where we ask the paths to be alternating for a fixed perfect matching $M$ is polynomial time equivalent to the Directed $2$-Disjoint Paths Problem which is still NP-complete.

The project so far

This paper is part of the larger project of extending the graph minors theory of Robertson and Seymour to bipartite graphs with perfect matchings.

A matching theoretic analogue of treewidth, called perfect matching width, was introduced by Norine [17].
Together with Hatzel and Rabinovich, the third author also derived a grid theorem for bipartite graphs with perfect matchings and perfect matching width from the related Directed Grid Theorem [8, 9].
This grid theorem was then further refined by the authors and Stephan Kreutzer in [6] which allowed for a structural characterisation of all classes of bipartite graphs with perfect matchings of bounded perfect matching width:
the exclusion of a planar and matching covered matching minor.
In [6] the complexity of $t$-DAPP on bipartite graphs of bounded perfect matching width was also discussed and a parametrised algorithm with $t$ and the perfect matching width as parameters was presented.

The current paper lays the necessary ground work towards a weak structure theorem or Flat Wall Theorem for bipartite graphs with perfect matchings.

Organisation of the Paper and Proof of <Ref>

Our proof of <ref> can be broken down into two essential pieces.
The first is <ref> which characterises the existence of matching crosses over conformal cycles in $K_{3,3}$-free braces.
<Ref> is dedicated to its proof, but the planar case, which is handled in <ref>, plays a major role.

Let $B$ be a $K_{3,3}$-free brace and $C$ a conformal cycle in $B$.
Then there is no matching cross over $C$ in $B$ if and only if there exists a planar $C$-reduction of $B$ in which $C$ bounds a face.

The second part is <ref>, which guarantees conformal crosses over $4$-cycles in braces that contain $K_{3,3}$, this proposition is proved in <ref>.

Let $B$ be a brace containing $K_{3,3}$ and $C$ a $4$-cycle in $B$, then there exists a conformal bisubdivision of $K_{3,3}$ with $C$ as a subgraph.

In <ref>, we establish some preliminary results which are needed for both the $K_{3,3}$-free case and the case where $B$ contains $K_{3,3}$, especially regarding paths and matching crosses over $4$-cycles.
An important role, in order to bridge between the existence of matching crosses and <ref> is held by the following lemma.

Let $B$ be a brace and $C$ a $4$-cycle in $B$, then there is a conformal cross over $C$ in $B$ if and only if $C$ is contained in a conformal bisubdivision of $K_{3,3}$.

As a last piece, we need to be able to make use of strong matching crosses over certain conformal cycles to find (not necessarily strong) matching crosses more easily.
This is especially useful when paired with <ref> but also finds applications in other places within this chapter.

Let $B$ be a brace and $C$ a conformal cycle.
If $C'\neq C$ is a conformal cycle such that there exists an edge $e\notin\E{C}$ with both endpoints on $C$ with $C'\subseteq C+e$, and there is a perfect matching $M$ of $B$ and $M$-alternating paths $L$ and $R$ that form a matching cross over $C'$, then there exists a matching cross over $C$ in $B$ that does not use $e$.

Since $C'\subseteq C+e$, $C'-e$ forms a subpath of $C$, hence the order of the endpoints of $L$ and $R$ on $C$ is the same as the order of these vertices on $C'$.
So in case $L$ and $R$ are internally disjoint from $C$, they immediately form a matching cross over $C$ as well.
Suppose exactly one of these paths, say $L$, intersects $C$.
Let $v_R$ and $w_R$ be the endpoints of $R$ and $v_1$, $w_1$ be the endpoints of $L$.
Then let $x_1$ be the last vertex of $L$ on $C$ we encounter when traversing along $L$ starting with $v_1$ such that $x_1$ is separated from $w_1$ on $C$ by $v_2$ and $w_2$.
Next let $x_2$ be the first vertex of $L$ that lies on $C$ we encounter after $x_1$.
Then $x_2$ must belong to the same component of $C-v_2-w_2$ as $w_1$ and thus $x_1$ and $x_2$ are separated on $C$ by $v_2$ and $w_2$.
Since $L$ and $R$ are disjoint and $M$-alternating, so are $x_1Lx_2$ and $R$ and thus we have found a matching cross over $C$.
So now assume that also $R$ intersects $C$.
In this case, let $y_1$ be the last vertex we encounter when traversing along $R$ starting in $v_2$ such that $y_1$ and $w_2$ belong to different components of $C-x_1-x_2$.
Then let $y_2$ be the first vertex of $C$ we encounter on $R$ after $y_1$.
By choice of $y_1$, $y_1$ and $y_2$ must belong to different components of $C-x_1-x_2$ and $y_1Ry_2$ is internally disjoint from $C$.
Hence $x_1Lx_2$ and $y_1Ry_2$ form a matching cross over $C$.
Moreover, since $e$ is not contained in either $L$ or $R$, we have found a matching cross over $C$ which does not contain $e$.

The four results above combined yield a short proof of <ref>.

Let $B$ be a brace and $C$ a conformal cycle in $B$.
Suppose $B$ is Pfaffian, then <ref> immediately yields both directions of our claim.
Hence we may assume $B$ to be non-Pfaffian.
If $C$ is a $4$-cycle, then <ref> guarantees the existence of a conformal cross over $C$ in $B$.
So we may assume $C$ to have length at least six.
Let $P$ be a subpath of $C$ of length three, so $P$ consists of exactly four vertices, two of each colour class.
Let $a\in V_1$ and $b\in V_2$ be the endpoints of $P$.
If the edge $ab$ does not exist in $B$ we introduce it, please note that introducing an edge does not change the status of $B$ being a brace, nor can $B+ab$ be $K_{3,3}$-free if $B$ is not.
Hence $C'\coloneqq P+ab$ is a $4$-cycle in $B+ab$ such that $C'\subseteq C+ab$.
By <ref> there is a conformal bisubdivision $L$ of $K_{3,3}$ in $B+ab$ that has $C'$ as a subgraph.
By <ref> there is a conformal cross over $C'$.
Please note that, with $L$ being a bisubdivision of $K_{3,3}$, we may choose a perfect matching $M$ of $B$ such that $L$ is $M$-conformal and $ab\notin M$.
An application of <ref> now yields a matching cross over $C$ in $B+ab$ that does not contain $ab$.
With $ab\notin M$ this means that there is a matching cross over $C$ in $B$.

With <ref>, we have a tool that can help us to obtain an algorithmic solution of $2$-MLP.
In fact, it is <ref>, which provides the important insight.
<Ref> can be used to solve the $2$-Linkage Problem by introducing a small local construction.
In <ref>, we describe how a similar construction can be used for the $2$-MLP.

§ MATCHING CROSSES IN PLANAR BRACES

In this section we establish the base case of <ref> in form of an exact characterisation of the existence of matching crosses over conformal cycles in planar braces.

Let $B$ be a brace and $C$ a conformal cycle in $B$, then there exists a strong matching cross over $C$ in $B$ if and only if $C$ does not bound a face.

Since every matching cross over $C$ in $B$ also is an ordinary[Ordinary here means a standard undirected cross in the sense of <ref>.] $C$-cross, the existence of such a cross immediately certifies that it is impossible to draw $B$ such that $C$ bounds a face.
We call a cycle $C$ in a planar graph $B$ separating if $B$ cannot be drawn in a way such that $C$ bounds a face.
So we only need to show the reverse direction.

Before we can continue, some additional information about braces is necessary.
Let $G$ be a graph and $k$ be a positive integer.
We say that $G$ is $k$-extendible if it is connected, has at least $2k+2$ vertices, and for every matching $F\subseteq\E{G}$ there exists a perfect matching $M$ of $G$ with $F\subseteq M$.

A bipartite graph $B$ is a brace if and only if it is either isomorphic to $C_4$, or it is $2$-extendible.

Let $k\in\N$ be a positive integer.
Then every $k$-extendible graph is $\Brace{k+1}$-connected.

Extendibility in bipartite graphs can be expressed in many different ways.
Of particular interest for us will be the existence of disjoint alternating paths, a property similar to strong $k$-connectivity in digraphs.

Let $B$ be a bipartite graph and $k\in\N$ a positive integer.
The following statements are equivalent.

* $B$ is $k$-extendible.

* $\Abs{V_1}=\Abs{V_2}$, and for all non-empty $S\subseteq V_1$, $\Abs{\NeighboursG{B}{S}}\geq \Abs{S}+k$.

* For all sets $S_1\subseteq V_1$ and $S_2\subseteq V_2$ with $\Abs{S_1}=\Abs{S_2}\leq k$ the graph $B-S_1-S_2$ has a perfect matching.

* There is a perfect matching $M\in\Perf{B}$ such that for every $v_1\in V_1$, every $v_2\in V_2$ there are $k$ pairwise internally disjoint internally $M$-conformal paths with endpoints $v_1$ and $v_2$.

* For every perfect matching $M\in\Perf{B}$, every $v_1\in V_1$, every $v_2\in V_2$ there are $k$ pairwise internally disjoint internally $M$-conformal paths with endpoints $v_1$ and $v_2$.

Since every brace is $3$-connected by <ref>, we can rely on the uniqueness of plane embeddings for $3$-connected graphs [28].

Let $B$ be a brace and $C$ a conformal and separating cycle in $B$.
Then by Whitney's Theorem [28] the interior and the exterior of $C$ are the same in every drawing of $B$ (up to the choice of the outer face).
In what follows, we always assume $B$ to come with a fixed drawing to avoid ambiguity.
We denote the subgraph of $B$ induced by the interior of $C$ together with $C$ itself by $\Inner{B}{C}$ and the subgraph of $B$ induced by the exterior of $C$ together with $C$ is denoted by $\Outer{B}{C}$.
In both graphs $C$ bounds a face and, since $C$ is conformal, both graphs have a perfect matching.

Let $B$ be a planar brace and $C$ a conformal and separating cycle in $B$, then $\Inner{B}{C}$ and $\Outer{B}{C}$ are matching covered.

It suffices to show the claim for one of the two graphs.
Moreover, by <ref> it suffices to show for a single perfect matching $M$, that any pair $a\in V_1$, $b\in V_2$ of vertices is linked by an internally $M$-conformal path.
So let us fix a perfect matching $M$ for which $C$ is $M$-conformal.
Since $B$ is a brace, given $a\in \Vi{1}{\Inner{B}{C}}$ and $b\in \Vi{2}{\Inner{B}{C}}$, <ref> guarantees the existence of two internally disjoint and internally $M$-conformal paths $P_1$, $P_2$ from $a$ to $b$.
If one of these paths is disjoint from $C$, there is nothing to show.
Hence we may assume that both meet $C$ and, by a similar argument, both of them need to contain an edge of $\Outer{B}{C}-\E{C}$.
Let $b_1$ be the first vertex of $P_1$ on $C$ when traversing $P_1$ from $a$ towards $b$ and let $a_1$ be the last vertex of $P_1$ on $C$.
Then $a_1$ and $b_1$ separate $C$ into two paths, one of them being $M$-conformal, let $P'$ be this path.
Moreover, $P_1b_1$ and $a_1P_1$ are internally $M$-conformal, and all three paths are contained in $\Inner{B}{C}$.
Hence $P_1b_1P'a_1P_1$ is an internally $M$-conformal $a$-$b$-path in $\Inner{B}{C}$, and we are done.

Towards the next partial result we need to introduce some additional information on tight cuts in bipartite graphs.

Let $B$ be a bipartite graph, and $X\subseteq\V{G}$.
If $\Abs{X\cap V_1}=\Abs{X\cap V_2}$ we say that $X$ is balanced, otherwise it is unbalanced.
Suppose $X$ is unbalanced, then there are $i,j\in[1,2]$, and $k\in\N$ such that $\Abs{X\cap V_i}=\Abs{X\cap V_j}+k$.
In this case we call $X\cap V_i$ the majority of $X$, denoted by $\Majority{X}$, and $X\cap V_j$ is the minority, denoted by $\Minority{X}$.

Let $B$ be a bipartite matching covered graph and $X\subseteq\V{B}$ where $\Abs{X}$ is odd.
Then $\Cut{X}$ is a tight cut if and only if $\Abs{\Majority{X}}-\Abs{\Minority{X}}=1$ and $\Neighbours{\Minority{X}}\subseteq\Majority{X}$.

Let $B$ be a planar brace, $C$ be a conformal and separating cycle in $B$ such that $B'\in\Set{\Inner{B}{C},\Outer{B}{C}}$ is not a brace, and $\CutG{B'}{X}$ be a non-trivial tight cut in $B'$.
Then $\Abs{X\cap\V{C}}\geq 3$ and $\Abs{\Complement{X}\cap\V{C}}\geq 3$.

Suppose $\Abs{X\cap\V{C}}\leq 1$.
By symmetry, it suffices to treat this case.
With <ref> we know that the minority of $X$ has no edge to a vertex of $\Complement{X}$, without loss of generality let us assume the majority of $\CutG{B}{X}$ to be in $V_1$.
If $\V{C}\cap X=\emptyset$, then clearly $\CutG{B}{X}$ must be a non-trivial tight cut in $B$, which is impossible.
Hence there must exist a unique vertex $a\in\V{C}\cap X$.
Moreover, since this vertex has neighbours in $C$ which do not belong to $X$, $a\in V_1$.
But also in this case $\CutG{B}{X}$ is non-trivially tight in $B$ and the claim follows.

Let $B$ be a planar brace, $C$ a conformal and separating $4$-cycle in $B$, then $\Inner{B}{C}$ and $\Outer{B}{C}$ are braces.

Please note that <ref> can be extended to non-planar braces as well.
The general version of <ref> is due to McCuaig.

Let $B_1$ and $B_2$ be bipartite graphs with a common $4$-cycle $C$ and otherwise disjoint. If $B_1+B_2$ is a brace, then so are $B_1$ and $B_2$.

The next lemma is a slightly restated version of Lemma 46 from [16] which can be derived with the methods presented there.

Let $B$ be a planar brace, $C$ be a facial cycle of $B$, $M$ be a perfect matching of $B$ for which $C$ is $M$-conformal, and $a\in \Vi{1}{C}$, $b\in \Vi{2}{C}$ be two vertices with $ab\notin\E{C}$.
Then there exists an internally $M$-conformal $a$-$b$-path $P$ in $B$ which is internally disjoint from $C$.

As a special case, we first assume that we are interested in a (strong) matching cross over some separating cycle $C$ in a planar brace $B$ where both $\Inner{B}{C}$ and $\Outer{B}{C}$ are braces.
We also need some deeper insight in how conformal bisubdivisions of $K_{3,3}$ and the cube can appear in braces with respect to cycles of length four.

Let $B$ be a brace and $C$ a $4$-cycle such that $B-\V{C}$ is connected, let $uv\in\E{C}$ and $x,y\in\V{B}\setminus\V{C}$ such that $ux,vy\in\E{B}$.
Then $B$ contains a conformal bisubdivision of the cube with $C+ux+vy$ as a subgraph, or $B$ contains a conformal bisubdivision of $K_{3,3}$ with $C$ as a subgraph.

The following is a slight weakening of the lemma above.

Let $B$ be a brace and $C$ a $4$-cycle such that $B-\V{C}$ is connected, then $B$ contains a conformal bisubdivision of the cube or $K_{3,3}$ with $C$ as a subgraph.

Please note that a version of <ref> can be found in [22], where the containment of a $4$-cycle in a conformal bisubdivision of the cube is referred to as being 'fat' while being a subgraph of some conformal bisubdivision of $K_{3,3}$ is called having a '$C$-cross'.
Sadly these two are not mutually exclusive as one can see in the example in <ref>.
However, there is a deeper connection between the existence of matching crosses, especially conformal ones, and the existence of conformal bisubdivisions of $K_{3,3}$ when it comes to $4$-cycles.
We revisit this topic in <ref>.

[v:mainempty] () at (-0.7,-0.7);
[v:main] () at (-0.7,0.7);
[v:main] () at (0.7,-0.7);
[v:mainempty] () at (0.7,0.7);
[v:main] () at (-0.4,0.1);
[v:mainempty] at (0.4,0.1);

[e:main] (-0.7,-0.7) rectangle (0.7,0.7);
[e:main] (-0.4,0.1) – (0.4,0.1);
[e:main] (-0.7,0.7)– (0.4,0.1);
[e:main] (0.7,0.7)– (-0.4,0.1);
[e:main] (-0.7,-0.7) – (-0.4,0.1);
[e:main] (0.7,-0.7) – (0.4,0.1);

[v:mainempty] () at (-0.5,-0.5);
[v:main] () at (-0.5,0.5);
[v:main] () at (0.5,-0.5);
[v:mainempty] () at (0.5,0.5);

[v:main] () at (-1,-1);
[v:mainempty] () at (1,-1);
[v:mainempty] () at (-1,1);
[v:main] () at (1,1);

[e:main] (-1,-1) rectangle (1,1);
[e:main] (-0.5,-0.5) rectangle (0.5,0.5);
[e:main] (-0.5,-0.5) – (-1,-1);
[e:main] (-0.5,0.5) – (-1,1);
[e:main] (0.5,-0.5) – (1,-1);
[e:main] (0.5,0.5) – (1,1);

[v:main] () at (-1,-1);
[v:mainempty] () at (1,-1);
[v:mainempty] () at (-1,1);
[v:main] () at (1,1);

[v:mainempty] () at (-0.7,-0.7);
[v:main] () at (-0.7,0.7);
[v:main] () at (0.7,-0.7);
[v:mainempty] () at (0.7,0.7);
[v:main] () at (-0.4,0.1);
[v:mainempty] at (0.4,0.1);

[e:marker] (-0.7,-0.7) rectangle (0.7,0.7);

[e:main] (-0.7,-0.7) rectangle (0.7,0.7);
[e:main] (-0.4,0.1) – (0.4,0.1);
[e:main] (-0.7,0.7)– (0.4,0.1);
[e:main] (0.7,0.7)– (-0.4,0.1);
[e:main] (-0.7,-0.7) – (-0.4,0.1);
[e:main] (0.7,-0.7) – (0.4,0.1);

[e:main] (-1,-1) rectangle (1,1);
[e:main] (-0.7,-0.7) – (-1,-1);
[e:main] (-0.7,0.7) – (-1,1);
[e:main] (0.7,-0.7) – (1,-1);
[e:main] (0.7,0.7) – (1,1);

From left to right: $K_{3,3}$, the cube, and a brace with a $4$-cycle (the marked one) which is contained in both, a conformal bisubdivision of $K_{3,3}$ and a conformal bisubdivision of the cube.
Please note that one could get rid of the fact that the marked cycle is separating by adding additional edges between vertices from different colour classes.

Let $B$ be a planar brace and $C$ a conformal and separating cycle in $B$ such that $\Inner{B}{C}$ and $\Outer{B}{C}$ both are braces.
Then there exists a strong matching cross over $C$ in $B$.

First, assume $C$ to have at least length $6$.
In this case, we can select a perfect matching $M$ of $B$ such that $C$ is $M$-conformal, which in turn implies that $M$ contains perfect matchings $M_{\text{int}}$ and $M_{\text{out}}$ of $\Inner{B}{C}$ and $\Outer{B}{C}$ respectively.
Now select vertices $s_1$, $s_2$, $t_1$, and $t_2$ such that they appear on $C$ in the order listed where $s_1,s_2\in V_1$, and $t_1,t_2\in V_2$.
According to <ref>, we may choose an internally $M$-conformal $s_1$-$t_1$-path $P_1$ in $\Inner{B}{C}$ and an internally $M$-conformal $s_2$-$t_2$-path $P_2$ in $\Outer{B}{C}$ such that each of the $P_i$ is internally disjoint from $C$.
Then $P_1$ and $P_2$ form a $C$-cross in $B$, and we are done.

What remains is the case where $C=\Brace{s_1,t_1,s_2,t_2}$ is a $4$-cycle.
By calling upon <ref> we can find a conformal bisubdivision of the cube in each of the two braces such that each of these cubes contains $C$ as a subgraph.
Let $H_1$ be a conformal bisubdivision of the cube in $\Inner{B}{C}$ and let $H_2$ be a conformal bisubdivision of the cube in $\Outer{B}{C}$.
One can easily see, that $H\coloneqq H_1+H_2$ is a conformal subgraph of $B$.
We can now use $H$ to find the required matching cross over $C$, as illustrated in <ref>.

[v:main] () at (-0.6,-0.6);
[v:mainempty] () at (-0.6,0.6);
[v:mainempty] () at (0.6,-0.6);
[v:main] () at (0.6,0.6);

[v:mainempty] () at (-1.3,-1.3);
[v:main] () at (-1.3,1.3);
[v:main] () at (1.3,-1.3);
[v:mainempty] () at (1.3,1.3);

[v:main] () at (-2,-2);
[v:mainempty] () at (-2,2);
[v:mainempty] () at (2,-2);
[v:main] () at (2,2);

[e:marker,color=myLightBlue] (-1.3,1.3) – (-2,2);
[e:marker,color=myLightBlue] (-2,2) – (2,2);
[e:marker,color=myLightBlue] (2,2) – (2,-2);
[e:marker,color=myLightBlue] (2,-2) – (1.3,-1.3);

[e:marker,color=myGreen] (1.3,1.3) – (0.6,0.6);
[e:marker,color=myGreen] (0.6,0.6) – (0.6,-0.6);
[e:marker,color=myGreen] (0.6,-0.6) – (-0.6,-0.6);
[e:marker,color=myGreen] (-0.6,-0.6) – (-1.3,-1.3);

[e:main] (-1.3,-1.3) rectangle (1.3,1.3);

[e:main,color=gray] (-2,2) – (-2,-2) – (2,-2) – (2,2);

[e:main,color=gray] (-0.6,-0.6) – (-0.6,0.6) – (0.6,0.6) – (0.6,-0.6);

[e:coloredborder] (0.6,-0.6) – (-0.6,-0.6);
[e:coloredthin,color=BostonUniversityRed] (0.6,-0.6) – (-0.6,-0.6);

[e:coloredborder] (2,2) – (-2,2);
[e:coloredthin,color=BostonUniversityRed] (2,2) – (-2,2);

[e:coloredborder] (-1.3,-1.3) – (-2,-2);
[e:coloredthin,color=BostonUniversityRed]  (-1.3,-1.3) – (-2,-2);

[e:coloredborder] (1.3,-1.3) – (2,-2);
[e:coloredthin,color=BostonUniversityRed]  (1.3,-1.3) – (2,-2);

[e:main,color=gray] (-1.3,-1.3) – (-0.6,-0.6);
[e:main,color=gray] (1.3,-1.3) – (0.6,-0.6);

[e:coloredborder] (-1.3,1.3) – (-0.6,0.6);
[e:coloredthin,color=BostonUniversityRed]  (-1.3,1.3) – (-0.6,0.6);

[e:coloredborder] (1.3,1.3) – (0.6,0.6);
[e:coloredthin,color=BostonUniversityRed]  (1.3,1.3) – (0.6,0.6);

[e:main,color=gray] (-1.3,1.3) – (-2,2);
[e:main,color=gray] (1.3,1.3) – (2,2);

The conformal subgraph $H$ in the proof of <ref> together with a strong matching cross over the separating $4$-cycle $C$.

Let $B$ be a bipartite matching covered graph, $M$ a perfect matching of $B$, $e=ab\in M$ with $a\in V_1$, $b\in V_2$, $X\subseteq V_1\setminus\Set{a}$ and $Y\subseteq V_2\setminus\Set{b}$ such that every internally $M$-conformal $X$-$Y$-path in $B$ contains $e$.
Then there exists a tight cut $\CutG{B}{Z}$ in $B$ with $X\subseteq Z$, $Y\subseteq\Complement{Z}$, and $e\in\CutG{B}{Z}$.

By assumption, there is no internally $M$-conformal $X$-$Y$-path in $B-a-b$ and thus no vertex of $X$ can share an elementary component with a vertex of $Y$.
Since, by definition, each elementary component would be matching covered and thus, <ref> would guarantee the existence of such a path.
Let $\Fkt{Up}{X}$ be the set of all vertices $w$ of $B-a-b$ such that there exist elementary components $K_X$ and $K_w$ with $K_X\leq_2 K_w$ where $K_X$ contains a vertex of $X$ and $w\in\V{K_w}$.
Then $Y\cap\Fkt{Up}{X}=\emptyset$.
Moreover, there is no $\Vi{1}{\Fkt{Up}{X}}$-$\Vi{2}{\Complement{\Fkt{Up}{X}}}$-path in $B-a-b$ at all.
Hence in $B$ $\Fkt{Up}{X}\cup\Set{b}$ is a set of odd cardinality, where no vertex of $V_1$ has a neighbour outside of it and the difference between the number of $V_1$-vertices and the number of $V_1$-vertices is exactly one.
So by <ref> $\CutG{B}{\Fkt{Up}{X}\cup\Set{b}}$ is a tight cut with $Y\subseteq\Complement{\Fkt{Up}{X}\cup\Set{b}}$.

Together with the upcoming lemma, <ref> imply <ref>.

Let $B$ be a planar brace, $C$ a conformal and separating cycle in $B$ such that $B'\in\Set{\Inner{B}{C},\Outer{B}{C}}$ is not a brace.
Then there exists a strong matching cross over $C$ in $B$.

By <ref> we may assume $\Abs{\V{C}}\geq 6$.
Let $\CutG{B'}{X}$ be a non-trivial tight cut in $B'$ maximising $\Abs{\Complement{X}}$.
Without loss of generality let us assume the majority of $X$ to be in $V_1$.
With <ref> and the fact that for every selection of three distinct edges of $C$, at least two of them belong to a common perfect matching, one can see that $\CutG{B'}{X}$ separates $C$ into two non-trivial paths $Q_1$ and $Q_2$ such that $\V{Q_1}\subseteq\Complement{X}$ and $\V{Q_2}\subseteq X$.
Let $a_1, a_2\in X\cap V_1$ be the endpoints of $Q_2$ and let $b_1,b_2\in\Complement{X}\cap V_2$ be the endpoints of $Q_1$ such that $a_ib_i\in\E{C}$ for both $i\in[1,2]$.
Next let $M$ be a perfect matching such that $C$ is $M$-conformal and $a_2b_2\in M$, moreover let $a'_1b_1, a_1b'_1\in M$.
Since the majority of $X$ is in $V_1$, there cannot exist an internally $M$-conformal path starting at some vertex of $V_1\cap\Complement{X}$ and ending in a vertex of $V_2\cap X$ that avoids $a_2b_2$.
However, with $B$ being a brace and <ref> there must be two internally disjoint and internally $M$-conformal $a'_1$-$x$-paths for every $x\in X\cap V_2$ and one of them must avoid $a_2b_2$.
Let us choose $b_X\in V_2\cap X$ such that there exists an internally $M$-conformal path $P$ from $a'_1$ to $b_X$ with the following properties:

* $a_{\Complement{X}}$ is the last vertex of $\V{C}\cap\Complement{X}$ along $P$ starting in $a'_1$, and

* $\V{P}\cap\V{Q_2}=\Set{b_X}$.

Then $a_{\Complement{X}}\in V_1$ and $P_1\coloneqq a_{\Complement{X}}P$ is an internally $M$-conformal path which is internally disjoint from $C$.
Moreover, $\V{P_1}\cap\V{B'}=\V{P_1}\cap\V{C}=\Set{a_{\Complement{X}},b_X}$.
Let $Q_3$ be the component of $Q_1-a_{\Complement{X}}$ containing $b_1$, let $Y$ be the component of $C-a_{\Complement{X}}-b_X$ containing $Q_3$ and at last let $\Complement{Y}$ denote the other component of $C-a_{\Complement{X}}-b_X$.
Let $e\in M$ be the edge covering $b_X$.
What follows is a case distinction on the existence of some internally $M$-conformal path from $\Complement{Y}\cap V_1$ to $Y\cap V_2$.

Case 1: There exists an internally $M$-conformal path $P'$ from $\Complement{Y}\cap V_1$ to $Y\cap V_2$ that avoids $e$.

If this is the case, let $b_Y$ be the first vertex of $Y$ encountered while traversing along $P'$ starting in $\Complement{Y}\cap V_1$.
Then let $a_{\Complement{Y}}$ be the last vertex of $P'$ in $\Complement{Y}\cap V_1$ encountered before $b_Y$.
Now $P_2\coloneqq a_{\Complement{Y}}P'b_Y$ is an internally $M$-conformal path with no inner vertex on $C$ that is disjoint from $P_1$.
Moreover, the vertices $a_{\Complement{Y}}$, $a_{\Complement{X}}$, $b_Y$, and $b_X$ appear on $C$ in the order listed and thus $P_1$ and $P_2$ form a strong matching cross over $C$ in $B$.

Case 2: All internally $M$-conformal $(\Complement{Y}\cap V_1)$-$(Y\cap V_2)$-paths contain $e$.

Then the deletion of both endpoints of $e$ in $B'$ leaves at least two elementary components, some containing vertices of $\Complement{Y}\cap V_1$ and some of the others containing vertices of $Y\cap V_2$ but never both.
Thus, by <ref>, there exists a tight cut $\CutG{B'}{Z}$ with $\Complement{Y}\cap V_1\subseteq Z$, $Y\cap V_2\subseteq\Complement{Z}$, and $e\in\CutG{B}{Z}$.
Indeed, the majority of $Z$ must be in $V_2$.
Since $X$ is odd, one of the two sets $X\cap Z$ and $X\setminus Z$ must be odd and therefore, by <ref>, one of these sets defines a tight cut in $B'$.
Clearly $a_2,b_X\in X\cap Z$ and thus $\Abs{X\cap Z}>1$.
Since $\CutG{B}{Z}$ cannot contain more than two edges of $C$ and by choice of $\CutG{B}{X}$ there cannot be a non-trivial tight cut $\CutG{B}{X'}$ with $X'\subset X$ in $B'$, hence
$X\setminus Z=\Set{a_1}$ and $a_1b_X\in M$.

With an argument similar to the one in Case 1 one can see that, in case there exists an internally $M$-conformal $(Y\cap V_1)$-$(\Complement{Y}\cap V_2)$-path avoiding the edge $e'\in M$ covering $a_{\Complement{X}}$, we are done again.
So we may also assume that $e'$ meets all internally $M$-conformal $(Y\cap V_1)$-$(\Complement{Y}\cap V_2)$-paths.
Then, with arguments as before, one derives the existence of a non-trivial tight cut $\CutG{B'}{Z'}$ such that $a_1\in Z'$ and $\Complement{Y}\cap B\subseteq\Complement{Z'}$.
In the end we arrive at the conclusion that $X\cap Z'=\Set{a_2}$ and $e'\in\CutG{B'}{Z'}$.

Moreover, this means that there is neither an internally $M$-conformal $(Y\cap V_1)$-$(\Complement{Y}\cap V_2)$-path, nor an internally $M$-conformal $(\Complement{Y}\cap V_1)$-$(Y\cap V_2)$-path in $B'$ after deleting the four vertices in $S\coloneqq e\cup e'$.
Hence there cannot be any $Y$-$\Complement{Y}$-path in $B'-S$.
This means there must be a face $C'$ of $B'$ containing both $a_{\Complement{X}}$ and $b_X$ that is distinct from $C$.
Since $B$ itself is a brace, there must be an internally $M$-conformal path $P'_1$ from $Y\cap V_1$ to $\Complement{Y}\cap V_2$ that avoids $e'$ and this path may be chosen to be internally disjoint from $B'$.
In particular, we may choose the endpoints of $P'_1$ to be disjoint from one of the two paths of $C'$, say $R$, connecting $a_{\Complement{X}}$ and $b_X$.
This is due to the fact that every internal vertex of a path in $C\cap C'$ must be of degree two in $B'$, $B'\neq C$, and $a_{\Complement{X}}b_X\notin\E{C}$.
Since $B'$ is matching covered by <ref>, <ref> guarantees the existence of a perfect matching $M'$ of $B'$ for which $C'$ is $M'$-conformal.
Let us choose $M'$ such that $R$ is an internally $M'$-conformal path and let $P'_2$ be an internally $M'$-conformal subpath of $R$ with both endpoints on $C$ and otherwise disjoint from $C$.
All of these choices are possible since $a_{\Complement{X}}$ and $b_X$ belong to different colour classes of $B'$.
At last let us set $M''\coloneqq M'\cup \Brace{M\setminus\E{B'}}$.
Since $B'$ is an $M$-conformal subgraph of $B$, $M''$ is a perfect matching of $B$, and by construction, both $P'_1$ and $P'_2$ are internally $M''$-conformal paths.
Now $P'_1$ and $P'_2$ form a strong matching cross over $C$ in $B$.

We are ready to derive the main result of this section.

Let $C$ be a conformal cycle in a planar brace.
If $C$ does not bound a face, it is separating and thus <ref> guarantee the existence of a strong matching cross over $C$ in $B$.
For the reverse suppose there exists a strong matching cross over $C$ in $B$, then exactly one path of the cross must lie in the interior of $C$ for every plane embedding of $B$.
Hence $C$ does not bound a face in any plane embedding of $B$.

§ PATHS AND MATCHING CROSSES THROUGH $4$-CYCLE SUMS

Cycles of length four play a key role in many aspects of bipartite matching theory as we have seen in <ref>.
In particular, the $4$-cycle sum operation and the fact, that no perfect matching $M$ for which an $M$-conformal matching cross over $C_4$ exists can contain a perfect matching of $C_4$ itself are things to be considered.
While tight cut contractions only preserve special types of matching crosses[It is possible to lose a matching cross if for example the tight cut separates the endpoints of both paths of the cross, in both tight cut contractions, what remains of the paths is now a set of two paths with a common endpoint.], at some point they are not applicable any more to further decompose a given graph.

In the spirit of the Two Paths Theorem, we want to decompose our graph further while maintaining a fixed subgraph, in most cases the conformal cycle for which we seek a matching cross.
This section, and the following ones, exist to describe exactly the interaction between conformal cycles, matching crosses over these cycles, $4$-cycle sums, and matching crosses over $4$-cycles in both $K_{3,3}$-free braces which are non-planar and in braces containing $K_{3,3}$.

A brace $B$ is a maximal $4$-cycle sum at the $4$-cycle $C$ of the braces $B_1,\dots,B_{\ell}$, $\ell\geq 3$, if there do not exist braces $H_1,\dots,H_n$, $n>\ell$ such that $B$ is a $4$-cycle sum of $H_1,\dots, H_m$ at $C$.

Let $B$ be a $K_{3,3}$-free brace, $n\geq 3$, $B_1,\dots,B_{\ell}$ braces such that $B$ is a maximal $4$-cycle sum of $B_1,\dots,B_{\ell}$ at the $4$-cycle $C=\Brace{a_1,b_1,a_2,b_2}$ and let $M$ be a perfect matching of $B_1$.
Then for every choice of $x\in\Set{a_1,a_2}$ and $y\in\Set{b_1,b_2}$ there exists a perfect matching $M'$ of $B$ such that $M\setminus\E{C}\subseteq M'$ and there are paths $P_1$ and $P_2$  in $B-\V{B_1-\V{C}}$ where

* $P_1$ has endpoints $x$ and $y$, while $P_2$ connects $\Set{a_1,a_2}\setminus\Set{x}$ to $\Set{b_1,b_2}\setminus\Set{y}$,

* $P_1$ and $P_2$ are disjoint, and

* $P_1+P_2+\CondSet{e\in M'}{e\cap \V{C}\neq\emptyset}$ is an $M'$-conformal subgraph of $B$.

By <ref> for every $i\in[2,\ell]$ $C$ is contained in a conformal bisubdivision of the cube, or of $K_{3,3}$.
Since $B$ is $K_{3,3}$-free the later can never be true and thus for every $j\in[2,\ell]$, $C$ is contained in a conformal bisubdivision $H$ of the cube in $B_j$.
As $H$ is conformal in $B_j$, every perfect matching of $H$ can be combined with a perfect matching of $B_j-\V{H}$ to a perfect matching of $B_j$.
In general, let $M_H$ be a perfect matching of $H$ such that for every $e\in\E{C}\cap M_H$ both endpoints of $e$ are covered by edges of $M\setminus\E{C}$ and a vertex of $C$ is covered by a non-$\E{C}$-edge in $M_H$ if and only if it is covered by an edge of $M\cap\E{C}$.
Moreover, let $M_j$ be a perfect matching of $B_j-\V{H}$, and for each $i\in2,\ell\setminus\Set{j}$ let $M_i$ be a perfect matching of $B_i-\V{C}$.
The matching $M_i$ clearly exists since the $B_i$ are braces, and therefore they are $2$-extendible.
Then $\Brace{M\setminus\E{C}}\cup\Brace{M_H\setminus\E{C}}\cup\bigcup_{i=1,i\neq j}^nM_i$ is a perfect matching of $B$.
Hence it suffices to show that, for any $M$ we are given, we can choose the matching $M_j$ such that we are able to find the desired paths within $H$.
What follows is a discussion of these paths depending on the number of edges in $M\cap\E{C}$.
We present these matchings together with the paths in <ref>.
As $H$ is a bisubdivision of the cube, each perfect matching of $H$ mirrors a perfect matching $M_H'$ of the cube in the sense that a bisubdivided edge of the cube is $M_H$-conformal if and only if the corresponding edge of the cube belongs to $M_H'$.

[v:mainempty] () at (-0.6,-0.6);
[v:main] () at (-0.6,0.6);
[v:main] () at (0.6,-0.6);
[v:mainempty] () at (0.6,0.6);
[v:main] () at (-1.2,-1.2);
[v:mainempty] () at (-1.2,1.2);
[v:mainempty] () at (1.2,-1.2);
[v:main] () at (1.2,1.2);

[v:mainemptygray] () at (0.4,1.2);
[v:maingray] () at (0.4,-1.2);
[v:maingray] () at (-0.4,1.2);
[v:mainemptygray] () at (-0.4,-1.2);

[v:mainemptygray] () at (-1.2,-0.4);
[v:maingray] () at (-1.2,0.4);
[v:maingray] () at (1.2,-0.4);
[v:mainemptygray] () at (1.2,0.4);

[v:maingray] () at (0.8,0.8);
[v:mainemptygray] () at (1,1);
[v:maingray] () at (-0.8,0.8);
[v:mainemptygray] () at (-1,1);
[v:maingray] () at (0.8,-0.8);
[v:mainemptygray] () at (1,-1);
[v:maingray] () at (-0.8,-0.8);
[v:mainemptygray] () at (-1,-1);

() at (0,0.3)$C$;
() at (-0.8,0.45)$a_{1}$;
() at (-0.8,-0.45)$b_{1}$;
() at (0.8,0.45)$b_{2}$;
() at (0.8,-0.45)$a_{2}$;

[e:marker,color=myLightBlue] (0.6,0.6) – (1.2,1.2);
[e:marker,color=myLightBlue] (1.2,1.2) – (1.2,-1.2);
[e:marker,color=myLightBlue] (1.2,-1.2) – (0.6,-0.6);
[e:marker,color=myGreen] (-0.6,0.6) – (-1.2,1.2);
[e:marker,color=myGreen] (-1.2,1.2) – (-1.2,-1.2);
[e:marker,color=myGreen] (-1.2,-1.2) – (-0.6,-0.6);

[e:main] (-0.6,-0.6) rectangle (0.6,0.6);

[e:main,color=gray] (0.8,0.8) – (1,1);
[e:main,color=gray] (-0.8,0.8) – (-1,1);
[e:main,color=gray] (0.8,-0.8) – (1,-1);
[e:main,color=gray] (-0.8,-0.8) – (-1,-1);

[e:coloredborder] (1,1) – (1.2,1.2);
[e:coloredthin,color=BostonUniversityRed] (1,1) – (1.2,1.2);
[e:coloredborder] (0.6,0.6) – (0.8,0.8);
[e:coloredthin,color=BostonUniversityRed] (0.6,0.6) – (0.8,0.8);
[e:coloredborder] (1,-1) – (1.2,-1.2);
[e:coloredthin,color=BostonUniversityRed] (1,-1) – (1.2,-1.2);
[e:coloredborder] (0.6,-0.6) – (0.8,-0.8);
[e:coloredthin,color=BostonUniversityRed] (0.6,-0.6) – (0.8,-0.8);
[e:coloredborder] (-1,1) – (-1.2,1.2);
[e:coloredthin,color=BostonUniversityRed] (-1,1) – (-1.2,1.2);
[e:coloredborder] (-0.6,0.6) – (-0.8,0.8);
[e:coloredthin,color=BostonUniversityRed] (-0.6,0.6) – (-0.8,0.8);
[e:coloredborder] (-1,-1) – (-1.2,-1.2);
[e:coloredthin,color=BostonUniversityRed] (-1,-1) – (-1.2,-1.2);
[e:coloredborder] (-0.6,-0.6) – (-0.8,-0.8);
[e:coloredthin,color=BostonUniversityRed] (-0.6,-0.6) – (-0.8,-0.8);

[e:coloredborder] (-0.4,1.2) – (0.4,1.2);
[e:coloredthin,color=BostonUniversityRed] (-0.4,1.2) – (0.4,1.2);
[e:coloredborder] (-0.4,-1.2) – (0.4,-1.2);
[e:coloredthin,color=BostonUniversityRed] (-0.4,-1.2) – (0.4,-1.2);
[e:coloredborder] (-1.2,-0.4) – (-1.2,0.4);
[e:coloredthin,color=BostonUniversityRed] (-1.2,-0.4) – (-1.2,0.4);
[e:coloredborder] (1.2,-0.4) – (1.2,0.4);
[e:coloredthin,color=BostonUniversityRed] (1.2,-0.4) – (1.2,0.4);

[e:main,color=gray] (-1.2,1.2) – (-0.4,1.2);
[e:main,color=gray] (1.2,1.2) – (0.4,1.2);
[e:main,color=gray] (-1.2,-1.2) – (-0.4,-1.2);
[e:main,color=gray] (1.2,-1.2) – (0.4,-1.2);
[e:main,color=gray] (-1.2,-1.2) – (-1.2,-0.4);
[e:main,color=gray] (-1.2,1.2) – (-1.2,0.4);
[e:main,color=gray] (1.2,-1.2) – (1.2,-0.4);
[e:main,color=gray] (1.2,1.2) – (1.2,0.4);

A perfect matching $M_H$ of $H$ where all four vertices of $C$ are matched to vertices of $H-\V{C}$. Two paths, as requested by the assertion of <ref>, are marked.

[v:mainempty] () at (-0.6,-0.6);
[v:main] () at (-0.6,0.6);
[v:main] () at (0.6,-0.6);
[v:mainempty] () at (0.6,0.6);
[v:main] () at (-1.2,-1.2);
[v:mainempty] () at (-1.2,1.2);
[v:mainempty] () at (1.2,-1.2);
[v:main] () at (1.2,1.2);

[v:mainemptygray] () at (0.4,1.2);
[v:maingray] () at (0.4,-1.2);
[v:maingray] () at (-0.4,1.2);
[v:mainemptygray] () at (-0.4,-1.2);

[v:mainemptygray] () at (-1.2,-0.4);
[v:maingray] () at (-1.2,0.4);
[v:maingray] () at (1.2,-0.4);
[v:mainemptygray] () at (1.2,0.4);

[v:maingray] () at (0.8,0.8);
[v:mainemptygray] () at (1,1);
[v:maingray] () at (-0.8,0.8);
[v:mainemptygray] () at (-1,1);
[v:maingray] () at (0.8,-0.8);
[v:mainemptygray] () at (1,-1);
[v:maingray] () at (-0.8,-0.8);
[v:mainemptygray] () at (-1,-1);

() at (0,0.3)$C$;
() at (-0.8,0.45)$a_{1}$;
() at (-0.8,-0.45)$b_{1}$;
() at (0.8,0.45)$b_{2}$;
() at (0.8,-0.45)$a_{2}$;

[e:marker,color=myGreen] (0.6,0.6) – (1.2,1.2);
[e:marker,color=myGreen] (1.2,1.2) – (1.2,-1.2);
[e:marker,color=myGreen] (1.2,-1.2) – (0.6,-0.6);
[e:marker,color=myLightBlue] (-0.6,0.6) – (-1.2,1.2);
[e:marker,color=myLightBlue] (-1.2,1.2) – (-1.2,-1.2);
[e:marker,color=myLightBlue] (-1.2,-1.2) – (-0.6,-0.6);

[e:main,color=gray] (-1.2,1.2) – (-0.4,1.2);
[e:main,color=gray] (1.2,1.2) – (0.4,1.2);
[e:main,color=gray] (-1.2,-1.2) – (-0.4,-1.2);
[e:main,color=gray] (1.2,-1.2) – (0.4,-1.2);
[e:main,color=gray] (-1.2,-0.4) – (-1.2,0.4);
[e:main,color=gray] (1.2,-0.4) – (1.2,0.4);

[e:main] (0.6,0.6) – (0.6,-0.6) – (-0.6,-0.6) – (-0.6,0.6);

[e:main,color=gray] (1,1) – (1.2,1.2);
[e:main,color=gray] (0.6,0.6) – (0.8,0.8);
[e:main,color=gray] (1,-1) – (1.2,-1.2);
[e:main,color=gray] (0.6,-0.6) – (0.8,-0.8);
[e:main,color=gray] (-1,1) – (-1.2,1.2);
[e:main,color=gray] (-0.6,0.6) – (-0.8,0.8);
[e:main,color=gray] (-1,-1) – (-1.2,-1.2);
[e:main,color=gray] (-0.6,-0.6) – (-0.8,-0.8);

[e:coloredborder] (-0.4,1.2) – (0.4,1.2);
[e:coloredthin,color=BostonUniversityRed] (-0.4,1.2) – (0.4,1.2);
[e:coloredborder] (-0.4,-1.2) – (0.4,-1.2);
[e:coloredthin,color=BostonUniversityRed] (-0.4,-1.2) – (0.4,-1.2);

[e:coloredborder] (-0.6,0.6) – (0.6,0.6);
[e:coloredthin,color=BostonUniversityRed] (-0.6,0.6) – (0.6,0.6);

[e:coloredborder](-1.2,-1.2) – (-1.2,-0.4);
[e:coloredthin,color=BostonUniversityRed](-1.2,-1.2) – (-1.2,-0.4);

[e:coloredborder](-1.2,1.2) – (-1.2,0.4);
[e:coloredthin,color=BostonUniversityRed](-1.2,1.2) – (-1.2,0.4);

[e:coloredborder](1.2,-1.2) – (1.2,-0.4);
[e:coloredthin,color=BostonUniversityRed](1.2,-1.2) – (1.2,-0.4);

[e:coloredborder](1.2,1.2) – (1.2,0.4);
[e:coloredthin,color=BostonUniversityRed](1.2,1.2) – (1.2,0.4);

[e:coloredborder] (0.8,0.8) – (1,1);
[e:coloredthin,color=BostonUniversityRed](0.8,0.8) – (1,1);
[e:coloredborder] (-0.8,0.8) – (-1,1);
[e:coloredthin,color=BostonUniversityRed](-0.8,0.8) – (-1,1);
[e:coloredborder] (0.8,-0.8) – (1,-1);
[e:coloredthin,color=BostonUniversityRed](0.8,-0.8) – (1,-1);
[e:coloredborder] (-0.8,-0.8) – (-1,-1);
[e:coloredthin,color=BostonUniversityRed](-0.8,-0.8) – (-1,-1);

[v:mainempty] () at (-0.6,-0.6);
[v:main] () at (-0.6,0.6);
[v:main] () at (0.6,-0.6);
[v:mainempty] () at (0.6,0.6);
[v:main] () at (-1.2,-1.2);
[v:mainempty] () at (-1.2,1.2);
[v:mainempty] () at (1.2,-1.2);
[v:main] () at (1.2,1.2);

[v:mainemptygray] () at (0.4,1.2);
[v:maingray] () at (0.4,-1.2);
[v:maingray] () at (-0.4,1.2);
[v:mainemptygray] () at (-0.4,-1.2);

[v:mainemptygray] () at (-1.2,-0.4);
[v:maingray] () at (-1.2,0.4);
[v:maingray] () at (1.2,-0.4);
[v:mainemptygray] () at (1.2,0.4);

[v:maingray] () at (0.8,0.8);
[v:mainemptygray] () at (1,1);
[v:maingray] () at (-0.8,0.8);
[v:mainemptygray] () at (-1,1);
[v:maingray] () at (0.8,-0.8);
[v:mainemptygray] () at (1,-1);
[v:maingray] () at (-0.8,-0.8);
[v:mainemptygray] () at (-1,-1);

() at (0,0.3)$C$;
() at (-0.8,0.45)$a_{1}$;
() at (-0.8,-0.45)$b_{1}$;
() at (0.8,0.45)$b_{2}$;
() at (0.8,-0.45)$a_{2}$;

[e:marker,color=myLightBlue] (-0.6,0.6) – (-1.2,1.2);
[e:marker,color=myLightBlue] (-1.2,1.2) – (1.2,1.2);
[e:marker,color=myLightBlue] (1.2,1.2) – (0.6,0.6);

[e:marker,color=myGreen] (-0.6,-0.6) – (-1.2,-1.2);
[e:marker,color=myGreen] (-1.2,-1.2) – (1.2,-1.2);
[e:marker,color=myGreen] (1.2,-1.2) – (0.6,-0.6);

[e:main,color=gray] (-1.2,-1.2) – (-0.4,-1.2);
[e:main,color=gray] (1.2,-1.2) – (0.4,-1.2);

[e:main,color=gray] (-1.2,-1.2) – (-1.2,0.4);
[e:main,color=gray] (-1.2,1.2) – (-1.2,-0.4);
[e:main,color=gray] (1.2,-1.2) – (1.2,0.4);
[e:main,color=gray] (1.2,1.2) – (1.2,-0.4);

[e:main] (0.6,0.6) – (0.6,-0.6) – (-0.6,-0.6) – (-0.6,0.6);

[e:main,color=gray] (1,1) – (1.2,1.2);
[e:main,color=gray] (0.6,0.6) – (0.8,0.8);

[e:main,color=gray] (-1,1) – (-1.2,1.2);
[e:main,color=gray] (-0.6,0.6) – (-0.8,0.8);

[e:main,color=gray] (-0.4,1.2) – (0.4,1.2);

[e:coloredborder] (-0.6,0.6) – (0.6,0.6);
[e:coloredthin,color=BostonUniversityRed] (-0.6,0.6) – (0.6,0.6);

[e:coloredborder] (-0.4,-1.2) – (0.4,-1.2);
[e:coloredthin,color=BostonUniversityRed] (-0.4,-1.2) – (0.4,-1.2);

[e:coloredborder] (-1.2,-0.4) – (-1.2,0.4);
[e:coloredthin,color=BostonUniversityRed] (-1.2,-0.4) – (-1.2,0.4);
[e:coloredborder] (1.2,-0.4) – (1.2,0.4);
[e:coloredthin,color=BostonUniversityRed] (1.2,-0.4) – (1.2,0.4);

[e:coloredborder] (-1,-1) – (-1.2,-1.2);
[e:coloredthin,color=BostonUniversityRed] (-1,-1) – (-1.2,-1.2);
[e:coloredborder] (-0.6,-0.6) – (-0.8,-0.8);
[e:coloredthin,color=BostonUniversityRed] (-0.6,-0.6) – (-0.8,-0.8);
[e:coloredborder] (0.6,-0.6) – (0.8,-0.8);
[e:coloredthin,color=BostonUniversityRed] (0.6,-0.6) – (0.8,-0.8);
[e:coloredborder] (1,-1) – (1.2,-1.2);
[e:coloredthin,color=BostonUniversityRed] (1,-1) – (1.2,-1.2);

[e:coloredborder] (0.8,0.8) – (1,1);
[e:coloredthin,color=BostonUniversityRed](0.8,0.8) – (1,1);
[e:coloredborder] (-0.8,0.8) – (-1,1);
[e:coloredthin,color=BostonUniversityRed](-0.8,0.8) – (-1,1);

[e:coloredborder] (-1.2,1.2) – (-0.4,1.2);
[e:coloredthin,color=BostonUniversityRed] (-1.2,1.2) – (-0.4,1.2);
[e:coloredborder] (1.2,1.2) – (0.4,1.2);
[e:coloredthin,color=BostonUniversityRed] (1.2,1.2) – (0.4,1.2);

[v:mainempty] () at (-0.6,-0.6);
[v:main] () at (-0.6,0.6);
[v:main] () at (0.6,-0.6);
[v:mainempty] () at (0.6,0.6);
[v:main] () at (-1.2,-1.2);
[v:mainempty] () at (-1.2,1.2);
[v:mainempty] () at (1.2,-1.2);
[v:main] () at (1.2,1.2);

[v:mainemptygray] () at (0.4,1.2);
[v:maingray] () at (0.4,-1.2);
[v:maingray] () at (-0.4,1.2);
[v:mainemptygray] () at (-0.4,-1.2);

[v:mainemptygray] () at (-1.2,-0.4);
[v:maingray] () at (-1.2,0.4);
[v:maingray] () at (1.2,-0.4);
[v:mainemptygray] () at (1.2,0.4);

[v:maingray] () at (0.8,0.8);
[v:mainemptygray] () at (1,1);
[v:maingray] () at (-0.8,0.8);
[v:mainemptygray] () at (-1,1);
[v:maingray] () at (0.8,-0.8);
[v:mainemptygray] () at (1,-1);
[v:maingray] () at (-0.8,-0.8);
[v:mainemptygray] () at (-1,-1);

() at (0,0.3)$C$;
() at (-0.8,0.45)$a_{1}$;
() at (-0.8,-0.45)$b_{1}$;
() at (0.8,0.45)$b_{2}$;
() at (0.8,-0.45)$a_{2}$;

[e:marker,color=myGreen] (0.6,0.6) – (1.2,1.2);
[e:marker,color=myGreen] (1.2,1.2) – (1.2,-1.2);
[e:marker,color=myGreen] (1.2,-1.2) – (0.6,-0.6);
[e:marker,color=myLightBlue] (-0.6,0.6) – (-1.2,1.2);
[e:marker,color=myLightBlue] (-1.2,1.2) – (-1.2,-1.2);
[e:marker,color=myLightBlue] (-1.2,-1.2) – (-0.6,-0.6);

[e:main,color=gray] (-1.2,1.2) – (-0.4,1.2);
[e:main,color=gray] (1.2,1.2) – (0.4,1.2);
[e:main,color=gray] (-1.2,-1.2) – (-0.4,-1.2);
[e:main,color=gray] (1.2,-1.2) – (0.4,-1.2);
[e:main,color=gray] (-1.2,-0.4) – (-1.2,0.4);
[e:main,color=gray] (1.2,-0.4) – (1.2,0.4);

[e:main] (0.6,0.6) – (0.6,-0.6);
[e:main] (-0.6,-0.6) – (-0.6,0.6);

[e:main,color=gray] (1,1) – (1.2,1.2);
[e:main,color=gray] (0.6,0.6) – (0.8,0.8);
[e:main,color=gray] (1,-1) – (1.2,-1.2);
[e:main,color=gray] (0.6,-0.6) – (0.8,-0.8);
[e:main,color=gray] (-1,1) – (-1.2,1.2);
[e:main,color=gray] (-0.6,0.6) – (-0.8,0.8);
[e:main,color=gray] (-1,-1) – (-1.2,-1.2);
[e:main,color=gray] (-0.6,-0.6) – (-0.8,-0.8);

[e:coloredborder] (-0.4,1.2) – (0.4,1.2);
[e:coloredthin,color=BostonUniversityRed] (-0.4,1.2) – (0.4,1.2);
[e:coloredborder] (-0.4,-1.2) – (0.4,-1.2);
[e:coloredthin,color=BostonUniversityRed] (-0.4,-1.2) – (0.4,-1.2);

[e:coloredborder] (-0.6,0.6) – (0.6,0.6);
[e:coloredthin,color=BostonUniversityRed] (-0.6,0.6) – (0.6,0.6);
[e:coloredborder] (-0.6,-0.6) – (0.6,-0.6);
[e:coloredthin,color=BostonUniversityRed] (-0.6,-0.6) – (0.6,-0.6);

[e:coloredborder](-1.2,-1.2) – (-1.2,-0.4);
[e:coloredthin,color=BostonUniversityRed](-1.2,-1.2) – (-1.2,-0.4);

[e:coloredborder](-1.2,1.2) – (-1.2,0.4);
[e:coloredthin,color=BostonUniversityRed](-1.2,1.2) – (-1.2,0.4);

[e:coloredborder](1.2,-1.2) – (1.2,-0.4);
[e:coloredthin,color=BostonUniversityRed](1.2,-1.2) – (1.2,-0.4);

[e:coloredborder](1.2,1.2) – (1.2,0.4);
[e:coloredthin,color=BostonUniversityRed](1.2,1.2) – (1.2,0.4);

[e:coloredborder] (0.8,0.8) – (1,1);
[e:coloredthin,color=BostonUniversityRed](0.8,0.8) – (1,1);
[e:coloredborder] (-0.8,0.8) – (-1,1);
[e:coloredthin,color=BostonUniversityRed](-0.8,0.8) – (-1,1);
[e:coloredborder] (0.8,-0.8) – (1,-1);
[e:coloredthin,color=BostonUniversityRed](0.8,-0.8) – (1,-1);
[e:coloredborder] (-0.8,-0.8) – (-1,-1);
[e:coloredthin,color=BostonUniversityRed](-0.8,-0.8) – (-1,-1);

[v:mainempty] () at (-0.6,-0.6);
[v:main] () at (-0.6,0.6);
[v:main] () at (0.6,-0.6);
[v:mainempty] () at (0.6,0.6);
[v:main] () at (-1.2,-1.2);
[v:mainempty] () at (-1.2,1.2);
[v:mainempty] () at (1.2,-1.2);
[v:main] () at (1.2,1.2);

[v:mainemptygray] () at (0.4,1.2);
[v:maingray] () at (0.4,-1.2);
[v:maingray] () at (-0.4,1.2);
[v:mainemptygray] () at (-0.4,-1.2);

[v:mainemptygray] () at (-1.2,-0.4);
[v:maingray] () at (-1.2,0.4);
[v:maingray] () at (1.2,-0.4);
[v:mainemptygray] () at (1.2,0.4);

[v:maingray] () at (0.8,0.8);
[v:mainemptygray] () at (1,1);
[v:maingray] () at (-0.8,0.8);
[v:mainemptygray] () at (-1,1);
[v:maingray] () at (0.8,-0.8);
[v:mainemptygray] () at (1,-1);
[v:maingray] () at (-0.8,-0.8);
[v:mainemptygray] () at (-1,-1);

() at (0,0.3)$C$;
() at (-0.8,0.45)$a_{1}$;
() at (-0.8,-0.45)$b_{1}$;
() at (0.8,0.45)$b_{2}$;
() at (0.8,-0.45)$a_{2}$;

[e:marker,color=myLightBlue] (-0.6,0.6) – (-1.2,1.2);
[e:marker,color=myLightBlue] (-1.2,1.2) – (1.2,1.2);
[e:marker,color=myLightBlue] (1.2,1.2) – (0.6,0.6);

[e:marker,color=myGreen] (-0.6,-0.6) – (-1.2,-1.2);
[e:marker,color=myGreen] (-1.2,-1.2) – (1.2,-1.2);
[e:marker,color=myGreen] (1.2,-1.2) – (0.6,-0.6);

[e:main] (0.6,0.6) – (0.6,-0.6);
[e:main] (-0.6,-0.6) – (-0.6,0.6);

[e:main,color=gray] (1,1) – (1.2,1.2);
[e:main,color=gray] (0.6,0.6) – (0.8,0.8);
[e:main,color=gray] (1,-1) – (1.2,-1.2);
[e:main,color=gray] (0.6,-0.6) – (0.8,-0.8);
[e:main,color=gray] (-1,1) – (-1.2,1.2);
[e:main,color=gray] (-0.6,0.6) – (-0.8,0.8);
[e:main,color=gray] (-1,-1) – (-1.2,-1.2);
[e:main,color=gray] (-0.6,-0.6) – (-0.8,-0.8);

[e:main,color=gray] (-0.4,1.2) – (0.4,1.2);
[e:main,color=gray] (-0.4,-1.2) – (0.4,-1.2);

[e:main,color=gray] (1.2,1.2) – (1.2,0.4);
[e:main,color=gray] (1.2,-1.2) – (1.2,-0.4);
[e:main,color=gray] (-1.2,-1.2) – (-1.2,0.4);
[e:main,color=gray] (-1.2,-1.2) – (-1.2,-0.4);

[e:coloredborder] (-0.6,0.6) – (0.6,0.6);
[e:coloredthin,color=BostonUniversityRed] (-0.6,0.6) – (0.6,0.6);
[e:coloredborder] (-0.6,-0.6) – (0.6,-0.6);
[e:coloredthin,color=BostonUniversityRed] (-0.6,-0.6) – (0.6,-0.6);

[e:coloredborder] (0.8,0.8) – (1,1);
[e:coloredthin,color=BostonUniversityRed](0.8,0.8) – (1,1);
[e:coloredborder] (-0.8,0.8) – (-1,1);
[e:coloredthin,color=BostonUniversityRed](-0.8,0.8) – (-1,1);
[e:coloredborder] (0.8,-0.8) – (1,-1);
[e:coloredthin,color=BostonUniversityRed](0.8,-0.8) – (1,-1);
[e:coloredborder] (-0.8,-0.8) – (-1,-1);
[e:coloredthin,color=BostonUniversityRed](-0.8,-0.8) – (-1,-1);

[e:coloredborder] (-1.2,-0.4) – (-1.2,0.4);
[e:coloredthin,color=BostonUniversityRed] (-1.2,-0.4) – (-1.2,0.4);
[e:coloredborder] (1.2,-0.4) – (1.2,0.4);
[e:coloredthin,color=BostonUniversityRed] (1.2,-0.4) – (1.2,0.4);

[e:coloredborder] (-1.2,1.2) – (-0.4,1.2);
[e:coloredthin,color=BostonUniversityRed] (-1.2,1.2) – (-0.4,1.2);
[e:coloredborder] (1.2,1.2) – (0.4,1.2);
[e:coloredthin,color=BostonUniversityRed] (1.2,1.2) – (0.4,1.2);

[e:coloredborder] (-1.2,-1.2) – (-0.4,-1.2);
[e:coloredthin,color=BostonUniversityRed] (-1.2,-1.2) – (-0.4,-1.2);
[e:coloredborder] (1.2,-1.2) – (0.4,-1.2);
[e:coloredthin,color=BostonUniversityRed] (1.2,-1.2) – (0.4,-1.2);

A bisubdivision of the cube together with a perfect matching.
Two paths, as requested by the assertion of <ref>, are marked.
All other cases, in particular the ones regarding to the exact identity of $e$, can be derived from this by symmetry.

Let $B$ be a brace and $C$ a $4$-cycle in $B$ as well as $P_1$, $P_2$ two paths that form a conformal cross over $C$.
Then for every $e\in C$, $C+P_1+P_2$ has a perfect matching $M_e$ such that $\Set{e}= M_e\cap\E{C}$.

By definition, since $P_1$ and $P_2$ form a conformal cross over $C$, there is a perfect matching $M$ of $H\coloneqq C+P_1+P_2$.
Since $C$ is a $4$-cycle, $P_1$ and $P_2$ each must connect two vertices of the same colour on $C$ and thus $M$ must contain exactly one edge, say $e'$, of $C$ since the cross is conformal, hence $M_{e'}\coloneqq M$.
Let $e\in\E{C}\setminus\Set{e'}$ be another edge of $C$.
If $e$ and $e'$ are disjoint, then $P_1+P_2+e+e'$ is an $M$-conformal cycle which contains all edges of $M_{e'}$ and thus, $M_e\coloneqq \E{P_1+P_2+e+e'}\setminus M_{e'}$ is a perfect matching containing $e$.
Otherwise, $e$ and $e'$ share exactly one endpoint and $e$ is incident with an endpoint of $P_i$ for some $i\in[1,2]$.
Then $P_i+e+e'$ is an $M_{e'}$-conformal cycle and thus $M_e\coloneqq \Brace{M_{e'}\setminus\E{P_i+e+e'}}\cup\Brace{\E{P_i+e+e'}\setminus M_{e'}}$ is a perfect matching as required.

A last and essential tool before we dive into the more specific cases of $K_{3,3}$-free and $K_{3,3}$-containing braces is the observation on conformal crosses over $4$-cycles in form of <ref>.

[v:main] () at (-1.2,-1.2);
[v:mainempty] () at (-1.2,1.2);

[v:mainempty] () at (1.2,-1.2);
[v:main] () at (1.2,1.2);

() at (-1.45,1)$C$;
[v:main] () at (0.6,0);
[v:mainempty] () at (-0.6,0);

[v:mainemptygray] () at (0.2,0);
[v:maingray] () at (-0.2,0);

[v:mainemptygray] () at (0.8,-0.4);
[v:maingray] () at (1,-0.8);
[v:maingray] () at (-0.8,-0.4);
[v:mainemptygray] () at (-1,-0.8);

[v:maingray] () at (-0.2,0.27);
[v:mainemptygray] () at (0.6,0.8);
[v:mainemptygray] () at (0.2,0.27);
[v:maingray] at (-0.6,0.8);

[e:marker,color=myLightBlue] (-1.2,1.2) – (0.6,0);
[e:marker,color=myLightBlue] (0.6,0) – (1.2,-1.2);

[e:marker,color=myGreen] (1.2,1.2) – (-0.6,0);
[e:marker,color=myGreen ] (-0.6,0) – (-1.2,-1.2);

[e:main] (-1.2,-1.2) – (-1.2,1.2) – (1.2,1.2) – (1.2,-1.2);

[e:main,color=gray] (-1.2,-1.2) – (-1,-0.8);
[e:main,color=gray] (1.2,-1.2) – (1,-0.8);

[e:main,color=gray] (-0.8,-0.4) – (-0.6,0);
[e:main,color=gray] (0.8,-0.4) – (0.6,0);

[e:main,color=gray] (-0.6,0) – (-0.2,0);
[e:main,color=gray] (0.6,0) – (0.2,0);

[e:main,color=gray] (-0.2,0.27) – (0.6,0.8);
[e:main,color=gray] (0.2,0.27) – (-0.6,0.8);

[e:coloredborder] (-1.2,-1.2) – (1.2,-1.2);
[e:coloredthin,color=BostonUniversityRed] (-1.2,-1.2) – (1.2,-1.2);

[e:coloredborder] (-0.2,0) – (0.2,0);
[e:coloredthin,color=BostonUniversityRed] (-0.2,0) – (0.2,0);

[e:coloredborder] (-1,-0.8) – (-0.8,-0.4);
[e:coloredthin,color=BostonUniversityRed] (-1,-0.8) – (-0.8,-0.4);
[e:coloredborder] (1,-0.8) – (0.8,-0.4);
[e:coloredthin,color=BostonUniversityRed] (1,-0.8) – (0.8,-0.4);

[e:coloredborder] (0.6,0) – (0.2,0.27);
[e:coloredthin,color=BostonUniversityRed] (0.6,0) – (0.2,0.27);
[e:coloredborder] (-0.6,0) – (-0.2,0.27);
[e:coloredthin,color=BostonUniversityRed] (-0.6,0) – (-0.2,0.27);

[e:coloredborder] (0.6,0.8) – (1.2,1.2);
[e:coloredthin,color=BostonUniversityRed] (0.6,0.8) – (1.2,1.2);

[e:coloredborder] (-0.6,0.8) – (-1.2,1.2);
[e:coloredthin,color=BostonUniversityRed] (-0.6,0.8) – (-1.2,1.2);

A bisubdivision of $K_{3,3}$ together with a perfect matching $M$ and two disjoint $M$-alternating paths that form a conformal cross over the $4$-cycle $C$.

First, assume that there exists a conformal bisubdivision $H$ of $K_{3,3}$ that has $C$ as a subgraph.
Then we may choose a perfect matching $M$ of $B$ such that it contains a perfect matching of $H$ as seen in <ref>.
The conformal cross over $C$ is also presented in the same figure.

For the reverse let $M$ be a perfect matching of $B$ and $P_1$, $P_2$ two $M$-alternating paths that form an $M$-conformal cross over $C$.
Since both, $P_1$ and $P_2$, must be of even length, they contain at least one inner vertex $v_1$ and $v_2$ respectively.
Moreover, no endpoint of $P_1$ belongs to the same colour class of an endpoint of $P_2$, and thus $v_1$ and $v_2$ can be chosen such that they belong to different colour classes, and, for each $i$, the colour class of $v_i$ is different from the colour class of the endpoints of $P_i$.
Let $e\in M\cap\E{C}$ be the unique edge of $C$ that belongs to $M$.
With $B$ being a brace and <ref> there must exist an internally $M$-conformal $v_1$-$v_2$ path $Q$ that avoids $e$.
For each $i\in[1,2]$ let $x_i$ be the endpoint of $P_i$ that is not incident with $e$.
We claim that $Q$ contains a subpath $R$ which is internally disjoint from $P_1$ and $P_2$, and for each $i\in[1,2]$ an endpoint of $R$ is an inner vertex of $P_i$.
Let $u_1$ be the last vertex of $Q$ when traversing along $Q$ starting in $v_1$, which belongs to $P_1$.
Then, as $P_1$ is $M$-alternating and the only vertex of $P_1$ not covered by $M$ within the path belongs to $e$, $u_1$ must be incident to an edge of $M\cap \E{P_1}$, and thus $P_1u_1$ is of even length.
Hence $u_1$ and $v_1$ belong to the same colour class of $B$, which is different from the colour of $x_1$.
Clearly, $u_1Q$ still contains an inner vertex of $P_2$, let $u_2$ be the first vertex of $P_2$ we encounter when traversing along $u_1Q$ starting in $u_1$.
By the same arguments as above, the edge of $M$ that covers $u_2$ must belong to $P_2$ and thus $u_2$ and $v_2$ have the same colour, which is different from $v_2$.
Consequently, $u_2\neq v_2$ and $R\coloneqq u_1Qu_2$ is an internally $M$-conformal path as required.
Since $R$ is internally $M$-conformal and $C+P_1+P_2$ is $M$-conformal, $H\coloneqq C+P_1+P_2+R$ is also $M$-conformal.
Moreover, for each $i\in[1,2]$, the vertex $u_i$ divides $P_i$ into two subpaths, and as $u_i$ has a different colour than any of the two endpoints of $P_i$, both of these paths must be of odd length.
Hence $H$ is a conformal bisubdivision of $K_{3,3}$ in $B$.

§ MATCHING CROSSES IN $K_{3,3}$-FREE BRACES

To proceed towards the proof of <ref>, we need to describe how the structure of $K_{3,3}$-free braces, especially the non-planar $K_{3,3}$-free braces we obtain via the trisum operation from <ref> by using planar braces as the base building blocks, behave regarding the existence of matching crosses.
The purpose of this section is to establish the $K_{3,3}$-free part of <ref> in the form of <ref>.

In what follows we are concerned with $K_{3,3}$-free braces that are not planar.
By <ref> there is a single exception to the $K_{3,3}$-free braces constructed from planar braces by the trisum operation, namely the Heawood graph.
While the Heawood graph does not contain a single $4$-cycle, in order to prove <ref>, we have to discuss its cycles.

Let $C$ be a conformal cycle of the Heawood graph $H_{14}$, then there exists a conformal cross over $C$ in $H_{14}$.

in 2,4,6,8,10,12,14

[v:main] () at (*25.71:15mm);

in 1,3,5,7,9,11,13

[v:mainempty] () at (*25.71:15mm);

[e:marker,myGreen] (257.1:15mm) – (25.65:15mm);
[e:marker,myGreen] (257.1:15mm) – (282.81:15mm);
[e:marker,myGreen] (282.81:15mm) – (308.52:15mm);

[e:marker,myLightBlue] (0:15mm) – (128.55:15mm);
[e:marker,myLightBlue] (77.13:15mm) – (102.84:15mm);
[e:marker,myLightBlue] (102.84:15mm) – (128.55:15mm);
[e:marker] (308.52:15mm) – (334.23:15mm);
[e:marker] (334.23:15mm) – (0:15mm);
[e:marker] (0:15mm) – (25.71:15mm);
[e:marker] (25.71:15mm) – (51.42:15mm);
[e:marker] (51.42:15mm) – (77.13:15mm);
[e:marker] (308.52:15mm) – (77.13:15mm);

in 2,4,6,8,10,12,14

[e:main] (*25.71:15mm) to (128.55+*25.71:15mm);

in 1,3,5,7,9,11,13

[e:main] (*25.71:15mm) to (25.71+*25.71:15mm);

in 2,4,6,8,10,12,14

[e:coloredborder] (*25.71:15mm) to (25.71+*25.71:15mm);
[e:coloredthin,color=BostonUniversityRed] (*25.71:15mm) to (25.71+*25.71:15mm);

in 2,4,6,8,10,12,14

[v:main] () at (*25.71:15mm);

in 1,3,5,7,9,11,13

[v:mainempty] () at (*25.71:15mm);

[e:marker,myGreen] (231.39:15mm) – (257.1:15mm);
[e:marker,myGreen] (257.1:15mm) – (282.81:15mm);
[e:marker,myGreen] (282.81:15mm) – (308.52:15mm);

[e:marker,myLightBlue] (334.23:15mm) – (0:15mm);
[e:marker,myLightBlue] (0:15mm) – (25.71:15mm);
[e:marker,myLightBlue] (25.71:15mm) – (51.42:15mm);

[e:marker] (308.52:15mm) – (77.13:15mm);
[e:marker] (308.52:15mm) – (334.23:15mm);
[e:marker] (334.23:15mm) – (205.68:15mm);
[e:marker] (205.68:15mm) – (231.39:15mm);
[e:marker] (231.39:15mm) – (102.84:15mm);
[e:marker] (102.85:15mm) – (128.55:15mm);
[e:marker] (128.55:15mm) – (154.26:15mm);
[e:marker] (154.26:15mm) – (179.97:15mm);
[e:marker] (179.97:15mm) – (51.42:15mm);
[e:marker] (51.42:15mm) – (77.13:15mm);

in 2,4,6,8,10,12,14

[e:main] (*25.71:15mm) to (128.55+*25.71:15mm);

in 1,3,5,7,9,11,13

[e:main] (*25.71:15mm) to (25.71+*25.71:15mm);

in 2,4,6,8,10,12,14

[e:coloredborder] (*25.71:15mm) to (25.71+*25.71:15mm);
[e:coloredthin,color=BostonUniversityRed] (*25.71:15mm) to (25.71+*25.71:15mm);

in 2,4,6,8,10,12,14

[v:main] () at (*25.71:15mm);

in 1,3,5,7,9,11,13

[v:mainempty] () at (*25.71:15mm);

[e:marker,myLightBlue] (0:15mm) – (128.55:15mm);
[e:marker,myGreen] (257.1:15mm) – (25.65:15mm);

in 1,...,14

[e:marker] (*25.71:15mm) to (25.71+*25.71:15mm);

in 2,4,6,8,10,12,14

[e:main] (*25.71:15mm) to (128.55+*25.71:15mm);

in 1,3,5,7,9,11,13

[e:main] (*25.71:15mm) to (25.71+*25.71:15mm);

in 2,4,6,8,10,12,14

[e:coloredborder] (*25.71:15mm) to (25.71+*25.71:15mm);
[e:coloredthin,color=BostonUniversityRed] (*25.71:15mm) to (25.71+*25.71:15mm);

The Heawood graph $H_{14}$ together with a perfect matching and the three, up to symmetry, different conformal cycles in $H_{14}$.
For each of these cycles, we provide a conformal cross.

It is known that the Heawood graph has, up to automorphisms, exactly one perfect matching, as, for example, the complement of every perfect matching of $H_{14}$ is a Hamilton cycle.
See [1] for a discussion on the matter.
Moreover, no conformal cycle in $H_{14}$ is of a length that is a multiple of four, for more details on that please consult [14].
Indeed, for every fixed length $\ell$ of a conformal cycle in $H_{14}$ it suffices to find a cross for one of them, as, again, the graph is highly symmetric.
Hence in order to prove the assertion, it suffices to fix a perfect matching and check conformal cycles of length $14$, $10$, and $6$.
This is done in <ref>.

Let us first observe that the $4$-cycle itself, on which a trisum operation has been performed, must have a strong matching cross.

Let $B$ be a $K_{3,3}$-free brace that is not the Heawood graph, $\ell\geq 3$, $B_1,\dots,B_{\ell}$ braces such that $B$ is a maximal $4$-cycle sum of $B_1,\dots,B_{\ell}$ at the $4$-cycle $C=\Brace{a_1,b_1,a_2,b_2}$.
Then for any choice of two distinct values $i,j\in[1,\ell]$ and any choice of $x\in\Set{a_1,a_2}$ and $y\in\Set{b_1,b_2}$, there is a strong matching cross over $C$ with paths $P_1$ and $P_2$ in $B_i+B_j$ such that the $P_i$ are $M$-alternating for some perfect matching $M$ of $B$ for which $x$ and $y$ are the two vertices of $C$ which are covered by edges of $M\cap\Brace{\E{P_1}\cup\E{P_2}\cup\E{C}}$.

The lemma is almost identical to <ref>.
In both $B_i$ and $B_j$ there exists a conformal bisubdivision of the cube which has $C$ as a subgraph by <ref> since $B$ cannot contain a conformal bisubdivision of $K_{3,3}$.
Hence $B_i+B_j$ contains a conformal bisubdivision of the graph $H_{12}$, which is obtained by identifying two cubes on one $4$-cycle.
See <ref> for an illustration of a bisubdivision of $H_{12}$.
The figure also shows perfect matchings of the respective conformal bisubdivision one can find, together with two paths that make up a strong matching cross as desired.
The exact matching cross depending on the choices of $x$ and $y$ can be obtained from the paths illustrated in <ref> by symmetry.
By adjusting the perfect matching such that exactly the requested edges belong to $M$, our proof is complete.

In the following sections we sometimes encounter the situation of two $M$-alternating paths which are not disjoint.
However, one can observe that whenever two $M$-alternating paths meet such that their union contains an $M$-conformal cycle $C$, we may switch $M$ along this cycle $C$ to obtain a new perfect matching $N$ together with two $N$-alternating paths that intersect slightly less then before.
This relatively simple observation yields a general and powerful tool for bipartite graphs.

Let $G$ be a bipartite graph with a perfect matching $M$ and $a_1,a_2\in V_1$ and $b_1b_1\in V_2$ four distinct vertices in $G$.
Let further $P_1$ and $P_2$ be two internally $M$-conformal paths such that $P_i$ has endpoints $a_i$ and $b_i$.
Then there exists a perfect matching $M'\in\Perf{G}$ together with two internally $M'$-conformal paths $P'_1$ and $P'_2$ such that:

* $P'_i$ has endpoints $a_i$ and $b_i$ for both $i\in\Set{1,2}$,

* $P'_1+P'_2$ is a subgraph of $P_1+P_2$,

* $M\setminus\E{P_1+P_2}=M'\setminus\E{P_1+P_2}$, and

* either $P'_1\cap P'_2$ is an $M'$-conformal path or $P'_1$ and $P'_2$ are disjoint.

Let $B$ be a $K_{3,3}$-free brace, $\ell\geq 3$, $B_1,\dots,B_{\ell}$ braces such that $B$ is a maximal $4$-cycle sum of $B_1,\dots,B_{\ell}$ at the $4$-cycle $C$ and let $C'$ be a conformal cycle in $B_1$ that also exists in $B$ such that $C'\cap C$ is a non-trivial path.
Then there is a strong matching cross over $C'$ in $B$.

Since $C\cap C'$ is a non-trivial path, let us call it $K$, the two cycles share at least two vertices, and $K$ contains at least one vertex of every colour class.

Let us first assume, $K$ has length one.
Let $M$ be a perfect matching of $B_1$ such that $C'$ is $M$-conformal and $M$ contains as many edges of $C$ as possible.
Let $a\in V_1$ and $b\in V_2$ be the two vertices of $C-\V{K}$.
We divide this case into two subcases, one, where $C$ is $M$-conformal as well and the second, where for every possible choice of $M$ we have $ab\notin M$.

So first let us assume $ab\in M$.
Then, since $B_1$ is a brace and by <ref>, there exists an internally $M$-conformal path $L'$ with $a$ as one endpoint and a vertex $b'$ of $\Vi{2}{C'}$ as its other endpoints such that $L'$ is internally disjoint from $C'$ and avoids $K$.
Similarly, there is an internally $M$-conformal path $R'$ with endpoints $b$ and $a'\in\Vi{1}{C'}\setminus\V{K}$.
By <ref> there exists a perfect matching $M'$ which coincides with $M$ everywhere outside of $L'+R'$ together with internally $M'$-conformal paths $L$ and $R$ linking $a$ and $b'$, and $b$ and $a'$ respectively such that either $L$ and $R$ are disjoint, or $L\cap R$ is an $M'$-conformal path.
In the later case, $ab$ together with the $a$-$b$-subpath of $L+R$ forms an $M'$-conformal cycle $O$.
In a slight abuse of notation let us adjust $M'$[We still keep the name '$M'$' for better readability] in these cases to be the perfect matching $M'\Delta\E{O}$.
Let $u$ and $v$ be the endpoints of $L\cap R$ such that $u$ appears on $L$ before $v$ when traversing along $L$ starting from $a$.
We also might have to adjust our definition of $L$ and $R$ slightly, depending on the way, $L$ and $R$ currently connect to $C'$.
In case we adjust the perfect matching, we also adjust $L$ to be the $M'$-alternating path $LuR$, while $R$ is adjusted to be the path $RvL$.

() at (0,1.05)$C$;
[gray] () at (-1.5,-2.5)$C'$;

() at (-0.4,-0.65)$a$;
() at (0.4,-0.6)$b$;
() at (1.8,-2)$a'$;
() at (-1.8,-2.1)$b'$;

[v:mainemptygray] () at (-0.3,-0.3);
[v:maingray] () at (-0.3,0.3);
[v:maingray] () at (0.3,-0.3);
[v:mainemptygray] () at (0.3,0.3);

[v:main] () at (-0.8,-0.8);
[v:mainempty] () at (-0.8,0.8);
[v:mainempty] () at (0.8,-0.8);
[v:main] () at (0.8,0.8);

[v:mainemptygray] () at (-1.3,-1.3);
[v:maingray] () at (-1.3,1.3);
[v:maingray] () at (1.3,-1.3);
[v:mainemptygray] () at (1.3,1.3);

[v:maingray] () at (-1.8,0);
[v:mainemptygray] () at (1.8,0);

[v:mainemptygray] () at (-1.8,-1.8);
[v:maingray] () at (1.8,-1.8);

[v:maingray] () at (-0.8,-2.6);
[v:mainemptygray] () at (0.8,-2.6);

[e:marker,myLightBlue] (0.8,0.8) – (0.3,0.3);
[e:marker,myLightBlue] (0.3,0.3) – (0.3,-0.3);
[e:marker,myLightBlue] (0.3,-0.3) – (-0.3,-0.3);
[e:marker,myLightBlue] (-0.3,-0.3) – (-0.8,-0.8);
[e:marker,bend left=45,myLightBlue] (-0.8,-0.8) to (-1.8,-1.8);

[e:marker,myGreen] (-0.8,0.8) – (-1.3,1.3);
[e:marker,myGreen] (-1.3,1.3) – (-1.3,-1.3);
[e:marker,myGreen] (-1.3,-1.3) – (1.3,-1.3);
[e:marker,myGreen] (1.3,-1.3) – (0.8,-0.8);
[e:marker,myGreen,bend right=45] (0.8,-0.8) to (1.8,-1.8);

[e:main] (-0.8,-0.8) rectangle (0.8,0.8);

[e:main,color=gray] (-1.3,1.3) – (1.3,1.3) – (1.3,-1.3) – (-1.3,-1.3);

[e:main,color=gray] (0.3,0.3) – (-0.3,0.3) – (-0.3,-0.3) – (0.3,-0.3);

[e:main,color=gray] (0.3,0.3) – (0.8,0.8);
[e:main,color=gray] (0.3,-0.3) – (0.8,-0.8);

[e:main,color=gray] (-0.8,0.8) – (-1.3,1.3);
[e:main,color=gray] (-0.8,-0.8) – (-1.3,-1.3);

[e:main,color=gray] (-0.8,0.8) – (-1.8,0);
[e:main,color=gray] (0.8,0.8) – (1.8,0);

[e:main,color=gray] (-1.8,-1.8) – (-0.8,-2.6);
[e:main,color=gray] (1.8,-1.8) – (0.8,-2.6);

[e:main,color=gray,bend left=45] (-0.8,-0.8) to (-1.8,-1.8);
[e:main,color=gray,bend right=45] (0.8,-0.8) to (1.8,-1.8);

[e:coloredborder] (0.3,-0.3) – (0.3,0.3);
[e:coloredthin,color=BostonUniversityRed] (0.3,-0.3) – (0.3,0.3);

[e:coloredborder] (-0.3,0.3) – (-0.8,0.8);
[e:coloredthin,color=BostonUniversityRed] (-0.3,0.3) – (-0.8,0.8);
[e:coloredborder] (-0.3,-0.3) – (-0.8,-0.8);
[e:coloredthin,color=BostonUniversityRed] (-0.3,-0.3) – (-0.8,-0.8);

[e:coloredborder] (-1.3,-1.3) – (-1.3,1.3);
[e:coloredthin,color=BostonUniversityRed]  (-1.3,-1.3) – (-1.3,1.3);

[e:coloredborder] (0.8,0.8) – (1.3,1.3);
[e:coloredthin,color=BostonUniversityRed] (0.8,0.8) – (1.3,1.3);
[e:coloredborder] (0.8,-0.8) – (1.3,-1.3);
[e:coloredthin,color=BostonUniversityRed] (0.8,-0.8) – (1.3,-1.3);

[e:coloredborder] (-1.8,0) – (-1.8,-1.8);
[e:coloredthin,color=BostonUniversityRed] (-1.8,0) – (-1.8,-1.8);
[e:coloredborder] (1.8,0) – (1.8,-1.8);
[e:coloredthin,color=BostonUniversityRed] (1.8,0) – (1.8,-1.8);

[e:coloredborder] (-0.8,-2.6) – (0.8,-2.6);
[e:coloredthin,color=BostonUniversityRed] (-0.8,-2.6) – (0.8,-2.6);

() at (0,1.05)$C$;
[gray] () at (-1.5,-2.5)$C'$;

() at (-0.4,-0.65)$a$;
() at (0.4,-0.6)$b$;

() at (1.8,-2)$a'$;
() at (-1.8,-2.1)$b'$;

[v:mainemptygray] () at (-0.3,-0.3);
[v:maingray] () at (-0.3,0.3);
[v:maingray] () at (0.3,-0.3);
[v:mainemptygray] () at (0.3,0.3);

[v:main] () at (-0.8,-0.8);
[v:mainempty] () at (-0.8,0.8);
[v:mainempty] () at (0.8,-0.8);
[v:main] () at (0.8,0.8);

[v:mainemptygray] () at (-1.3,-1.3);
[v:maingray] () at (-1.3,1.3);
[v:maingray] () at (1.3,-1.3);
[v:mainemptygray] () at (1.3,1.3);

[v:maingray] () at (-1.8,0);
[v:mainemptygray] () at (1.8,0);

[v:mainemptygray] () at (-1.8,-1.8);
[v:maingray] () at (1.8,-1.8);

[v:maingray] () at (-0.8,-2.6);
[v:mainemptygray] () at (0.8,-2.6);

[v:maingray]() at (0,-1.5);
() at (0.2,-1.7)$v$;

[v:mainemptygray] at (0,-2.1);
[] () at (0,-2.4)$u$;

[e:marker,myGreen,bend right=13] (-1.8,-1.8) to (0,-1.5);
[e:marker,myGreen,bend right=20] (0,-1.5) to (0.8,-0.8);
[e:marker,myGreen] (0.8,-0.8) – (1.3,-1.3);
[e:marker,myGreen] (1.3,-1.3) – (1.3,1.3);
[e:marker,myGreen] (1.3,1.3) – (0.8,0.8);

[e:marker,bend left=10,myLightBlue] (1.8,-1.8) to (0,-2.1);
[e:marker,bend left=20,myLightBlue] (0,-2.1) to (-0.8,-0.8);
[e:marker,myLightBlue] (-0.8,-0.8) – (-0.3,-0.3);
[e:marker,myLightBlue] (-0.3,-0.3) – (-0.3,0.3);
[e:marker,myLightBlue] (-0.3,0.3) – (-0.8,0.8);

[e:main,color=gray] (-1.3,1.3) – (1.3,1.3);
[e:main,color=gray] (1.3,-1.3) – (-1.3,-1.3);

[e:main] (0.8,0.8) – (0.8,-0.8) – (-0.8,-0.8) – (-0.8,0.8);

[e:main,color=gray] (0.3,0.3) – (0.8,0.8);
[e:main,color=gray] (0.3,-0.3) – (0.8,-0.8);

[e:main,color=gray] (0.3,0.3) – (-0.3,0.3);
[e:main,color=gray] (-0.3,-0.3) – (0.3,-0.3);

[e:main,color=gray] (-0.3,0.3) – (-0.8,0.8);
[e:main,color=gray] (-0.3,-0.3) – (-0.8,-0.8);

[e:main,color=gray] (-0.8,0.8) – (-1.3,1.3);
[e:main,color=gray] (-0.8,-0.8) – (-1.3,-1.3);

[e:main,color=gray] (0.8,0.8) – (1.3,1.3);
[e:main,color=gray] (0.8,-0.8) – (1.3,-1.3);

[e:main,color=gray] (-0.8,0.8) – (-1.8,0);
[e:main,color=gray] (0.8,0.8) – (1.8,0);

[e:main,color=gray] (-1.8,-1.8) – (-0.8,-2.6);
[e:main,color=gray] (1.8,-1.8) – (0.8,-2.6);

[e:main,color=gray,bend right=13] (-1.8,-1.8) to (0,-1.5);
[e:main,color=gray,bend left=10] (1.8,-1.8) to (0,-2.1);

[e:main,color=gray,densely dashed] (0,-1.5) – (0,-2.1);

[e:coloredborder] (0.3,-0.3) – (0.3,0.3);
[e:coloredthin,color=BostonUniversityRed] (0.3,-0.3) – (0.3,0.3);

[e:coloredborder] (-0.3,0.3) – (-0.3,-0.3);
[e:coloredthin,color=BostonUniversityRed] (-0.3,0.3) – (-0.3,-0.3);

[e:coloredborder] (-0.8,0.8) – (0.8,0.8);
[e:coloredthin,color=BostonUniversityRed] (-0.8,0.8) – (0.8,0.8);

[e:coloredborder] (-1.3,-1.3) – (-1.3,1.3);
[e:coloredthin,color=BostonUniversityRed]  (-1.3,-1.3) – (-1.3,1.3);

[e:coloredborder] (1.3,1.3) – (1.3,-1.3);
[e:coloredthin,color=BostonUniversityRed] (1.3,1.3) – (1.3,-1.3);

[e:coloredborder] (-1.8,0) – (-1.8,-1.8);
[e:coloredthin,color=BostonUniversityRed] (-1.8,0) – (-1.8,-1.8);
[e:coloredborder] (1.8,0) – (1.8,-1.8);
[e:coloredthin,color=BostonUniversityRed] (1.8,0) – (1.8,-1.8);

[e:coloredborder] (-0.8,-2.6) – (0.8,-2.6);
[e:coloredthin,color=BostonUniversityRed] (-0.8,-2.6) – (0.8,-2.6);

[e:coloredborder,bend right=20] (0,-1.5) to (0.8,-0.8);
[e:coloredthin,color=BostonUniversityRed,bend right=20] (0,-1.5) to (0.8,-0.8);

[e:coloredborder,bend left=20] (0,-2.1) to (-0.8,-0.8);
[e:coloredthin,color=BostonUniversityRed,,bend left=20] (0,-2.1) to (-0.8,-0.8);

() at (0,1.05)$C$;
[gray] () at (-1.5,-2.5)$C'$;

() at (-0.4,-0.65)$a$;
() at (0.4,-0.6)$b$;

[v:mainemptygray] () at (-0.3,-0.3);
[v:maingray] () at (-0.3,0.3);
[v:maingray] () at (0.3,-0.3);
[v:mainemptygray] () at (0.3,0.3);

[v:main] () at (-0.8,-0.8);
[v:mainempty] () at (-0.8,0.8);
[v:mainempty] () at (0.8,-0.8);
[v:main] () at (0.8,0.8);

[v:mainemptygray] () at (-1.3,-1.3);
[v:maingray] () at (-1.3,1.3);
[v:maingray] () at (1.3,-1.3);
[v:mainemptygray] () at (1.3,1.3);

[v:maingray] () at (-1.8,0);
[v:mainemptygray] () at (1.8,0);

[v:mainemptygray] () at (-1.8,-1.8);
[v:maingray] () at (1.8,-1.8);

[v:maingray] () at (-0.8,-2.6);
() at (-0.4,-2.3)$a'$;
[v:mainemptygray] () at (0.8,-2.6);
() at (0.3,-2.3)$b'$;

[e:marker,color=myGreen] (0.8,0.8) – (1.3,1.3);
[e:marker,color=myGreen] (1.3,1.3) – (1.3,-1.3);
[e:marker,color=myGreen] (1.3,-1.3) – (0.8,-0.8);
[e:marker,color=myGreen,bend left=20] (0.8,-0.8) to (0,-1.7);
[e:marker,color=myGreen,bend right=20] (0,-1.7) to (-0.8,-2.6);

[e:marker,myLightBlue] (-0.8,0.8) – (-0.3,0.3);
[e:marker,myLightBlue] (-0.3,0.3) – (-0.3,-0.3);
[e:marker,myLightBlue] (-0.3,-0.3) – (-0.8,-0.8);
[e:marker,bend right=20,myLightBlue] (-0.8,-0.8) to (0,-1.7);
[e:marker,bend left=20,myLightBlue] (0,-1.7) to (0.8,-2.6);

[e:main] (-0.8,-0.8) rectangle (0.8,0.8);

[e:main,color=gray] (-1.3,1.3) – (1.3,1.3) – (1.3,-1.3) – (-1.3,-1.3);

[e:main,color=gray] (0.3,0.3) – (0.8,0.8);
[e:main,color=gray] (0.3,-0.3) – (0.8,-0.8);

[e:main,color=gray] (0.3,0.3) – (-0.3,0.3) – (-0.3,-0.3) – (0.3,-0.3);

[e:main,color=gray] (-0.8,0.8) – (-1.3,1.3);
[e:main,color=gray] (-0.8,-0.8) – (-1.3,-1.3);

[e:main,color=gray] (-0.8,0.8) – (-1.8,0);
[e:main,color=gray] (0.8,0.8) – (1.8,0);

[e:main,color=gray] (-1.8,-1.8) – (-0.8,-2.6);
[e:main,color=gray] (1.8,-1.8) – (0.8,-2.6);

[e:main,color=gray,bend right=20] (-0.8,-0.8) to (0,-1.7);
[e:main,color=gray,bend left=20] (0,-1.7) to (0.8,-2.6);

[e:main,color=gray,bend left=20] (0.8,-0.8) to (0,-1.7);
[e:main,color=gray,bend right=20] (0,-1.7) to (-0.8,-2.6);

[e:coloredborder] (-0.3,0.3) – (-0.8,0.8);
[e:coloredthin,color=BostonUniversityRed] (-0.3,0.3) – (-0.8,0.8);
[e:coloredborder] (-0.3,-0.3) – (-0.8,-0.8);
[e:coloredthin,color=BostonUniversityRed] (-0.3,-0.3) – (-0.8,-0.8);

[e:coloredborder] (0.3,-0.3) – (0.3,0.3);
[e:coloredthin,color=BostonUniversityRed] (0.3,-0.3) – (0.3,0.3);

[e:coloredborder] (-1.3,-1.3) – (-1.3,1.3);
[e:coloredthin,color=BostonUniversityRed]  (-1.3,-1.3) – (-1.3,1.3);

[e:coloredborder] (0.8,0.8) – (1.3,1.3);
[e:coloredthin,color=BostonUniversityRed] (0.8,0.8) – (1.3,1.3);
[e:coloredborder] (0.8,-0.8) – (1.3,-1.3);
[e:coloredthin,color=BostonUniversityRed] (0.8,-0.8) – (1.3,-1.3);

[e:coloredborder] (-1.8,0) – (-1.8,-1.8);
[e:coloredthin,color=BostonUniversityRed] (-1.8,0) – (-1.8,-1.8);
[e:coloredborder] (1.8,0) – (1.8,-1.8);
[e:coloredthin,color=BostonUniversityRed] (1.8,0) – (1.8,-1.8);

[e:coloredborder] (-0.8,-2.6) – (0.8,-2.6);
[e:coloredthin,color=BostonUniversityRed] (-0.8,-2.6) – (0.8,-2.6);

() at (0,1.05)$C$;
[gray] () at (-1.5,-2.5)$C'$;

() at (-0.4,-0.65)$a$;
() at (0.4,-0.6)$b$;

[v:mainemptygray] () at (-0.3,-0.3);
[v:maingray] () at (-0.3,0.3);
[v:maingray] () at (0.3,-0.3);
[v:mainemptygray] () at (0.3,0.3);

[v:main] () at (-0.8,-0.8);
[v:mainempty] () at (-0.8,0.8);
[v:mainempty] () at (0.8,-0.8);
[v:main] () at (0.8,0.8);

[v:mainemptygray] () at (-1.3,-1.3);
[v:maingray] () at (-1.3,1.3);
[v:maingray] () at (1.3,-1.3);
[v:mainemptygray] () at (1.3,1.3);

[v:maingray] () at (-1.8,0);
[v:mainemptygray] () at (1.8,0);

[v:mainemptygray] () at (-1.8,-1.8);
[v:maingray] () at (1.8,-1.8);

[v:maingray] () at (-0.8,-2.6);
() at (-0.3,-2.3)$a'$;
[v:mainemptygray] () at (0.8,-2.6);
() at (0.2,-2.3)$b'$;

[v:mainemptygray] () at (-0.3,-1.7);
() at (-0.7,-1.7)$u$;
[v:maingray] () at (0.3,-1.7);
() at (0.7,-1.7)$v$;

[e:marker,myGreen] (-0.8,0.8) – (-1.3,1.3);
[e:marker,myGreen] (-1.3,1.3) – (-1.3,-1.3);
[e:marker,myGreen] (-1.3,-1.3) – (1.3,-1.3);
[e:marker,myGreen] (1.3,-1.3) – (0.8,-0.8);
[e:marker,myGreen,bend right=20] (0.8,-0.8) to (0.3,-1.7);
[e:marker,myGreen,bend right=20] (0.3,-1.7) to (0.8,-2.6);

[e:marker,myLightBlue] (0.8,0.8) – (0.3,0.3);
[e:marker,myLightBlue] (0.3,0.3) – (0.3,-0.3);
[e:marker,myLightBlue] (0.3,-0.3) – (-0.3,-0.3);
[e:marker,myLightBlue] (-0.3,-0.3) – (-0.8,-0.8);
[e:marker,bend left=20,myLightBlue] (-0.8,-0.8) to (-0.3,-1.7);
[e:marker,bend left=20,myLightBlue] (-0.3,-1.7) to (-0.8,-2.6);

[e:main] (-0.8,-0.8) rectangle (0.8,0.8);

[e:main,color=gray] (-1.3,1.3) – (1.3,1.3) – (1.3,-1.3) – (-1.3,-1.3);

[e:main,color=gray] (0.3,0.3) – (0.8,0.8);
[e:main,color=gray] (0.3,-0.3) – (0.8,-0.8);

[e:main,color=gray] (0.3,0.3) – (-0.3,0.3) – (-0.3,-0.3) – (0.3,-0.3);

[e:main,color=gray] (-0.8,0.8) – (-1.3,1.3);
[e:main,color=gray] (-0.8,-0.8) – (-1.3,-1.3);

[e:main,color=gray] (-0.8,0.8) – (-1.8,0);
[e:main,color=gray] (0.8,0.8) – (1.8,0);

[e:main,color=gray] (-1.8,-1.8) – (-0.8,-2.6);
[e:main,color=gray] (1.8,-1.8) – (0.8,-2.6);

[e:main,color=gray] (-0.8,-2.6) – (0.8,-2.6);

[e:main,color=gray,densely dashed] (-0.3,-1.7) – (0.3,-1.7);

[e:main,color=gray,bend left=20] (-0.8,-0.8) to (-0.3,-1.7);
[e:main,color=gray,bend right=20] (0.8,-0.8) to (0.3,-1.7);

[e:coloredborder] (-0.3,0.3) – (-0.8,0.8);
[e:coloredthin,color=BostonUniversityRed] (-0.3,0.3) – (-0.8,0.8);
[e:coloredborder] (-0.3,-0.3) – (-0.8,-0.8);
[e:coloredthin,color=BostonUniversityRed] (-0.3,-0.3) – (-0.8,-0.8);

[e:coloredborder] (0.3,-0.3) – (0.3,0.3);
[e:coloredthin,color=BostonUniversityRed] (0.3,-0.3) – (0.3,0.3);

[e:coloredborder] (-1.3,-1.3) – (-1.3,1.3);
[e:coloredthin,color=BostonUniversityRed]  (-1.3,-1.3) – (-1.3,1.3);

[e:coloredborder] (0.8,0.8) – (1.3,1.3);
[e:coloredthin,color=BostonUniversityRed] (0.8,0.8) – (1.3,1.3);
[e:coloredborder] (0.8,-0.8) – (1.3,-1.3);
[e:coloredthin,color=BostonUniversityRed] (0.8,-0.8) – (1.3,-1.3);

[e:coloredborder] (-1.8,0) – (-1.8,-1.8);
[e:coloredthin,color=BostonUniversityRed] (-1.8,0) – (-1.8,-1.8);
[e:coloredborder] (1.8,0) – (1.8,-1.8);
[e:coloredthin,color=BostonUniversityRed] (1.8,0) – (1.8,-1.8);

[e:coloredborder,bend left=20] (-0.3,-1.7) to (-0.8,-2.6);
[e:coloredthin,color=BostonUniversityRed,bend left=20] (-0.3,-1.7) to (-0.8,-2.6);

[e:coloredborder,bend right=20] (0.3,-1.7) to (0.8,-2.6);
[e:coloredthin,color=BostonUniversityRed,bend right=20] (0.3,-1.7) to (0.8,-2.6);

The four different ways to obtain a strong matching cross over $C'$ in the first subcase of the first case in the proof of <ref>.

In either case, we can now use <ref> to find a perfect matching $M''$ of $B$ such that $\E{M'}\setminus\E{C}\subseteq M''$ together with two $M''$-alternating paths $P_1$ and $P_2$ such that these paths are disjoint and are completely contained in $B-\V{B_1-\V{C}}$.
Indeed, $P_1$ and $P_2$ can be chosen such that $P_1L$ and $P_2R$ form a strong matching cross over $C'$ in $B$ as illustrated in <ref>.

() at (0,1.05)$C$;
[gray] () at (-1.5,-2.5)$C'$;

() at (-0.4,-0.65)$a$;
() at (0.4,-0.6)$b$;
() at (-0.5,-2.3)$a'$;
() at (0.5,-2.3)$b'$;

[v:mainemptygray] () at (-0.3,-0.3);
[v:maingray] () at (-0.3,0.3);
[v:maingray] () at (0.3,-0.3);
[v:mainemptygray] () at (0.3,0.3);

[v:main] () at (-0.8,-0.8);
[v:mainempty] () at (-0.8,0.8);
[v:mainempty] () at (0.8,-0.8);
[v:main] () at (0.8,0.8);

[v:mainemptygray] () at (-1.3,-1.3);
[v:maingray] () at (-1.3,1.3);
[v:maingray] () at (1.3,-1.3);
[v:mainemptygray] () at (1.3,1.3);

[v:maingray] () at (-1.8,0);
[v:mainemptygray] () at (1.8,0);

[v:mainemptygray] () at (-1.8,-1.8);
[v:maingray] () at (1.8,-1.8);

[v:maingray] () at (-0.8,-2.6);
[v:mainemptygray] () at (0.8,-2.6);

[v:mainemptygray] () at (-0.8,-1.7);
() at (-1.1,-1.7)$x$;

[v:maingray] () at (0.8,-1.7);
() at (1.1,-1.75)$y$;

[e:marker,myLightBlue] (0.8,0.8) – (0.3,0.3);
[e:marker,myLightBlue] (0.3,0.3) – (0.3,-0.3);
[e:marker,myLightBlue] (0.3,-0.3) – (-0.3,-0.3);
[e:marker,myLightBlue] (-0.3,-0.3) – (-0.8,-0.8);
[e:marker,myLightBlue] (-0.8,-0.8) – (-0.8,-1.7);
[e:marker,myLightBlue] (-0.8,-2.6) – (-0.8,-1.7);

[e:marker,myGreen] (-0.8,0.8) – (-1.3,1.3);
[e:marker,myGreen] (-1.3,1.3) – (-1.3,-1.3);
[e:marker,myGreen] (-1.3,-1.3) – (1.3,-1.3);
[e:marker,myGreen] (1.3,-1.3) – (0.8,-0.8);
[e:marker,myGreen] (0.8,-0.8) – (0.8,-1.7);
[e:marker,myGreen] (0.8,-2.6) – (0.8,-1.7);

[e:main] (-0.8,-0.8) rectangle (0.8,0.8);

[e:main,color=gray] (-1.3,1.3) – (1.3,1.3) – (1.3,-1.3) – (-1.3,-1.3);

[e:main,color=gray] (0.3,0.3) – (-0.3,0.3) – (-0.3,-0.3) – (0.3,-0.3);

[e:main,color=gray] (0.3,0.3) – (0.8,0.8);
[e:main,color=gray] (0.3,-0.3) – (0.8,-0.8);

[e:main,color=gray] (-0.8,0.8) – (-1.3,1.3);
[e:main,color=gray] (-0.8,-0.8) – (-1.3,-1.3);

[e:main,color=gray] (-0.8,0.8) – (-1.8,0);
[e:main,color=gray] (0.8,0.8) – (1.8,0);

[e:main,color=gray] (-1.8,-1.8) – (-0.8,-2.6);
[e:main,color=gray] (1.8,-1.8) – (0.8,-2.6);

[e:main,color=gray] (-0.8,-2.6) – (0.8,-2.6);

[e:main,color=gray] (-0.8,-0.8) – (-0.8,-1.7);
[e:main,color=gray] (0.8,-0.8) – (0.8,-1.7);

[e:coloredborder] (0.3,-0.3) – (0.3,0.3);
[e:coloredthin,color=BostonUniversityRed] (0.3,-0.3) – (0.3,0.3);

[e:coloredborder] (-0.3,0.3) – (-0.8,0.8);
[e:coloredthin,color=BostonUniversityRed] (-0.3,0.3) – (-0.8,0.8);
[e:coloredborder] (-0.3,-0.3) – (-0.8,-0.8);
[e:coloredthin,color=BostonUniversityRed] (-0.3,-0.3) – (-0.8,-0.8);

[e:coloredborder] (-1.3,-1.3) – (-1.3,1.3);
[e:coloredthin,color=BostonUniversityRed]  (-1.3,-1.3) – (-1.3,1.3);

[e:coloredborder] (0.8,0.8) – (1.3,1.3);
[e:coloredthin,color=BostonUniversityRed] (0.8,0.8) – (1.3,1.3);
[e:coloredborder] (0.8,-0.8) – (1.3,-1.3);
[e:coloredthin,color=BostonUniversityRed] (0.8,-0.8) – (1.3,-1.3);

[e:coloredborder] (-1.8,0) – (-1.8,-1.8);
[e:coloredthin,color=BostonUniversityRed] (-1.8,0) – (-1.8,-1.8);
[e:coloredborder] (1.8,0) – (1.8,-1.8);
[e:coloredthin,color=BostonUniversityRed] (1.8,0) – (1.8,-1.8);

[e:coloredborder] (0.8,-1.7) – (0.8,-2.6);
[e:coloredthin,color=BostonUniversityRed] (0.8,-1.7) – (0.8,-2.6);

[e:coloredborder] (-0.8,-1.7) – (-0.8,-2.6);
[e:coloredthin,color=BostonUniversityRed] (-0.8,-1.7) – (-0.8,-2.6);

() at (0,1.05)$C$;
[gray] () at (-1.5,-2.5)$C'$;

() at (-0.4,-0.65)$a$;
() at (0.4,-0.6)$b$;

[v:mainemptygray] () at (-0.3,-0.3);
[v:maingray] () at (-0.3,0.3);
[v:maingray] () at (0.3,-0.3);
[v:mainemptygray] () at (0.3,0.3);

[v:main] () at (-0.8,-0.8);
[v:mainempty] () at (-0.8,0.8);
[v:mainempty] () at (0.8,-0.8);
[v:main] () at (0.8,0.8);

[v:mainemptygray] () at (-1.3,-1.3);
[v:maingray] () at (-1.3,1.3);
[v:maingray] () at (1.3,-1.3);
[v:mainemptygray] () at (1.3,1.3);

[v:maingray] () at (-1.8,0);
[v:mainemptygray] () at (1.8,0);

[v:mainemptygray] () at (-1.8,-1.8);
[v:maingray] () at (1.8,-1.8);

[v:maingray] () at (-0.8,-2.6);
() at (-0.7,-2.25)$a'$;
[v:mainemptygray] () at (0.8,-2.6);
() at (0.8,-2.25)$b'$;

[v:mainemptygray] () at (-0.8,-1.7);
() at (-1.1,-1.7)$x$;

[v:maingray] () at (0.8,-1.7);
() at (1.1,-1.75)$y$;

[v:maingray] at (-0.3,-2);
() at (-0.3,-1.7)$u$;

[v:mainemptygray] at (0.3,-2);
() at (0.3,-1.7)$v$;

[e:marker,color=myGreen] (0.8,0.8) – (1.3,1.3);
[e:marker,color=myGreen] (1.3,1.3) – (1.3,-1.3);
[e:marker,color=myGreen] (1.3,-1.3) – (0.8,-0.8);
[e:marker,myGreen] (0.8,-0.8) – (0.8,-1.7);
[e:marker,myGreen] (0.8,-1.7) – (0.3,-2);
[e:marker,myGreen] (0.3,-2) – (-0.8,-2.6);

[e:marker,myLightBlue] (-0.8,0.8) – (-0.3,0.3);
[e:marker,myLightBlue] (-0.3,0.3) – (-0.3,-0.3);
[e:marker,myLightBlue] (-0.3,-0.3) – (-0.8,-0.8);
[e:marker,myLightBlue] (-0.8,-0.8) – (-0.8,-1.7);
[e:marker,myLightBlue] (-0.8,-1.7) – (-0.3,-2);
[e:marker,myLightBlue] (-0.3,-2) – (0.8,-2.6);

[e:main] (-0.8,-0.8) rectangle (0.8,0.8);

[e:main,color=gray] (-1.3,1.3) – (1.3,1.3) – (1.3,-1.3) – (-1.3,-1.3);

[e:main,color=gray] (0.3,0.3) – (0.8,0.8);
[e:main,color=gray] (0.3,-0.3) – (0.8,-0.8);

[e:main,color=gray] (0.3,0.3) – (-0.3,0.3) – (-0.3,-0.3) – (0.3,-0.3);

[e:main,color=gray] (-0.8,0.8) – (-1.3,1.3);
[e:main,color=gray] (-0.8,-0.8) – (-1.3,-1.3);

[e:main,color=gray] (-0.8,0.8) – (-1.8,0);
[e:main,color=gray] (0.8,0.8) – (1.8,0);

[e:main,color=gray] (-1.8,-1.8) – (-0.8,-2.6);
[e:main,color=gray] (1.8,-1.8) – (0.8,-2.6);

[e:main,color=gray] (-0.8,-0.8) – (-0.8,-1.7);
[e:main,color=gray] (0.8,-0.8) – (0.8,-1.7);

[e:main,color=gray,densely dashed] (-0.3,-2) – (0.3,-2);

[e:main,color=gray] (-0.3,-2) – (0.8,-2.6);
[e:main,color=gray] (0.3,-2) – (-0.8,-2.6);

[e:coloredborder] (-0.3,0.3) – (-0.8,0.8);
[e:coloredthin,color=BostonUniversityRed] (-0.3,0.3) – (-0.8,0.8);
[e:coloredborder] (-0.3,-0.3) – (-0.8,-0.8);
[e:coloredthin,color=BostonUniversityRed] (-0.3,-0.3) – (-0.8,-0.8);

[e:coloredborder] (0.3,-0.3) – (0.3,0.3);
[e:coloredthin,color=BostonUniversityRed] (0.3,-0.3) – (0.3,0.3);

[e:coloredborder] (-1.3,-1.3) – (-1.3,1.3);
[e:coloredthin,color=BostonUniversityRed]  (-1.3,-1.3) – (-1.3,1.3);

[e:coloredborder] (0.8,0.8) – (1.3,1.3);
[e:coloredthin,color=BostonUniversityRed] (0.8,0.8) – (1.3,1.3);
[e:coloredborder] (0.8,-0.8) – (1.3,-1.3);
[e:coloredthin,color=BostonUniversityRed] (0.8,-0.8) – (1.3,-1.3);

[e:coloredborder] (-1.8,0) – (-1.8,-1.8);
[e:coloredthin,color=BostonUniversityRed] (-1.8,0) – (-1.8,-1.8);
[e:coloredborder] (1.8,0) – (1.8,-1.8);
[e:coloredthin,color=BostonUniversityRed] (1.8,0) – (1.8,-1.8);

[e:coloredborder] (-0.8,-2.6) – (0.8,-2.6);
[e:coloredthin,color=BostonUniversityRed] (-0.8,-2.6) – (0.8,-2.6);

[e:coloredborder] (-0.8,-1.7) – (-0.3,-2);
[e:coloredthin,color=BostonUniversityRed](-0.8,-1.7) – (-0.3,-2);

[e:coloredborder] (0.8,-1.7) – (0.3,-2);
[e:coloredthin,color=BostonUniversityRed](0.8,-1.7) – (0.3,-2);

() at (0,1.05)$C$;
[gray] () at (-1.5,-2.5)$C'$;

() at (-0.4,-0.65)$a$;
() at (0.4,-0.6)$b$;

() at (1.8,-2)$a'$;
() at (-1.8,-2.1)$b'$;

[v:mainemptygray] () at (-0.3,-0.3);
[v:maingray] () at (-0.3,0.3);
[v:maingray] () at (0.3,-0.3);
[v:mainemptygray] () at (0.3,0.3);

[v:main] () at (-0.8,-0.8);
[v:mainempty] () at (-0.8,0.8);
[v:mainempty] () at (0.8,-0.8);
[v:main] () at (0.8,0.8);

[v:mainemptygray] () at (-1.3,-1.3);
[v:maingray] () at (-1.3,1.3);
[v:maingray] () at (1.3,-1.3);
[v:mainemptygray] () at (1.3,1.3);

[v:maingray] () at (-1.8,0);
[v:mainemptygray] () at (1.8,0);

[v:mainemptygray] () at (-1.8,-1.8);
[v:maingray] () at (1.8,-1.8);

[v:maingray] () at (-0.8,-2.6);
[v:mainemptygray] () at (0.8,-2.6);

[v:mainemptygray] () at (-0.8,-1.7);
() at (-1.1,-1.7)$x$;

[v:maingray] () at (0.8,-1.7);
() at (1.1,-1.75)$y$;

[e:marker,myGreen] (0.8,-0.8) – (1.3,-1.3);
[e:marker,myGreen] (1.3,-1.3) – (1.3,1.3);
[e:marker,myGreen] (1.3,1.3) – (0.8,0.8);
[e:marker,myGreen] (0.8,-0.8) – (0.8,-1.7);
[e:marker,myGreen,bend left=30] (0.8,-1.7) to (-1.8,-1.8);

[e:marker,myLightBlue] (-0.8,-0.8) – (-0.3,-0.3);
[e:marker,myLightBlue] (-0.3,-0.3) – (-0.3,0.3);
[e:marker,myLightBlue] (-0.3,0.3) – (-0.8,0.8);
[e:marker,myLightBlue] (-0.8,-0.8) – (-0.8,-1.7);
[e:marker,bend right=30,myLightBlue] (-0.8,-1.7) to (1.8,-1.8);

[e:main,color=gray] (-1.3,1.3) – (1.3,1.3);
[e:main,color=gray] (1.3,-1.3) – (-1.3,-1.3);

[e:main] (0.8,0.8) – (0.8,-0.8) – (-0.8,-0.8) – (-0.8,0.8);

[e:main,color=gray] (0.3,0.3) – (0.8,0.8);
[e:main,color=gray] (0.3,-0.3) – (0.8,-0.8);

[e:main,color=gray] (0.3,0.3) – (-0.3,0.3);
[e:main,color=gray] (-0.3,-0.3) – (0.3,-0.3);

[e:main,color=gray] (-0.3,0.3) – (-0.8,0.8);
[e:main,color=gray] (-0.3,-0.3) – (-0.8,-0.8);

[e:main,color=gray] (-0.8,0.8) – (-1.3,1.3);
[e:main,color=gray] (-0.8,-0.8) – (-1.3,-1.3);

[e:main,color=gray] (0.8,0.8) – (1.3,1.3);
[e:main,color=gray] (0.8,-0.8) – (1.3,-1.3);

[e:main,color=gray] (-0.8,0.8) – (-1.8,0);
[e:main,color=gray] (0.8,0.8) – (1.8,0);

[e:main,color=gray] (-1.8,-1.8) – (-0.8,-2.6);
[e:main,color=gray] (1.8,-1.8) – (0.8,-2.6);

[e:main,color=gray,bend right=30] (-0.8,-1.7) to (1.8,-1.8);
[e:main,color=gray,bend left=30] (0.8,-1.7) to (-1.8,-1.8);

[e:coloredborder] (0.3,-0.3) – (0.3,0.3);
[e:coloredthin,color=BostonUniversityRed] (0.3,-0.3) – (0.3,0.3);

[e:coloredborder] (-0.3,0.3) – (-0.3,-0.3);
[e:coloredthin,color=BostonUniversityRed] (-0.3,0.3) – (-0.3,-0.3);

[e:coloredborder] (-0.8,0.8) – (0.8,0.8);
[e:coloredthin,color=BostonUniversityRed] (-0.8,0.8) – (0.8,0.8);

[e:coloredborder] (-1.3,-1.3) – (-1.3,1.3);
[e:coloredthin,color=BostonUniversityRed]  (-1.3,-1.3) – (-1.3,1.3);

[e:coloredborder] (1.3,1.3) – (1.3,-1.3);
[e:coloredthin,color=BostonUniversityRed] (1.3,1.3) – (1.3,-1.3);

[e:coloredborder] (-1.8,0) – (-1.8,-1.8);
[e:coloredthin,color=BostonUniversityRed] (-1.8,0) – (-1.8,-1.8);
[e:coloredborder] (1.8,0) – (1.8,-1.8);
[e:coloredthin,color=BostonUniversityRed] (1.8,0) – (1.8,-1.8);

[e:coloredborder] (-0.8,-2.6) – (0.8,-2.6);
[e:coloredthin,color=BostonUniversityRed] (-0.8,-2.6) – (0.8,-2.6);

[e:coloredborder] (-0.8,-0.8) – (-0.8,-1.7);
[e:coloredthin,color=BostonUniversityRed] (-0.8,-0.8) – (-0.8,-1.7);

[e:coloredborder] (0.8,-0.8) – (0.8,-1.7);
[e:coloredthin,color=BostonUniversityRed] (0.8,-0.8) – (0.8,-1.7);

() at (0,1.05)$C$;
[gray] () at (-1.5,-2.5)$C'$;

() at (-0.4,-0.65)$a$;
() at (0.4,-0.6)$b$;

() at (1.8,-2)$a'$;
() at (-1.8,-2.1)$b'$;

[v:mainemptygray] () at (-0.3,-0.3);
[v:maingray] () at (-0.3,0.3);
[v:maingray] () at (0.3,-0.3);
[v:mainemptygray] () at (0.3,0.3);

[v:main] () at (-0.8,-0.8);
[v:mainempty] () at (-0.8,0.8);
[v:mainempty] () at (0.8,-0.8);
[v:main] () at (0.8,0.8);

[v:mainemptygray] () at (-1.3,-1.3);
[v:maingray] () at (-1.3,1.3);
[v:maingray] () at (1.3,-1.3);
[v:mainemptygray] () at (1.3,1.3);

[v:maingray] () at (-1.8,0);
[v:mainemptygray] () at (1.8,0);

[v:mainemptygray] () at (-1.8,-1.8);
[v:maingray] () at (1.8,-1.8);

[v:maingray] () at (-0.8,-2.6);
[v:mainemptygray] () at (0.8,-2.6);

[v:mainemptygray] () at (-0.8,-1.7);
() at (-1.1,-1.7)$x$;

[v:maingray] () at (0.8,-1.7);
() at (1.1,-1.75)$y$;

[v:maingray] at (-0.3,-2);
() at (-0.3,-1.7)$u$;

[v:mainemptygray] at (0.3,-2);
() at (0.3,-1.7)$v$;

[e:marker,myGreen] (-0.8,0.8) – (-1.3,1.3);
[e:marker,myGreen] (-1.3,1.3) – (-1.3,-1.3);
[e:marker,myGreen] (-1.3,-1.3) – (1.3,-1.3);
[e:marker,myGreen] (1.3,-1.3) – (0.8,-0.8);
[e:marker,myGreen] (0.8,-0.8) – (0.8,-1.7);
[e:marker,myGreen] (0.8,-1.7) – (0.3,-2);
[e:marker,myGreen,bend right=20] (0.3,-2) to (1.8,-1.8);

[e:marker,myLightBlue] (0.8,0.8) – (0.3,0.3);
[e:marker,myLightBlue] (0.3,0.3) – (0.3,-0.3);
[e:marker,myLightBlue] (0.3,-0.3) – (-0.3,-0.3);
[e:marker,myLightBlue] (-0.3,-0.3) – (-0.8,-0.8);
[e:marker,myLightBlue] (-0.8,-0.8) – (-0.8,-1.7);
[e:marker,myLightBlue] (-0.8,-1.7) – (-0.3,-2);
[e:marker, bend left=20,myLightBlue] (-0.3,-2) to (-1.8,-1.8);

[e:main] (-0.8,-0.8) rectangle (0.8,0.8);

[e:main,color=gray] (-1.3,1.3) – (1.3,1.3) – (1.3,-1.3) – (-1.3,-1.3);

[e:main,color=gray] (0.3,0.3) – (0.8,0.8);
[e:main,color=gray] (0.3,-0.3) – (0.8,-0.8);

[e:main,color=gray] (0.3,0.3) – (-0.3,0.3) – (-0.3,-0.3) – (0.3,-0.3);

[e:main,color=gray] (-0.8,0.8) – (-1.3,1.3);
[e:main,color=gray] (-0.8,-0.8) – (-1.3,-1.3);

[e:main,color=gray] (-0.8,0.8) – (-1.8,0);
[e:main,color=gray] (0.8,0.8) – (1.8,0);

[e:main,color=gray] (-1.8,-1.8) – (-0.8,-2.6);
[e:main,color=gray] (1.8,-1.8) – (0.8,-2.6);

[e:main,color=gray] (-0.8,-2.6) – (0.8,-2.6);

[e:main,color=gray] (-0.8,-0.8) – (-0.8,-1.7);
[e:main,color=gray] (0.8,-0.8) – (0.8,-1.7);

[e:main,color=gray,densely dashed] (-0.3,-2) – (0.3,-2);

[e:main,color=gray,bend left=20] (-0.3,-2) to (-1.8,-1.8);
[e:main,color=gray,bend right=20] (0.3,-2) to (1.8,-1.8);

[e:coloredborder] (-0.3,0.3) – (-0.8,0.8);
[e:coloredthin,color=BostonUniversityRed] (-0.3,0.3) – (-0.8,0.8);
[e:coloredborder] (-0.3,-0.3) – (-0.8,-0.8);
[e:coloredthin,color=BostonUniversityRed] (-0.3,-0.3) – (-0.8,-0.8);

[e:coloredborder] (0.3,-0.3) – (0.3,0.3);
[e:coloredthin,color=BostonUniversityRed] (0.3,-0.3) – (0.3,0.3);

[e:coloredborder] (-1.3,-1.3) – (-1.3,1.3);
[e:coloredthin,color=BostonUniversityRed]  (-1.3,-1.3) – (-1.3,1.3);

[e:coloredborder] (0.8,0.8) – (1.3,1.3);
[e:coloredthin,color=BostonUniversityRed] (0.8,0.8) – (1.3,1.3);
[e:coloredborder] (0.8,-0.8) – (1.3,-1.3);
[e:coloredthin,color=BostonUniversityRed] (0.8,-0.8) – (1.3,-1.3);

[e:coloredborder] (-1.8,0) – (-1.8,-1.8);
[e:coloredthin,color=BostonUniversityRed] (-1.8,0) – (-1.8,-1.8);
[e:coloredborder] (1.8,0) – (1.8,-1.8);
[e:coloredthin,color=BostonUniversityRed] (1.8,0) – (1.8,-1.8);

[e:coloredborder] (-0.8,-1.7) – (-0.3,-2);
[e:coloredthin,color=BostonUniversityRed](-0.8,-1.7) – (-0.3,-2);

[e:coloredborder] (0.8,-1.7) – (0.3,-2);
[e:coloredthin,color=BostonUniversityRed](0.8,-1.7) – (0.3,-2);

The four different ways to obtain a strong matching cross over $C'$ in the second subcase of the first case in the proof of <ref>.

So now let us assume $ab\notin M$.
Then there exist vertices $x,y\in\V{B}\setminus\Brace{\V{C}\cup\V{C'}}$ such that $ax,by\in M$.
Let $P$ be an internally $M$-conformal path connecting $x$ to some vertex $a'$ of $\Vi{1}{C'}$ such that $P$ is internally disjoint from $C'$ and avoids $K$.
Similarly we choose $Q$ to be an internally $M$-conformal path connecting $y$ to some vertex $b'\in\Vi{2}{C'}$ while avoiding $K$ and being internally disjoint from $C'$.
Suppose one of the two paths contains the initial matching edge of the other.
Since these cases are symmetric, it suffices to consider one of them, so let us assume $P$ contains $by$.
Then $Py$ is an internally $M$-conformal path that is disjoint from $C'$ and does not meet $K$ at all.
Hence $Pyba$ is an $M$-conformal cycle and thus $M\Delta\E{Pyba}$ is a perfect matching of $B_1$ for which $C'+K$ is conformal.
Since we ruled this possibility out by the first case this cannot happen, and thus $P$ does not contain $by$, and neither does $Q$ contain the edge $ax$.
By calling upon <ref> again, we find a perfect matching $M'$ and paths $L'$ and $R'$ such that $L'\cap R'$ is either empty or an $M'$-conformal path, $M'$ equals $M$ outside of $axP$ and $byQ$, and $R$ and $L$ connect $\Set{a,b}$ to $\Set{a',b'}$ while being internally disjoint from $C'$ and avoiding $K$.
If case $L'\cap R'$ is empty let $L\coloneqq L'$ and $R\coloneqq R'$.
Otherwise, let $u$ and $v$ be the endpoints of $L'\cap R'$ such that $u$ is the first vertex of $R'$ one encounters when traversing $L'$ starting in $a$.
Then $ab$ together with the unique $a$-$b$-subpath of $L'+R'$ forms an $M'$-conformal cycle $O$.
By adjusting $M'$ to be the perfect matching $M'\Delta\E{O}$ and setting $L\coloneqq L'uR'$, $R\coloneqq R'vL'$ we have found two disjoint $M'$-alternating paths that can be extended to form a strong matching cross over $C'$ in $B$ by using <ref> as before.
Please note that this step might alter the perfect matching $M'$ again with regards to the edges of $C$.
See <ref> for an illustration of the cases that might arise.

[v:ghost] () at (0,-2.2);
() at (0,1.05)$C$;
() at (-1.55,0.4)$C'$;

[v:maingray] () at (-0.3,-0.3);
[v:mainemptygray] () at (-0.3,0.3);
[v:mainemptygray] () at (0.3,-0.3);
[v:maingray] () at (0.3,0.3);

[v:mainempty] () at (-0.8,-0.8);
[v:main] () at (-0.8,0.8);
[v:main] () at (0.8,-0.8);
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Strong matching crosses over $C'$ in the second and third case in the proof of <ref>.

So now let us assume $K$ to be of length two.
In this case, there is a unique vertex $u\in\V{C}$ that does not belong to  $C'$.
For this case, let us choose $M$ to be a perfect matching of $B_1$ for which $C$ is $M$-conformal.
Since $B$ is a brace, by <ref>, there exists an internally $M$-conformal path $P$ connecting $v$ to a vertex of $C'$ while avoiding any vertex in $\V{K}$.
We then use <ref> to find paths $L'$ and $R$ together with a perfect matching $M'$ of $B$ such that $L\coloneqq L'P$ and $R$ form a strong matching cross over $C'$ as illustrated in <ref>.

At last, consider the case where $K$ contains all of $C$.
Here <ref> yields the strong matching cross over $C'$ in $B$ immediately.

The above lemma illustrates why we cannot allow to always reduce a brace along a $4$-cycle sum.
In some cases, even the small separator given by the $4$-cycle is enough to provide a matching cross.
The following lemmas aim to make this observation more general and exact.

Let $B$ be a $K_{3,3}$-free brace, $\ell\geq 3$, $B_1,\dots,B_{\ell}$ braces such that $B$ is a maximal $4$-cycle sum of $B_1,\dots,B_{\ell}$ at the $4$-cycle $C$ and let $C'$ be a conformal cycle in $B_1$ that also exists in $B$ such that $\Abs{\V{C'}\cap\V{C}}\geq 2$ and $\V{C}\cap\V{C'}$ contains vertices of both colour classes, then there is a matching cross over $C'$ in $B$.

We divide this proof into three cases.

1: $\Abs{\V{C}\cap\V{C'}}=2$ and the vertices in $\V{C}\cap\V{C'}$ are adjacent on $C$.

2: $\Abs{\V{C}\cap\V{C'}}=3$.

3: $\Abs{\V{C}\cap\V{C'}}=4$.

Case 1: $\Abs{\V{C}\cap\V{C'}}=2$ and the vertices in $\V{C}\cap\V{C'}$ are adjacent on $C$.

In this case, let $x$ and $y$ be the two adjacent vertices of $C$ belong to $C'$.
In case $xy\in\E{C'}$, we are done immediately by <ref>.
If $xy\notin\E{C'}$, then $x$ and $y$ divide $C'$ into two paths of odd length, say $P_1$ and $P_2$, both with endpoints $x$ and $y$.
If $M$ is a perfect matching of $B$ such that $C'$ is $M$-conformal, then exactly one of the two paths, say $P_1$ is also $M$-conformal and thus $P_1+xy$ is an $M$-conformal cycle as well.
By <ref> $P_1+xy$ has a strong matching cross and thus, by <ref>, $C'$ must have a matching cross in $B$.

Case 2: $\Abs{\V{C}\cap\V{C'}}=3$.

In case $C\cap C'$ is a subpath of $C'$, we are done immediately by <ref>.
Hence we may assume that this is not the case.
Next, suppose $C'$ contains exactly one edge of $C$ and there is $xy\in\E{C}$ such that $x,y\in\V{C'}$, but $xy\notin\E{C'}$.
Let $z$ be the remaining vertex of $C$ on $C'$, then $x$ and $y$ separate $C'$ into two paths, where one of them, say $P$, does not contain $z$.
We may choose a perfect matching $M$ of $B$ such that $P$ is internally $M$-conformal.
Then $K\coloneqq C'-P+xy$ is also an $M$-conformal cycle, and by our assumption, $K\cap C$ is a subpath of $C$.
Hence we may apply <ref> together with <ref> to obtain a matching cross over $C'$.
At last assume that $C'$ does not contain an edge of $C$.
If we call the vertices of $C$ on $C'$ $x$, $y$, and $z$ again such that $x$ and $z$ belong to the same colour class, we again find the path $P$ avoiding $z$ but connecting $x$ and $y$ as before.
But we also find a path $Q\subseteq C'$ that connects $z$ and $y$ and avoids $x$.
By choosing a perfect matching $M$ of $B$ such that $C'$ is $M$-conformal and $Q$ is internally $M$-conformal, we have found a perfect matching for which $K'\coloneqq C'-Q+yz$ is an $M$-conformal cycle.
For this cycle, we find a matching cross as discussed above, and by applying <ref> again, we obtain a matching cross for $C'$ as well.

Case 3: $\Abs{\V{C}\cap\V{C'}}=4$.

Let $C=\Brace{a_1,b_1,a_2,b_2}$.
If the vertices of $C$ appear on $C'$ in the same order as they do on $C$, we can use <ref> to find a strong matching cross over $C$ whose paths are internally disjoint from $B_1$.
Hence we have found a strong matching cross over $C'$ in $B$.

Hence the vertices of $C$ do not appear on $C'$ in the order listed.
The only way this is possible is, if they appear on $C'$ in the order $a_1$, $a_2$, $b_1$, $b_2$, or $a_1$, $a_2$, $b_2$, $b_1$.
In both cases we can use <ref> to obtain a perfect matching $M$ of $B$ and internally $M$-conformal paths $P_1$ and $P_2$ which are internally disjoint from $B_1$ such that $P_1$ connects $a_1$ and $b_1$ while $P_2$ connects $a_2$ and $b_2$ in case the order of appearance is $a_1$, $a_2$, $b_1$, $b_2$.
Otherwise, $P_1$ and $P_2$ may be chosen such that $P_1$ connects $a_1$ and $b_2$ while $P_2$ connects $a_2$ and $b_1$.
Either way, the two paths form a strong matching cross over $C'$ in $B$.

Let $B$ be a $K_{3,3}$-free brace, $C$ a $4$-cycle in $B$ such that $B-C$ is not connected, and $C'$ be any conformal cycle in $B$.
Suppose there is a matching cross over $C'$ in $B$ such that at least one of the paths of this cross uses $C$ and extends into another component of $B-C$.
Let $B'$ be a brace with $C'\subseteq B'$ such that $B$ is made from a set of $K_{3,3}$-free braces including $B'$ via a $4$-cycle sum at $C$.
Then some information on the matching cross over $C'$ in $B$ should also exist in $B'$.
Our goal is to use this information to show that even $B'$ cannot be planar while $C'$ bounds a face.

Let $B$ be a $K_{3,3}$-free brace, $C$ a $4$-cycle in $B$, $M$ a perfect matching and $C'$ a conformal cycle in $B$.
A tuple $\Brace{P_1,P_2,Q,M}$ is a diffuse $C'$-$M$-precross through $C$ if $P_1$, $P_2$ and $Q$ are pairwise disjoint, internally disjoint from $C'$, all are $M$-alternating paths[In particular, $P_1$ and $P_2$ are allowed to be single vertices of $C'$.], and for each $i\in[1,2]$, $P_i$ is a $\V{C'}$-$\V{C}$-path such that the endpoints of the $P_i$ on $C'$ belong to different components of $C'-\V{Q}$.
Additionally, we require among the endpoints of the $P_i$ on $C$ at least one of each colour class to be covered by an edge of $\E{C}\cap M$.
A diffuse $C'$-$M$-precross through $C$ is daring if the endpoints of the $P_i$ on $C$ belong to the same colour class.

Let $B$ be a $K_{3,3}$-free brace, $\ell\geq 3$, $B_1,\dots,B_{\ell}$ braces such that $B$ is a maximal $4$-cycle sum of $B_1,\dots,B_{\ell}$ at the $4$-cycle $C$ and let $C'$ be a conformal cycle in $B_1$ that also exists in $B$.
If there is a diffuse $C'$-$M_1$-precross $H$ through $C$ in $B_1$, for some perfect matching $M_1$ of $B_1$, which is not daring, then there are matching crosses over $C'$ in $B_1$ and $B$.

Let $H=\Brace{P_1,P_2,Q,M_1}$.
Since $H$ is not daring, we may apply <ref> to find a perfect matching $M$ of $B$ with $M_1\setminus\E{C}\subseteq M$ together with a path $R$ in $B_j$ for some $j\in[2,\ell]$ such that $P_1RP_2^{-1}$ is $M$-alternating and disjoint from $Q$.
Note that such a matching exists in particular because either $C$ is $M$-conformal, or the edge connecting the endpoints of the $P_i$ on $C$ must belong to $M$.
Hence in this case we have found our matching cross over $C'$ in $B$.
Since $R$ is a subpath of an $M$-alternating path, it itself is $M$-alternating.
Moreover, with $H$ not being daring, the endpoints of $R$ belong to different colour classes, and thus $R$ is either $M$-conformal or internally $M$-conformal.
Either way, the endpoints, let us call them $x$ and $y$, are adjacent on $C$ and there exists a perfect matching $M'$ of $B_1$ such that $\E{B_1-\V{C}}\cap M\subseteq M'$ and $xy\in M'$ if and only if $R$ is $M$-conformal.
Hence $P_1xyP_2^{-1}$ is an $M'$-alternating path which forms, together with $Q$, a matching cross over $C'$ in $B_1$.

Hence diffuse precrosses can always be extended to actual crosses if they are not daring.
Next we inspect daring precrosses more closely.

Let $B$ be a $K_{3,3}$-free brace, $C$ a $4$-cycle in $B$, $M$ a perfect matching and $C'$ an $M$-conformal cycle in $B$.
A diffuse $C'$-$M$-precross through $C$ $\Brace{P_1,P_2,Q,M}$ is successful if it is daring, and either $P_1$ is internally $M$-conformal while $P_2$ is $M$-conformal or $P_1$ and $P_2$ are both of even length, and the endpoint of $P_1$ on $C'$ is covered by an edge of $\E{P_1}\cap M$ if and only if the endpoint of $P_2$ on $C'$ is not covered by an edge of $\E{P_2}\cap M$.

Let $B$ be a $K_{3,3}$-free brace, $\ell\geq 3$, $B_1,\dots,B_{\ell}$ braces such that $B$ is a maximal $4$-cycle sum of $B_1,\dots,B_{\ell}$ at the $4$-cycle $C$ and let $C'$ be a conformal cycle in $B_1$ that also exists in $B$ such that there is $i\in[1,2]$ with $\V{C}\cap\V{C'}\subseteq V_i$.
Then there is a matching cross over $C'$ in $B$ if and only if one of the following is true

* there is a matching cross over $C'$ in $B_1$,

* there is a perfect matching $M_1$ of $B_1$ such that there exists a diffuse $C'$-$M_1$-precross through $C$, which is successful.

Let us first prove that in case one of our conditions holds true we find a matching cross over $C'$ in $B$.
First let $L$ and $R$ be the two paths of a matching cross over $C'$ in $B_1$ and let $M$ be a perfect matching such that $L$ and $R$ are $M$-alternating.
If $L$ and $R$ also exist in $B$, we are done.
So let us assume exactly one of them, say $L$, contains an edge of $C$ which does not exist in $B$, note that this is the only possibility why $L$ is not an alternating path in $B$.
Let us traverse along $L$ starting in one of its endpoints, and let $x$ be the first vertex of $C$ we encounter, while $y$ is the last vertex of $C$ on $L$.
In case $xy\in\E{C}$, we either have $xy\in M$, or $xy\notin M$, but $xLy$ is $M$-conformal, hence $xLyx$ is an $M$-conformal cycle, and we can adjust $M$ to include the edge $xy$.
Then, after possibly adjusting the matching as above, $\Brace{Lx,yL,R,M}$ is a diffuse $C'$-$M$-precross through $C$ which is not daring.
Hence we are done by <ref>.
So we may assume $x$ and $y$ not to be adjacent on $C$ and thus they belong to the same colour class.
In this case, $xLy$ is an $M$-alternating path of even length and thus can neither be internally $M$-conformal, nor $M$-conformal.
However, since $L$ contains an edge of $C$, there must be a vertex $z\in\V{C}\setminus\Set{x,y}$ such that one of the edges $xz$, $yz$ belongs to $L$.
Without loss of generality let us assume $xz\in\E{L}$.
In case $xz\in M$, $\Brace{Lx,yL,R,M}$ is a successful diffuse $C'$-$M$-precross through $C$ and we are done by <ref>.
So assume $xz\notin M$.
Then $zLy$ must be an $M$-conformal path, and thus $zLyz$ is an $M$-conformal cycle.
Hence we may adjust $M$ such that $yz\in M$ and again $\Brace{Lx,yL,R,M}$ is a successful diffuse $C'$-$M$-precross through $C$ and we are done by <ref>.
Hence we may now assume $L$ and $R$ to contain an edge of $C$ each.
Since $C$ only has four vertices, this means there is a unique edge $x_Ly_L\in\E{L}\cap\E{C}$ and a unique edge $x_Ry_R\in\E{R}\cap\E{C}$.
Now <ref> provides us with the two paths in $B_2$ that are necessary to be combined with $Lx_L$, $y_LL$ and $Rx_R$, $y_RR$ respectively in order to obtain a strong matching cross over $C'$ in $B$.

So now we have to show that the existence of a matching cross in $B$ implies the existence of one of the two structures above.
Let $R$ and $L$ be two $M$-alternating paths for some perfect matching $M$ that form a matching cross over $C'$.
If neither $L$ nor $R$ contains a vertex of $C$, $L$ and $R$ must be completely contained in $B_1$ and thus form a strong matching cross over $C'$ in $B_1$ as well.
So let us assume that exactly one of the two paths contains a vertex of $C$ and further assume, without loss of generality, that $L$ is that path.
In case $L$ contains exactly one vertex of $C$, it cannot contain any vertex of $B_i-\V{C}$ for any $i\in[2,\ell]$ since $C$ separates the $B_j$ from each other.
Hence in this case, $L$ and $R$ again also exist in $B_1$.
So assume that $L$ contains exactly two vertices of $C$, say $v$ and $w$.
Then either $vw\in\E{C}$, or $v$ and $w$ belong to the same colour class.
In the first case, either $vw\in\E{L}$, or $vLwv$ is an $M$-conformal cycle, and we may adjust $M$ such that $LvwL$ is $M$-alternating.
In any case, after possibly adjusting $M$, $LvwL$ and $R$ form a matching cross over $C'$ in $B_1$ for some perfect matching $M_1$ where $vw\in M_1$ if and only if $vLw$ is $M$-conformal.
In the second case, $vLw$ must be of even length.
In case exactly one of $Lv$ and $wL$ is of even length, there must be a vertex $u\in\V{C}\setminus\Set{v,w}$ such that $u$ is not covered by an edge of $M\cap\E{B_1}$, or $vLw=vuw$.
Then we may choose a perfect matching $M_1$ of $B_1$ such that $M_1\setminus\E{C}\subseteq M$ and  $LvuwL$ is an $M_1$ alternating path of the same type as $L$ in $B_1$.
Hence there is a matching cross over $C'$ in $B_1$.
Thus we may assume $Lv$ and $wL$ to either both be of odd or both be of even length.
In total, this means that $L$ is of even length.
Hence, by choosing $M_1$ as before, $\Brace{Lv,wL,R,M_1}$ is a successful diffuse $C'$-$M_1$-precross over $C'$ in $B_1$.
In case $L$ contains more than two vertices of $C$, let $v$ be the first vertex of $C$ on $L$ and $w$ be the last one.
Let $Q$ be a shortest $v$-$w$-path on $C$, then $Q$ and $vLw$ are of the same parity, and we can choose a perfect matching $M_1$ such that both $R$ and $LvQwL$ are $M_1$ alternating in $B_1$, thus forming a matching cross over $C'$ in $B_1$.

With this, we may now assume that both $L$ and $R$ contain vertices of $C$.
If $R$ contains exactly one vertex of $C$ and the edge of $M$ covering this vertex belongs to $B_1$, this case can be handled the same way as the cases where $R$ does not contain any vertex of $C$.
If, on the other hand, $R$ contains exactly one vertex of $C$ and the edge of $M$ covering this vertex does not belong to $B_1$ this means this vertex of $C$ is an endpoint of $R$ and thus belongs to $C'$.
Let us assume that both $L$ and $R$ contain exactly one vertex of $C$ each and the edges of $M$ covering these vertices do not belong to $B_1$.
This means that these endpoints of $L$ and $R$ on $C$ must belong to the same colour class by our assumption and thus no edge of $M$ that covers a vertex of $C$ can belong to $B_1$.