Thin Edges in Braces ∗

The bicontraction of a vertex v of degree two in a graph, with precisely two neighbours v1 and v2, consists of shrinking the set {v1, v, v2} to a single vertex. The retract of a matching covered graph G, denoted by Ĝ, is the graph obtained from G by repeatedly bicontracting vertices of degree two. Up to isomorphism, the retract of a matching covered graph G is unique. If G is a brace on six or more vertices, an edge e of G is thin if Ĝ− e is a brace. A thin edge e in a simple brace G is strictly thin if Ĝ− e is a simple brace. Theorems concerning the existence of strictly thin edges have been used (implicitly by McCuaig (Pólya’s Permanent Problem, Electron. J. Combin., 11, 2004 ) and explicitly by the authors (On the Number of Perfect Matchings in a Bipartite Graph, SIAM J. Discrete Math., 27, 940–958, 2013)) as inductive tools for establishing properties of braces. Let G and J be two distinct braces, where G is of order six or more and J is a simple matching minor of G. It follows from a theorem of McCuaig (Brace Generation, J. Graph Theory, 38, 124–169, 2001) that G has a thin edge e such that J is a matching minor of G − e. In Section 2, we give an alternative, and simpler proof, of this assertion. Our method of proof lends itself to proving stronger results concerning thin edges. Let G denote the family of braces consisting of all prisms, all Möbius ladders, all biwheels, and all extended biwheels. Strengthening another result of McCuaig on brace generation, we show that every simple brace of order six or more which is not a member of G has at least two strictly thin edges. We also give examples to show that this result is best possible.


Matching Covered Graphs
Graphs considered here are loopless, but they may have multiple edges.For graph theoretical notation and terminology, we essentially follow Bondy and Murty [1].One notable exception is that here we denote the subgraph of a graph G obtained by deleting an edge e from it by G − e, in [1] it is denoted by G \ e.
McCuaig's paper [11] provides an excellent introduction to the study of procedures for generating braces, and the motivation that led to his work.For the convenience of the reader, in the first part of this section we briefly review the relevant terminology, definitions and results from the theory of matching covered graphs.The later parts of this section include several useful basic results concerning tight cuts and removable edges in bipartite matching covered graphs.
A graph G is matching covered if it is connected, has at least two vertices and each edge lies in a perfect matching.Some authors refer to matching covered graphs as 1extendable graphs.Every 2-edge-connected cubic graph is matching covered.The treatise by Lovász and Plummer [10] contains the basic theory of matching covered graphs.One simple property stated in that book is the following result: Proposition 1 ([10, 5.1 (3)]).Every matching covered graph is 2-connected.
We shall denote a bipartite graph G with bipartition (A, B) by G[A, B], and assume throughout that |A| = |B| 1.The following result provides a characterization of bipartite matching covered graphs.It follows immediately from Theorem 4.1.1 in Lovász and Plummer's book [10].
Proposition 2. Let G := G[A, B] be a bipartite graph on four or more vertices.Then, G is matching covered if and only if for every partition (A ′ , A ′′ ) of A into two nonempty sets, and every partition (B ′ , B ′′ ) of B such that |A ′ | = |B ′ |, graph G has at least one edge that joins a vertex in B ′ to a vertex in A ′′ .

Tight cuts
Let X be a subset of the vertex set of a graph G.We denote by ∂(X) the set of all edges with one end in X and one end in X = V \ X.Clearly, ∂(X) is an edge cut of G; we shall simply refer to such sets of edges as cuts.If G is connected and ∂(X) = ∂(Y ), then either Y = X or Y = X; these two sets are then referred to as the shores of ∂(X).A cut is trivial if it has a shore consisting of exactly one vertex.
Given any cut C := ∂(X) in a connected graph G, where X is a nonempty proper subset of V (G), one may obtain two other graphs, namely G/X and G/X, by contracting the shores of C to single vertices.These two graphs are called the C-contractions of G.When it is necessary to name the contraction vertices (that is, the vertices resulting from the contractions of shores), we shall use an alternative notation to represent Ccontractions.Thus, G/(X → x) and G/(X → x) denote G/X and G/X, respectively, where x and x are the corresponding contraction vertices.Proposition 3. Let ∂(X) be a cut of a graph G.If both G/X and G/X are matching covered then G is also matching covered.Now let G be a matching covered graph.A cut C of G is tight if |C ∩ M| = 1, for every perfect matching M of G. Simplest examples of tight cuts are the trivial cuts.A basic fact concerning matching covered graphs is that, if C is a tight cut of G, then both C-contractions of G are also matching covered.By Proposition 1, we then have the following simple proposition.Proposition 4. The subgraphs of a matching covered graph induced by the shores of a tight cut are both connected.
Let G := G[A, B] be a bipartite matching covered graph, and let X be a set of vertices of G such that |X| is odd.Then |X ∩ A| and |X ∩ B| are clearly distinct; one with smaller cardinality is called the minority part and is denoted X − , and the other, with larger cardinality, is called the majority part of X and is denoted X + .The following property, which is easily proved, gives a description of tight cuts in bipartite matching covered graphs.
Proposition 5 (See Lemma 1.4 in [9]).Let G be a bipartite matching covered graph, C := ∂(X) a cut of G, |X| odd.Then, C is tight if and only if (i) |X + | = |X − | + 1 and (ii) every edge of C is incident with a vertex of X + .

Uncrossing tight cuts
Let G be a matching covered graph.Consider two cuts C := ∂(X) and D := ∂(Y ) of G.The four sets X ∩ Y , X ∩ Y , X ∩ Y and X ∩ Y are the quadrants defined by C and D. The two cuts C and D cross if each of the four quadrants is nonnull.A collection C of cuts of G is laminar if no two of its cuts cross.The following result, proved in [9], is a fundamental property of tight cuts in graphs.The following corollary will play a useful role in this paper.
Corollary 7. Let G be a matching covered graph, C := ∂(X) and D := ∂(Y ) be tight cuts of G that cross.If |X ∩ Y | is odd then the graphs G/X/(X ∩ Y ) and G/Y /(X ∩ Y ) are isomorphic, up to multiple edges (Figure 1).Proof.The vertex set of each of the graphs G/X/(X ∩ Y ) and G/Y /(X ∩ Y ) consists of X ∩ Y and two contraction vertices.Since there are no edges between X ∩ Y and X ∩ Y (by Proposition 6), to establish the required isomorphism it suffices to show that the contraction vertices are adjacent in both graphs.But this follows from the fact that, by Proposition 4, G has edges joining vertices of X ∩ Y to vertices of X ∩ Y and also to vertices of X ∩ Y .
The edges that make the two contraction vertices adjacent in the graphs in Corollary 7 -dashed line indicates the possibility of one or more edges Let G := G[A, B] be a matching covered graph.In our illustrations, we shall represent vertices in A by hollow, white discs and the vertices in B by black discs.In view of Proposition 5, all edges in a tight cut C := ∂(X) must emanate from vertices of the same colour in X.

Tight cut decompositions
A matching covered graph without nontrivial tight cuts is called a brace if it is bipartite, and a brick if it is nonbipartite.The complete bipartite graphs K 2 , C 4 and K 3,3 are the unique simple braces on two, four and six vertices, respectively.Every brace on eight vertices contains the cube as a spanning subgraph.
Given any matching covered graph G, we may apply to it a procedure, called a tight cut decomposition of G, which produces a list of bricks and braces.If G itself is a brick or a brace then the list consists of just G. Otherwise, let C be any nontrivial tight cut of G.Then, both C-contractions of G are matching covered.One may recursively apply the tight cut decomposition procedure to each C-contraction of G, and then combine the resulting lists to produce a tight cut decomposition of G.We remark that, associated with a tight cut decomposition of G there is a maximal laminar collection C of nontrivial tight cuts of G.
Based on the modularity property (Proposition 6), Lovász [9] proved the following remarkable result on tight cut decompositions.Theorem 8. Any two applications of the tight cut decomposition procedure to a matching covered graph produce the same list of bricks and braces, up to multiple edges.
In particular, the numbers of bricks and braces are numerical invariants of matching covered graphs.The following result is a consequence of Proposition 5.

Bicontractions and retracts
The bicontraction of a vertex v of degree two in a graph G, with precisely two neighbours v 1 and v 2 , is the graph G/X, where X := {v 1 , v, v 2 }.The retract of a matching covered graph G, denoted by G, is the graph obtained from G by repeatedly bicontracting vertices of degree two.Up to isomorphism, the retract of a matching covered graph G is unique (see [4,Proposition 3.11]).If G is not a cycle, then, in its retract, each vertex has degree three or more.

Braces
Recall that a bipartite matching covered graph is a brace if it has no nontrivial tight cuts.The following characterization of braces will play an important role in this paper.For any graph G and any set X of vertices of G, N G (X) denotes the set of neighbours of X in G.We omit the subscript G if it is understood and write simply N(X).
Theorem 10 ([9, 1.4], [10]).Let G be a matching covered graph with bipartition (A, B).The following are equivalent: The above theorem implies that every vertex of a brace on at least six vertices has at least three distinct neighbours.
Lemma 11.Let G := G(A, B) be a brace, let S be a set of three vertices of G not all in the same part of G.Then, G − S is connected.
Proof.The assertion holds immediately if G has order four.Assume thus that G has order six or more.Adjust notation so that S = {a 1 , a 2 , b 1 }, where a 1 , a 2 ∈ A, and b 1 ∈ B. As G − S has an odd number of vertices, it must have an odd component, say K. Suppose, contrary to the assertion, that G − S is not connected, and let L be a component of a perfect matching, say M. The restriction of M to E(K) would then be a perfect matching of K.This is impossible because K has an odd number of vertices.

Prisms, Möbius ladders, and biwheels
We now describe the four families of braces mentioned in the abstract.Their relevance to the theory of braces was first established by McCuaig [11].
A Möbius ladder M 4n+2 , n 1, is the graph obtained from a cycle of length 4n + 2, (v 1 , v 2 , . . ., v 4n+2 , v 1 ), by the addition of the 2n + 1 chords v i v i+2n+1 , 1 i 2n + 1 of the cycle, where the addition in the suffixes is understood to be taken modulo 4n + 2. The family of Möbius ladders is denoted M. A biwheel B 2n , n 4, is the graph obtained from a cycle (v 1 , v 2 , . . ., v 2n−2 , v 1 ) of length 2n − 2, called the rim of B 2n , by the addition of two vertices, h 1 and h 2 , called the hubs of B 2n , and by the addition of edges The family of biwheels is denoted B. Figure 2c shows the biwheel B 10 .
Apart from the three families defined above, there is a fourth family of braces, related to biwheels, which appears in McCuaig's work.For n 4, the extended biwheel B + 2n is obtained from the biwheel B 2n by adding an edge joining the two hubs.In addition, we take K 3,3 to be the extended biwheel B + 6 .The family of extended biwheels is denoted B + .Note that K 3,3 is both an extended biwheel and a Möbius ladder, whereas the cube is both prism P 8 and biwheel B 8 .

A lemma concerning crossing cuts
We shall now present a lemma which plays a crucial role in this paper.If S is any set and e is an element of S, we shall simply write S − e for the set S \ {e}.
it follows that its end b 1 lies in X + , which is a subset of B (Figure 3).Proof.Let u and v denote the ends of edge e. Adjust notation so that X ∩ Y contains the end u of e.As the cuts C − e and D − e are both nontrivial and tight in G − e, it follows that u ∈ X − ∩ Y − and v lies in (X) − ∩ (Y ) − .Now consider the quadrants X ∩ Y and X ∩ Y .If either of them is empty, then one of X and Y is a subset of the other.In this case, the assertion holds immediately.We may thus assume that C and D cross.By Lemma 12, ∂(X ∩ Y ) − e is nontrivial and tight in G − e.As the end v of e lies in (X) − ∩ (Y ) − , it also follows that ∂(X ∩ Y ) − e is nontrivial and tight in G − e.

Removable edges
An edge e in a matching covered graph G is removable if G − e is also matching covered.Proposition 2 implies the following useful result concerning nonremovable edges in bipartite matching covered graphs.Lemma 14.Let G[A, B] be a bipartite matching covered graph, and let e be a nonremovable edge of G.If G has two or more edges, then there exist partitions (A ′ , A ′′ ) of A and , and e is the only edge with one end in B ′ and one end in A ′′ .
One may easily deduce the following theorem concerning braces from Lemma 14 and Theorem 10.
Theorem 15.In a brace on six or more vertices, every edge is removable.
Corollary 16.Let G[A, B] be a bipartite matching covered graph on four or more vertices, and let x be a vertex of G.Then, either there is an edge of G, not incident with x, which is removable in G; or there is a subset Proof.If G has only four vertices then the assertion holds immediately.We may thus assume that G has order six or more.If G is a brace, the statement follows from Theorem 15.If not, let Y be a minimal subset of V − x such that ∂(Y ) is a nontrivial tight cut of G.Then, G/Y is a brace, by the minimality of Y .If this brace has order at least six, one can again appeal to Theorem 15.Otherwise |Y | = 3, and the assertion follows.

Graphs obtained by deleting an edge from a brace
Let G be a brace on six or more vertices, and let e = uv be an edge of G.Then, by Theorem 15, G − e is matching covered.We shall now establish two special properties of G − e which play crucial roles in this paper.The first concerns tight cut decompositions of G − e, and the second concerns removable edges in G − e.

Tight cut decompositions of G − e
Suppose that G−e is not a brace, then u and v belong to different shores of any nontrivial tight cut of G − e.Let C be the family of tight cuts corresponding to a tight cut decomposition of G − e.Then the shores of the cuts in C containing u form a nested family of subsets of V , and may be described in the following simple manner.Corollary 13 implies that there is a unique minimal subset of V containing u, say X 1 , such that ∂(X 1 ) is a nontrivial tight cut of G − e.By the minimality of X 1 , it follows that (G − e)/X 1 is a brace.The graph (G − e)/X 1 may or may not be a brace.If it is not, let X 2 denote a (not necessarily unique) minimal subset of V , properly containing X 1 , such that ∂(X 1 ) is a nontrivial tight cut of G − e.Then (G − e)/X 1 /X 2 is a brace (by the minimality of X 2 ).If (G − e)/X 2 is a brace, we have a tight cut decomposition of G − e. Otherwise (G − e)/X 2 /X 3 is a brace for a minimal subset X 3 of V , properly containing X 2 , for which ∂(X 3 ) is a nontrivial tight of G − e. Proceeding in this manner, we obtain a nested family of subsets of V as described in the following lemma: Lemma 17.Let G be a brace of order at least six, and let e = uv be any edge of G such that G − e is not a brace.Then there there exists a nested family 4), each containing u, such that As mentioned above, the first member X 1 of the nested family is unique.Analogously, X k is unique.This observation leads to the following corollary: Corollary 18.Let G be a brace of order at least six, and let e be an edge of G.If G − e has at most three braces, then G − e has a unique tight cut cut decomposition, that is, G − e has a unique maximal laminar family of nontrivial tight cuts.

Removable edges in G − e
We shall now establish a useful general result concerning removable edges in a graph obtained from deleting an edge from a brace.
Lemma 19.Let G[A, B] be a brace on six or more vertices, and let e be an edge of G.Then, every edge that lies in a nontrivial tight cut of G − e is removable in G − e.
Proof.Let C := ∂(X) be a nontrivial cut of G such that C − e is tight in G − e.Let f be any edge of C − e. Assume, to the contrary, that f is not removable in G − e.Then, f is not removable in some (C − e)-contraction of G − e. Adjust notation so that f is not removable in H := (G − e)/(X → x).Adjust notation so that X + ⊂ A. Then, edge e has an end, say v, in X − , in turn a subset of B (Figure 5).
By Lemma 14, the sets (A ∩ V (H)) ∪ {x} and B ∩ V (H) have partitions (A ′ , A ′′ ) and (B ′ , B ′′ ) such that |A ′ | = |B ′ | and f is the only edge of H that joins a vertex of B ′ to a vertex of A ′′ .Thus, f is not removable in G.This is a contradiction, as G is a brace on six or more vertices.
Lemma 20.Let G[A, B] be a brace on six or more vertices, and let e be an edge of G. Let u be a vertex of G having degree three or more in G − e.Then, at most one edge of ∂(u) − e is not removable in G − e.
Proof.If every edge of ∂(u) − e is removable then the assertion holds immediately.We may thus assume that ∂(u) − e contains an edge, f , that is not removable in G − e.Let us prove that f is the only edge of ∂(u) − e that is not removable in G − e.
Adjust notation so that vertex u lies in A. Edge f is not removable in G − e.Thus, there exists a partition (A ′ , A ′′ ) of A and a partition (B ′ , B ′′ ) such that |A ′ | = |B ′ | and f is the only edge of G − e that joins a vertex of B ′ to a vertex of A ′′ (which is u, see Figure 6).
the electronic journal of combinatorics 22(4) (2015), #P4.14 If A ′′ has two or more vertices then A ′ ∪ B ′ ∪ {u} is the shore of a nontrivial tight cut of G − e that contains all the edges of ∂(u) − e − f .In that case, every edge of ∂(u) − e − f is removable in G − e.We may thus assume that A ′′ is a singleton.In that case, B ′′ is also a singleton.As u has degree three or more in G − e, it follows that all the edges of ∂(u) − e − f are multiple edges in G − e, whence removable.In both alternatives, we deduce that every edge of ∂(u) − e − f is removable in G − e.

Thin edges and their indices
Let e be an edge of a graph G.The index of e in G is the number of ends of e having degree two in G − e.An edge e of a brace G is thin if G − e (the retract of G − e) is a brace.Figure 7 illustrates thin edges and non-thin edges of a brace.The three edges e 0 , e 1 and e 2 in Figure 7 are, respectively, thin edges of index zero, one and two in that brace.If an edge e of brace G is thin then G − e is a reduction of G.
e 0 e 1 e 2 f Figure 7: Edges e 0 , e 1 , and e 2 are thin, but f is not.
In the prism P 4n , as described in Section 1.2.1, each of the edges u i v i , 1 i 2n, is a thin edge of index two; no other edge is thin.In the Möbius ladder M 4n+2 , each of the the electronic journal of combinatorics 22(4) (2015), #P4.14 chords u i u i+2n+1 , 1 i 2n + 1, is a thin edge of index two; no other edge is thin.And, in the biwheel, each of the edges incident with a hub is a thin edge of index one; no other edge is thin.
We shall show in Section 2 that every brace on six or more vertices has a thin edge.

Matching minors
We now proceed to describe the notion of a matching minor which is central to this paper.
Although the idea of a matching minor appears in McCuaig's work [11], the term itself was introduced by Norine and Thomas [13].
A matching covered subgraph H of a matching covered graph G is well-fitted if the graph G−V (H) has a perfect matching.(Well-fitted subgraphs are referred to as nice subgraphs by Lovász and Plummer [10], and as central subgraphs by Norine and Thomas [13].)According to Norine and Thomas [13], a matching covered graph J is a matching minor of another matching covered graph G if there exists a well-fitted subgraph H of G such that a graph isomorphic to J is obtainable from H by means of a sequence of bicontractions of vertices of degree two.
Clearly K 2 is a matching minor of every matching covered graph.According to a theorem of Little, any two edges of a matching covered graph are contained together in a well-fitted cycle in that graph [8] (see also [10,Theorem 5.4.4]).It follows that C 4 is a matching minor of any matching covered graph of order at least four.
Using the theory of ear decompositions (see [10] or [2]), it can be shown that a matching covered subgraph H of a bipartite matching covered graph G is well-fitted if and only if H can be obtained from G by bicontractions of vertices of degree two, and deletions of removable edges.We thus have the following characterization of matching minors of bipartite matching covered graphs.
Lemma 21.Let G and J be two bipartite matching covered graphs.Then J is a matching minor of G if and only if there exists a sequence G 1 , G 2 , . . ., G r of graphs such that (i) G 1 = G, and G r ∼ = J, and (ii) for 1 i < r, G i+1 is obtained from G i either by the deletion of a removable edge of G i , or by the bicontraction of a vertex of degree two in G i .(An analogous result also holds for non-bipartite matching covered graphs but, in that case, the deletions of removable doubletons will also have to be permitted.) As an immediate consequence, we have: Corollary 22.Let G be a matching covered graph, and let J be a matching minor of G. Any matching minor of J is also a matching minor of G.
Another useful result is the following: Lemma 23.Let G be a bipartite matching covered graph, and let C := ∂(X) be a tight cut of G.Then, both C-contractions of G are matching minors of G.
Proof.By induction on the number of edges of G. Let us first show that G/X is a matching minor of G.
the electronic journal of combinatorics 22(4) (2015), #P4.14 If there is an edge e in G/X → x, not incident with x, which is removable, then G/X = (G − e)/X.By induction, (G − e)/X is a matching minor of G − e, and hence G/X is a matching minor of G. So, we may assume that no such edge exists.If X is a singleton then G/X and G are isomorphic, the assertion holds immediately.We may thus assume that |X| 3.By Corollary 16, there is a subset Y of X, with |Y | = 3, such that ∂(Y ) is a tight cut, where the minority vertex in Y has degree two in G.In this case, G/Y → y is a matching minor of G because it can be obtained from G by the bicontraction of the minority vertex in Y .However, C is tight in G/Y and G/X = (G/Y )/(X − Y + y).Thus, G/X is a C-contraction of G/Y .Since G/Y has fewer edges than G, we may use induction, and deduce that G/X is a matching minor of G.
A similar argument shows that G/X is a matching minor of G.
Corollary 24.Every brace of a bipartite matching covered graph G is a matching minor of G.
We conclude this section with a crucial result concerning matching minors of bipartite matching covered graphs.

Lemma 25. Let G be a bipartite matching covered graph. A simple brace is a matching minor of G if and only if it is a matching minor of some brace of G.
Proof.Let J be a simple brace.If J is a matching minor of some brace of G then J is a matching minor of G, by Corollary 24.
We prove the converse by induction on the number of edges of G. Assume that J is a matching minor of G.If G is a brace then J, a matching minor of G, is a matching minor of a brace of G.We may thus assume that G is not a brace.By hypothesis, J is a brace.Thus, J and G are distinct.As J is a matching minor of G, then either J is a matching minor of a bicontraction of G or J is a matching minor of G − e, where e is a removable edge of G.
Consider first the case in which G has a vertex v such that J is a matching minor of the graph H, obtained from G by the bicontraction of v.Then, v has degree two and has two distinct neighbours, v 1 and v 2 , where H = G/X 0 and X 0 = {v, v 1 , v 2 }.By induction, J is a matching minor of a brace of H. Cut ∂(X 0 ) is tight in G. Every brace of H is a brace of G. Thus, J is a matching minor of a brace of G.The assertion holds.
We may thus assume that G has a removable edge e such that J is a matching minor of G − e.By induction, J is a matching minor of a brace of G − e, say K. Let K 0 denote the underlying simple graph of K.As J is simple, J is a matching minor of K 0 .We have assumed that G is not a brace.Thus, G has a nontrivial tight cut, C := ∂(X).Every perfect matching of G − e is a perfect matching of G. Thus, C − e is a tight cut of G − e.By the uniqueness of the tight cut decomposition, G − e has a (C − e)-contraction that has a brace, K ′ , which is isomorphic to K, up to multiple edges.
Adjust notation so that K ′ is a brace of (G −e)/X.The simple brace K 0 is isomorphic to the underlying simple graph of K ′ .Thus, J, a matching minor of K 0 , is a matching minor of K ′ .By Corollary 24, J is a matching minor of (G − e)/X.However, either e is not an edge of G/X or e is a removable edge of G/X.In the former alternative, the electronic journal of combinatorics 22(4) (2015), #P4.14 G/X = (G − e)/X, whence J is a matching minor of G/X.In the latter alternative, (G − e)/X is a matching minor of G/X, whence J is a matching minor of G/X, by Corollary 22.In both alternatives, J is a matching minor of G/X.By induction, J is a matching minor of a brace of G/X.Every brace of G/X is a brace of G. Then J is a matching minor of a brace of G.
The brace C 4 is a matching minor of every brace of order four or more.There is a polynomial-time algorithm for deciding whether or not K 3,3 is a matching minor of a given input brace G. (This is due to McCuaig [12] and Robertson, Seymour and Thomas [14], and is related to their theory of Pfaffian orientations.)We are not aware of any work related to the complexity status of the problem of deciding whether or not a simple brace J of order eight or more is a matching minor of a given input brace G.
We conclude this section by noting that if G is member of the family G + , and a simple brace J of order six or more is a matching minor of G, then J is also a member of G + .

Existence of Thin Edges
Throughout this section, G and J denote two distinct braces where G has order at least six, J is simple and is a matching minor of G.We shall first establish the main result of this paper, which asserts that there exists a thin edge e of G such that J is a matching minor of G − e.As noted in the abstract, this result may be derived from McCuaig's work [11].But our approach is quite different and makes it possible for us to deduce that any brace of order six or more has at least two thin edges.
Our proof technique is constructive and is based on the notion of rank of an edge.Given an edge e of G which is not thin, we shall show that there exists an edge of higher rank than e.This leads us to the conclusion that an edge of G of maximum rank is a thin edge with the desired property.(We used a similar technique in [5] for showing that every brick different from K 4 , C 6 , and the Petersen graph has a 'thin' edge.)

The rank of an edge
Since J is a matching minor of G, by Lemma 21, a graph isomorphic to J is obtainable from G by a sequence of bicontractions and deletions of removable edges.As G and J are distinct, any such sequence has at least two members, and its second member is obtained from G either by the bicontraction of a vertex of degree two, or by the deletion of an edge of G.However, as G is a brace of order six or more, it has no vertices of degree two.Therefore there must exist an edge e of G such that J is a matching minor of G − e.
Let R denote the set of edges e of G such that J is a matching minor of G − e.Let e be an edge in R.Then, by Lemma 25, J is a matching minor of one of the braces of G − e.Our objective is to define the notion of the rank of an edge and then show that an edge of maximum rank in R is a thin edge.But, first, we introduce a closely related function which we shall refer to as 'pre-rank'.For an edge e in R, the pre-rank of e, denoted by r 0 (e), is defined to be the maximum of the orders of all braces of G − e that have J as a matching minor.As an example, consider the brace G shown in Figure 8.If J is K 3,3 , then it can be verified that R = {ab, ad, aj, aw, bc, cd, ck, id, ij, ik, xw, yw, yk}.The graph G − aw has two braces, one being the cube and the other being K 3,3 .The cube, being planar, does not contain K 3,3 as a matching minor.Thus r 0 (aw) = 6.The graph G − ab has two braces, one has four vertices and the other has ten vertices and has K 3,3 as a matching minor.Thus r 0 (ab) = 10.
In the brace shown in Figure 8, an edge in R with the largest possible pre-rank is a thin edge.However, this is not in general true.It turns out that, in defining the rank of an edge, in addition to considering the number of vertices in a largest brace of G − e, we need also to consider the number of contraction vertices that brace has.To illustrate this point, consider the brace G in Figure 9, where J is brace K 3,3 .C − e).Edge f is thin and the graph G − f has only one brace isomorphic to K 3,3 , up to multiple edges; it has two other braces, both of order four.Thus, both e and f lie in R and have pre-rank six.But edge e is not thin, whereas f is thin.Note that the braces of G − e have only one contraction vertex, whereas the brace of G − f on six vertices has two contraction vertices.So, in defining the rank of an edge e, we take into account the number of contraction vertices in a largest brace of G − e, giving preference to braces having two contraction vertices.
We thus define the rank r(e) of an edge e in R to be: G − e has a brace of order r 0 (e) with two contraction vertices r 0 (e), otherwise.
We remark that the same edge e may provide different choices for the brace of G − e having J as a matching minor.For example, in Figure 10, G−e has two braces isomorphic to K 3,3 up to multiple edges.One brace contains the vertices in {1, 2, 7, 8, 9}, plus a contraction vertex.The other brace contains vertices in {3, 4, a, b}, plus two contraction vertices.Thus, the latter is responsible for the value seven for r(e). .Let e = uv be an edge in R and suppose that e is not thin.Then there exist two edges f and g in R such that: (i) f and g are adjacent to each other, but not to e; (ii) r(f ) r(e), r(g) r(e); and (iii) either r(f ) > r(e) or r(g) > r(e).(i) G ⋆ has J as a matching minor; the electronic journal of combinatorics 22(4) (2015), #P4.14 (ii) G ⋆ has order r 0 (e); and (iii) if r(e) is odd then G ⋆ has two contraction vertices.
Let G be a tight cut decomposition of G − e that contains braces in G ⋆ .We must now choose a brace G ⋆ of G that lies in G ⋆ .As e is not thin, every brace in G has a contraction vertex that is the result of the contraction of a set having five or more vertices.In particular, every brace in G that lies in G ⋆ has a contraction vertex that is the result of the contraction of a set having five or more vertices.Let X be a maximal such set, let G ⋆ be the corresponding brace.
We note that r 0 (e) is at least four (all braces of any bipartite matching covered graph of order at least four have order at least four).Furthermore, clearly, with equality only if G ⋆ has precisely one contraction vertex.Note that edge e has its ends in X − and X − .Let u denote the end of e in X − , v its other end.Recall that (A, B) denotes the bipartition of G. Adjust notation so that u lies in A. (See Figure 11).
Figure 11: Edges e, f and g.
As |X| 5, it follows that X − contains two or more vertices.Let s ∈ X − be a vertex distinct from u.Then, s has degree at least three in G − e.By Lemma 20, there are edges f and g, both incident with s and removable in G − e.Then G − e − f and G − e − g are matching covered subgraphs of G − e.
Let t 1 and t 2 denote the ends of f and g in B. The vertices of G adjacent to s in G lie all in X + .Thus, both t 1 and t 2 lie in X + (Figure 11).This establishes that f and g are adjacent to each other, but not to e.We shall now prove that f lies in R and r(f ) r(e), and characterize under what conditions equality holds.For this, we consider some cases.In all cases we show that f lies in R. In all cases except the last, we show that r(f ) > r(e).In the last case we conclude that r(f ) = r(e).A similar reasoning holds for g.We then finally show that equalities r(f ) = r(e) and r(g) = r(e) cannot both hold.In this case, either X or X is contained in a shore of every tight cut in G − f .For each nontrivial tight cut D in G − f , the ends of f lie in the minority parts of the shores of D. As f has no end in X, it follows that X is contained in a shore of every nontrivial tight cut in G − f .Then X is contained in the vertex set of a brace F of G − f .As C is not tight in G − f , shore X is, in fact, strictly contained in a shore of every nontrivial tight cut D in G − f .It follows that F has at least |X| + 3 vertices.
Moreover, as (F − e)/X = (G − e)/X, the brace J is a matching minor of F .Thus, r(f ) |V (F )| |X| + 3 > r(e), establishing that f lies in R and r(f ) > r(e).
Case 2. There exist tight cuts in G − f that cross C.
For each nontrivial tight cut of G − f , edge f is incident with vertices in the minority parts of both shores of that cut.Among the (nontrivial) tight cuts of G − f that cross cut C, choose one so that its shore that contains the end s of f in X − is minimal.Let D denote the cut in G, let Y denote the shore that contains vertex s.See Figure 12.By definition, the end s of f belongs to X − , and also to Y ; more specifically to Y − , by Proposition 5. Thus, s belongs to X − ∩ Y − .By using Lemma 12, we deduce the following facts: Since the edge f has both its ends in X, it does not belong to ∂(X ∩ Y ) − e, implying that this cut is tight in G − e.
With the view to finding a lower bound for the rank of f , we now consider nontrivial tight cuts of (G − f )/(Y → y).
the electronic journal of combinatorics 22(4) (2015), #P4.14 the contraction vertex y then D ′ would be a nontrivial tight cut of G itself, a contradiction.Thus, Y ′ is a proper subset of Y .By the minimality of Y , it follows that D ′ and C do not cross.Thus, either Corollary 28.Let S denote the maximal subset of X ∩ Y which contains s and is the shore of a (possibly trivial) tight cut of We now consider two subcases depending on whether or not In this case, |X ∩ Y | = 1, and we have: If equality does not hold all the way through, then, r(f ) > r(e).Alternatively, if equality holds throughout then, by ( 2), (G − e)/X is a brace of G − e of order r(e) with one contraction vertex, whereas with two contraction vertices.By definition of the rank function, we deduce that r(f ) > r(e).In both alternatives, r(f ) > r(e).In this case we also conclude that f lies R and r(f ) > r(e).It follows that, in this case, (G − e)/X is not a brace, whence G ⋆ has two contraction vertices.Let X ′ be the subset of X such that G ⋆ = (G − e)/X/X ′ .
We shall prove that in this case X ′ = X ∩ Y .In other words, G ⋆ is precisely equal to (G − e)/X/(X ∩ Y ).We shall also prove that Observe now the following implications: Proof.Let C ′ := ∂(X ′ ).The cuts ∂(X ∩ Y ) − e and C ′ − e are nontrivial and tight in G − e.The end v of e lies in X, a superset of both X ′ and X ∩ Y .Thus, the end v of e lies in X ′ ∩ X ∩ Y .
Suppose that ∂(X ∩ Y ) and C ′ cross.By Corollary 13, ∂( ) − e would be a nontrivial tight cut of the brace G ⋆ .Thus X ′ ∪ (X ∩ Y ) = X (Figure 13). Let where the congruence, up to multiple edges, follows by Corollary 7. Thus, G ⋆ is isomorphic to L, up to multiple edges.This is a contradiction to the choice of X, as ) − e would be a nontrivial tight cut of brace G ⋆ .On the other hand, note that (G − e)/X/(X ∩ Y ) is a brace, by Lemma 29.If X ′ were a strict super set of X ∩ Y , then ∂(X ′ ) − e would be a nontrivial tight cut of this brace.We deduce that X ′ = X ∩ Y .
Finally, by Lemma 29, In sum, in all cases we have shown that edge f lies in R. We have also proved that r(f ) r(e), with equality only if G has a cut the end s of f lies in X and X ′ = X ∩ Y .Similar conclusions hold for edge g.It now remains to be shown that the last part of the statement holds.For this, assume, to the contrary, that r(f ) = r(e) = r(g).
We then have a shore Y of a nontrivial tight cut of G − f and a shore Z of a nontrivial tight cut of G − g such that the common end s of f and g lies in Y ∩ Z, the contraction set X ′ of G − e is nontrivial, and X ∩ Y = X ′ = X ∩ Z.
Let a 1 and a 2 denote two neighbours of v in G − e.The end v of e lies in B ∩ X ′ , therefore a 1 and a 2 lie all in A ∩ X ′ .As X ∩ Y = X ′ = X ∩ Z, it follows that a 1 and a 2 lie both in A ∩ Y ∩ Z.
The brace G ⋆ contains at least four vertices, two of which are contraction vertices lying in distinct parts of the bipartition of G ⋆ .As X ′ = X ∩ Y , the set of vertices of G ⋆ that are internal, distinct from the contraction vertices, is X ∩ Y .This implies that X ∩ Y contains as many vertices in A as it contains in Let b 2 denote a vertex adjacent to s and distinct from t 1 and t 2 .As s lies in X − , in turn a subset of A, and since ∂(Y In sum, a Likewise, vertices b 1 and b 2 lie in B ∩ Z = Z + , vertices a 1 and a 2 lie in Z, and ∂(Z) − g is a tight cut of G − g.Thus, edge g lies in M. We conclude that M contains both edges f and g.This is a contradiction, as f and g are adjacent.We have thus established the validity of the last part of Theorem 26.
The above theorem implies that an edge of maximum rank in R is thin.We thus have the main result we set out to prove.
Theorem 31 (The Thin Edge Theorem for Braces).Let G be a brace of order at least six, and let J be a simple brace distinct from G. If J is a matching minor of G, then G has a thin edge e such that J is a matching minor of G − e.

Multiple thin edges in braces
We now turn our attention to proving that every brace G on six or more vertices has at least two thin edges.We first note that if a brace J is a matching minor of G, then it is, in general, not true that G has two thin edges e and e ′ such that J is a matching minor of both G − e and G − e ′ .For instance, take J to be any cubic brace on at least eight vertices and let G be obtained from J by adding an edge e joining two nonadjacent vertices.Then e is the only thin edge of G such that J is a matching minor of G − e, because any edge e ′ = e is incident with a vertex of degree three, implying that the retract of G − e ′ has at most |V (G)| − 2 = |V (J)| − 2 vertices.Thus, J cannot be a matching minor of G − e ′ , for any edge e ′ = e.Nevertheless, it would be of interest to establish properties of braces without reference to matching minors of large orders.The following result is a strengthening of Theorem 31 in the case where J = C 4 .Every bipartite matching covered graph of order six or more has C 4 as a matching minor.Thus, when J = C 4 , every edge of the brace belongs to the set R as defined in the beginning of Section 2.1.
Theorem 32.Every brace of order six or more has at least two thin edges.
Proof.Let G be a brace of order six or more, and take J to be C 4 .By applying Theorem 31, we immediately deduce that G has a thin edge.Let e 0 be a thin edge of G.As noted above, all edges in E belong to R. Let e 1 be an edge in E − e 0 of maximum possible rank.If e 1 is thin, then there is nothing more to prove.So, assume that e 1 is not thin.By Theorem 26, with e 1 playing the role of e, there exist two edges f 1 and g 1 such that (i) f 1 and g 1 are adjacent to each other, but are not adjacent to e 1 , (ii) r(f 1 ) r(e 1 ), and r(g 1 ) r(e 1 ), and (iii) either r(f 1 ) > r(e 1 ) or r(g 1 ) > r(e 1 ) (or both).Assume without loss of generality that r(f 1 ) > r(e 1 ).If f 1 = e 0 , the maximality of the rank of e 1 would be violated.So, suppose that f 1 = e 0 .If r(g 1 ) > r(e 1 ), the maximality of the rank of e 1 would again be violated.So, suppose that r(g 1 ) = r(e 1 ).If g 1 is thin, there is nothing more to prove.Assume that it is not.Now, by Theorem 26, with g 1 playing the role of e, there exist two edges f 2 and g 2 such that (i) f 2 and g 2 are adjacent to each other, but are the electronic journal of combinatorics 22(4) (2015), #P4.14 not adjacent to g 1 , (ii) r(f 2 ) r(g 1 ), and r(g 2 ) r(g 1 ), and (iii) either r(f 2 ) > r(g 1 ) or r(g 2 ) > r(g 1 ).Since f 1 = e 0 is adjacent to g 1 , neither f 2 nor g 2 can be equal to e 0 .But then we have a contradiction as to the maximality of the rank of e 1 among the edges in E − e 0 .Hence g 1 is thin, and the desired assertion follows.
Most known families of braces have many thin edges.For instance, biwheels on 2n vertices have 2(n − 1) thin edges.Möbius ladders and prisms on 2n vertices have n thin edges.This leads us to surmise the following: Conjecture 33.There exists a positive constant c such that every brace on n vertices has cn thin edges.
New ideas seem to be necessary to approach even weaker conjectures.For example, we do not know if every brace of order six or more has two nonadjacent thin edges.Our hope is that, by gaining further knowledge about the existence of multiple thin edges, one might be able to settle questions on braces such as our biwheel conjecture.(It states that there exists an integer N such that, for all n N, a brace of order 2n has at least (n − 1) 2 perfect matchings, see [7]).

Strictly Thin Edges
McCuaig [11] implicitly used the notion of thin edges to devise recursive procedure for generating braces.In order to establish such procedures for generating simple braces, where all the intermediate graphs are also simple, one needs the notion of a strictly thin edge.Let G be a simple brace on six or more vertices.An edge e of G is strictly thin if e is thin and the retract of G − e is simple.
We saw in the last section that every brace of order six or more has a thin edge (Theorem 32).However, not every brace has strictly thin edges.For example, Möbius ladders, prisms, and biwheels do not have any strictly thin edges.McCuaig [11] showed that, among other things, a simple brace which does not belong to any one of the above mentioned families has a strictly thin edge.
We shall now proceed to show how Theorems 31 and 32 may be used to deduce the following strenghtening of the above mentioned statement.
Theorem 34 (The Main Theorem).Let G and J be distinct simple braces, where G is not in G + and has more than four vertices, and J is a matching minor of G. Then (i) G has a strictly thin edge e such that J is a matching minor of G − e.
(ii) G has two strictly thin edges.
Part (i) of the above theorem can be deduced from the main theorem in McCuaig [11].Note that, as is the case with thin edges, one cannot claim that there are two strictly thin edges e and f of G such that J is a matching minor of both G − e and G − f .

Multiple edges in retracts
Suppose that G is a simple brace, and e is a thin edge of G.We begin with a brief review of the conditions under which bicontractions of G−e result in a graph with multiple edges.The simplest case arises when e is a thin edge of index one.(If the index of e is zero, then the retract of G − e is itself, and it has no multiple edges.)Let e = x 0 y 0 be such an edge.Let vertex x 0 be the end of e of degree three in G, and y 1 and y 2 be its neighbours distinct from y 0 .If e 1 = y 1 w and e 2 = y 2 w are two edges incident with a common vertex w which belongs to V (G) \ {x 0 , y 1 , y 2 }, then e 1 and e 2 are multiple edges in the retract of G − e.Note that the degree of w is at least four in G. See Figure 14.(The rectangle with rounded corners includes all the non-contraction vertices in the retracts.)

An exchange property of thin edges
In this section, we shall investigate implications of a thin edge being not strictly thin.We shall adopt throughout the notation introduced below.
Notation 35.Let G, J be two distinct simple braces, where G is not in G + and has at least six vertices, and J is a matching minor of G.There are two subsets of E(G), namely T and T − , defined below, which are of special interest to us.
T : the set of thin edges e of G such that J is a matching minor of G − e, and T * : the subset of those edges in T which are also strictly thin.
By Theorem 31, the set T is nonempty.If G is a prism, or a Möbius ladder, or a biwheel, then T * = ∅.If G is an extended biwheel, then T * has just one member, namely the edge that joins the two hubs.Any edge of index zero in T is also in T * .Thus, the index of any edge e := x 0 y 0 in T \ T * will have to be either one or two.Consequently, at least one end of e has degree three.If the index of e is one, we shall adjust notation and assume that y 0 has degree three, and denote its two neighbours in G − e by x 1 and x 2 .If the index of e is two, then x 0 also has degree three, and we shall denote its two neighbours in G − e by y 1 and y 2 .In both cases, we shall let e 1 and e 2 denote two parallel edges in the retract of G − e.When e is of index two then two subsets of V (G), described below, will play a special role in Lemma 40: Having established the requisite notation, let us first note that G must have at least eight vertices because it is simple and, not being a member of G + , is different from K 3,3 .Consequently, H 0 has at least four vertices.By Lemma 25, J is a matching minor of the underlying simple graph of a brace of G. Thus: Proposition 36.The brace J is a matching minor of H 0 .
As our aim is to find a thin edge which is also strictly thin, we should look for thin edges other than e.The two edges e 1 and e 2 , which are thin in H, are obvious candidates for being thin in G as well.As a first step, we establish that these two edges are removable in G − e.
Lemma 37.For i = 1, 2, edge e i is removable in both G and G − e.Moreover, J is a matching minor of both G − e − e i and G − e i .
Proof.By hypothesis, G is a brace and has more than four vertices.Thus, every edge of G is removable.In particular, e i is removable in G.
Let us now prove that e i is removable in G − e. Edge e i is removable in H.If e i is not incident with a contraction vertex of H then e i is certainly removable in G − e.Alternatively, if e i is incident with a contraction vertex of H then e i is removable in G − e, by Lemma 19.In both alternatives, e i is removable in G − e.
In order to obtain H from G − e, index(e) bicontractions are performed.Applying to G − e − e i the same bicontractions, we obtain H − e i .Thus, H − e i is a matching minor of G − e − e i .Consequently, H 0 is a matching minor of G − e − e i .By Proposition 36, J is a matching minor of H 0 .It follows that J is a matching minor of both G − e − e i and G − e i .
Corollary 38.For i = 1, 2, if e i is thin, then e i ∈ T .
Lemma 39.Suppose that e belongs to T \ T * and that its index is one.Then both e 1 and e 2 are thin edges of index one and belong to T .
Proof.Since e is a thin edge of index one, then G − e has just two braces, one of then is H, and the other has order four.Clearly, for i = 1, 2, G − e − e i also has only two braces, one of them being H − e i , and the other being of order four.Graph G − e i , which is obtained by adding e to G − e − e i , is either a brace or has only two braces, one of order four, the other of order |V (H)|.It follows that e i is a thin edge of index one.The assertion follows from Corollary 38.
If the index of the edge e is two, and e 1 and e 2 are parallel edges in the retract of G−e, it is not necessary for both e 1 and e 2 to be thin in G. (For example, consider the brace shown in Figure 16, where e = x 0 y 0 is a thin edge of index two.The edges e 1 = y 1 x 3 and e 2 = y 2 x 3 are parallel edges in the retract H of G − e.The cut ∂({y 0 , y 1 , x 0 , x 1 , x 2 }) is a tight cut of G − e 1 both of whose shores have at least five vertices, implying that e 1 is not a thin edge of G.)Among other things, Lemma 40 asserts that at least one of the two edges e 1 and e 2 is thin in G.  (ii) edge e 2 belongs to T and has index zero or one; (iii) brace G has an edge f distinct from e 2 that belongs to T and such that e 2 is not a multiple edge in the retract of G − f .Let X denote the set {x 0 , x 1 , x 2 } and Y the set {y 0 , y 1 , y 2 }.Edge e has both ends in Z.Thus, either X ⊆ Z + or Y ⊆ Z + .Edge e 1 has one end in Z − , the other in Z, in the same part of G containing the vertices of Z + .Thus, the end w of e 1 in Z does not lie in X ∪ Y .Edge e 1 is incident with y 1 .Thus, y 1 is the end of e 1 in Z − .Edge e has both ends in X.Thus, Z − = {y 0 , y 1 } and Z + = X.Vertex y 1 is adjacent to at least two vertices of Z + .It certainly is adjacent to x 0 , by definition.Adjust notation so that y 1 is adjacent to x 1 .It is possible that y 1 is also adjacent to x 2 .
Edges e 1 and e 2 are parallel in H.Moreover, e 2 is incident with y 2 , whence e 2 = wy 2 .Assume, to the contrary, that e 2 is not thin.Repeating the reasoning done with e 1 , we deduce that N(y 2 ) ⊆ X ∪ {w}.Then, N(Y ) = X ∪ {w}.But G is a brace, |Y | = 3 and |X ∪ {w}| = 4, therefore G has only eight vertices.This is a contradiction, as |Z| 5 = |Z|.We deduce that e 2 is thin in G.We also know that J is a matching minor of G − e 2 .Thus, e 2 lies in T .The end w of e 2 has degree four or more.Thus, e 2 has index zero or one.Now let us turn to the proof of the third part of the assertion.

Edge
One of the (C − e 1 )-contractions of G ′′ is H − e 1 , a brace.The other (C − e 1 )-contraction of G ′′ is a bipartite graph whose underlying simple graph is a 6-cycle with one or two chords.It follows that G ′′ is the splicing of two matching covered graphs, whence it is matching covered.Moreover, C − e 1 is tight in G ′′ .Let us consider a tight cut decomposition of G ′′ in which we use C − e 1 as one of the tight cuts of G ′′ .Then, G ′′ has precisely three braces, two of order four, the third is the brace H − e 1 , up to multiple edges.Brace J is a matching minor of H − e 1 , whence it is also a matching minor of G ′′ .We deduce that J is a matching minor of G ′ .
To complete the proof, we must show that f is thin in G.The graph G ′′ has precisely three braces, two of which of order four, the third of order |V (G)| − 4. For every cut We may thus obtain a tight cut decomposition of G ′′ by starting with a tight cut decomposition of G ′ and then proceed by removing e 1 from each brace obtained and continue with the tight cut decomposition procedure.By doing this, we obtain two braces of order four plus a brace of order |V (G)| − 4.
Let G 1 be the graph obtained from G ′ by bicontracting vertex y 0 .This operation corresponds to a (possibly partial) tight cut decomposition of G − f , where the two graphs are a brace of order four and G 1 .Moreover, G 1 − e 1 has precisely two braces, one of order four, the other of order |V (G)| − 4. Thus, either G 1 is a brace or G 1 has precisely two braces, one of order four.
If G 1 is a brace then certainly f is thin, of index one.Assume thus that G 1 has precisely two braces, one of order four.Graph G 1 has order |V (G)| − 2, whence it has order eight or more.We deduce that G 1 has a vertex which is adjacent only to two vertices.Every vertex of H − e 1 , a brace on six or more vertices, is adjacent to three or more vertices.Thus, every vertex of Z is adjacent to three or more vertices in G 1 .Vertex y 1 is adjacent to three vertices in G 1 , namely, w, x 1 and the contraction vertex of G 1 .The contraction vertex of G 1 is adjacent to three or more vertices, otherwise N(x 0 , x 2 ) = Y , a contradiction to the fact that G is a brace on ten or more vertices.We deduce that x 1 is the vertex of G 1 that is adjacent only to two vertices.As G is simple, it follows that x 1 has degree two in G 1 .Thus, the retract of G 1 is a brace of order |V (G)| − 4. Indeed, f is thin in G.
To complete the proof, let us now show that e 2 is not a multiple edge of G − f .For this, assume the contrary.Then, an end of e 2 is adjacent to an end of f of degree three.The end w of e 2 is not adjacent to the end y 0 of f .Thus, y 2 is adjacent to x 1 , and x 1 must have degree three.In that case, N(x 0 , x 1 ) = Y , whence brace G has only six vertices, a contradiction.

Proof of the Main Theorem
Our proof of the Main Theorem relies on the following crucial result.
Lemma 41 (Key Lemma).Let G and J be distinct simple braces, where G is not in G + and has more than four vertices, and J is a matching minor of G. Suppose that G has a thin edge e such that J is a matching minor of G − e.If e is not strictly thin then G has two strictly thin edges f and g such that J is a matching minor of both G − f and G − g.

Proof.
By hypothesis, G is a simple brace of order six or more that is not a member of G + .The only simple brace on six vertices is K 3,3 , a Möbius ladder.Thus, G has order eight or more.In what follows, we adopt the notation introduced at the beginning of Section 3.2.
Case 1.There are edges of index one in T \ T * .
Let e = x 0 y 0 be an edge of index one in T \ T * , where y 0 has degree three, and let e 1 and e 2 , incident with x 1 and x 2 , respectively, be parallel edges in the retract H. Since e 1 and e 2 are parallel edges in the retract H of G − e, they must have a common end, say w, which has degree four or more in G (see Figure 14).
By Lemma 39, both e 1 and e 2 belong to T .If they are both strictly thin, then there would be nothing more to prove.On the other hand, if either of them, say e 2 , is not strictly thin, then it would be possible to apply Lemma 39 with e 2 playing the role of e and assert the existence of a configuration similar to the one in Figure 14, which is based on e 1 .If that does not yield two strictly thin edges in T , the procedure can be repeated with a new thin in T that is not strictly thin.In this manner, as we argue below, we would be able to show that either there exist two strictly thin edges of the required type, or we would be able to obtain a contradiction to the hypothesis that G / ∈ G + by showing that G is either a biwheel or an extended biwheel.
Let h 1 and h 2 denote two distinct vertices of G, let P := (v 1 , v 2 , . . ., v k ), k 3, be a maximal path in G − {h 1 , h 2 } such that the following properties are satisfied: (i) Each vertex of P is adjacent either to h 1 or to h 2 (but not both).
(ii) Each internal vertex of P has degree three, and the edge that joins it to h 1 or h 2 lies in T \ T * and has index one.
We denote the subgraph of G consisting of the path P , vertices h 1 and h 2 , together with the edges joining h 1 and h 2 to vertices of P by F .See Figure 18.(Such a maximal configuration must exist because, with appropriate relabelling, the subgraph of G shown in Figure 14 yields a configuration with k = 3.) Adjust notation so that v 1 is adjacent to h 1 , whereupon each vertex v i of P , with i odd, is adjacent to h 1 and each vertex of v i , with i even, is adjacent to h 2 .For i = 1, . . ., k, denote by f i the edge that joins v i to h 1 or to h 2 .
By definition, edge f 2 lies in T \T * and has index one.The edges f 1 andf 3 are multiple edges in the retract of G − f 2 .By Lemma 39, with f 2 playing the role of e, f 1 playing the role of e 1 , and f 3 the role of e 2 , we conclude that edge f 1 lies in T .Likewise, f k lies in T .
If f 1 and f k are both strictly thin then the assertion holds in this case.We may thus assume that one of f 1 and f k is not strictly thin.Adjust notation so that f 1 is not strictly thin.Brace G has eight or more vertices, therefore the retract of G − f 2 has six or more vertices.Thus, the common end h 1 of f 1 and f 3 has three or more neighbours in the retract of G − f 2 , whence h 1 has degree four or more in G. Thus, f 1 has index zero or one.We have assumed that f 1 is not strictly thin.Thus, it has index one, whence v 1 has degree three.
Let v denote the vertex of G distinct from h 1 and v 2 that is adjacent to v 1 .As G is bipartite and every internal vertex of P has degree three, either v = v k or v does not lie in V (F ).
Assume that v = v k , this implies that k is even.The graph If h 1 and h 2 are not adjacent then G is a biwheel, otherwise it is an extended biwheel.In any case, it is a contradiction, since by hypothesis G is not in G + .Thus, v does not lie in V (F ).Edge f 1 is thin of index one, but not strictly thin.Moreover, v 1 has degree three.Thus, v is adjacent to v 3 or to h 2 .By the maximality of P , vertex v is not adjacent to h 2 .Thus, v is adjacent to v 3 .Every internal vertex of P has degree three.Thus, k = 3 (see Figure 19).If v 3 has degree three then P is a connected component of G − {h 1 , h 2 , v}; in that case, by Lemma 11, G has only six vertices, a contradiction.Thus, v 3 has degree four or more.We have seen that f k lies in T .Also, h 1 has degree four or more.But k = 3.Thus, f k = h 1 v 3 .We deduce that f 3 has index zero, whence it is strictly thin.
Note that in G − f 2 , the edges vv 1 and vv 3 are parallel.Moreover, as G has eight or more vertices, G − f 2 has six or more vertices, whence vertex v must be adjacent to three or more vertices in G − f 2 .Thus, v has degree four or more in G.By Lemma 39, with f 2 playing the role of e, edge vv 1 playing the role of e 1 and edge vv 3 playing the role of e 2 , we deduce that vv 3 lies in T .In sum, vv 3 is an edge in T whose ends both have degree Figure 19: Edge f 1 is not strictly thin (and k = 3) four or more.Thus, vv 3 is strictly thin.We conclude that vv 3 and f 3 are both strictly thin edges in T ⋆ .The assertion holds.
Case 2. There are no edges of index one in T \ T * .
Let e = x 0 y 0 be an edge of index two in T \ T * , and let e 1 and e 2 be two multiple edges in the retract H. Assume without loss of generality that e 1 is incident with y 1 and e 2 is incident with y 2 .We shall divide the analysis of this case into three subcases depending on where the ends of e 1 and e 2 different from y 1 and y 2 are situated.Note that vertex x 1 has degree four or more, otherwise N({x 0 , x 1 }) = Y , and brace G would have only six vertices, a contradiction.Thus, e 1 and e 2 both belong to T and have index less than two.If e 1 is not strictly thin then it has index one, contrary to the hypothesis of the case under consideration.Thus, e 1 is strictly thin.Likewise, e 2 is strictly thin.The assertion holds in this case.Adjust notation so that e i = x i y i , for i = 1, 2. Note that G has a subgraph formed by the union of two disjoint paths (x 1 , y 0 , x 2 ) and (y 1 , x 0 , y 2 ), and the addition of an edge joining x i y i , for i = 0, 1, 2.Moreover, the edge x 0 y 0 lies in T \ T * .Let F be a maximal subgraph of G formed by the union of two disjoint paths P := (u 1 , u 2 , . . ., u k ) and Q := (v 1 , v 2 , . . ., v k ) (k 3) and the addition of the edges u i v i for i = 1, . . ., k, and such that the edges u i v i lie in T \ T * , for i = 2, . . ., k − 1 (Figure 20).
Note that every internal vertex of P and Q has degree three, as each edge u i v i , for i = 2, . . ., k − 1, lies in T \ T * , and so it is a thin edge of index two.
By definition, edge u 2 v 2 lies in T \ T * .The retract of G − u 2 v 2 has parallel edges u 1 v 1 and u 3 v 3 .By Lemma 40, edge If u 1 v 1 and u k v k are both strictly thin then the assertion holds.We may thus assume that at least one of them lies in T \ T * .Adjust notation so that u 1 v 1 lies in T \ T * .Then, u 1 v 1 has index two, whence u 1 and v 1 both have degree three.Let u be the vertex of Then, u is not an internal vertex of P , nor of Q.Likewise, v is not an internal vertex of P , nor of Q.
Proposition 42.Vertices u and v do not belong to V (F ).
Proof.Suppose that at least one of u and v is in V (F ).Adjust notation so that u is in V (F ).By definition, u = v 1 .We have seen that u is not an internal vertex of P , nor of Q.
whence G is a Möbius ladder.In both alternatives we get a contradiction to the hypothesis that G is not in G + .
We may thus assume that neither u nor v is in V (F ).We have assumed that u 1 v 1 belongs to T \ T * .Thus, the retract of G − u 1 v 1 has multiple edges.Thus, either u is adjacent to a vertex in {v, v 2 , u 3 } or v is adjacent to a vertex in {u, u 2 , v 3 }.Vertices u and v 2 are not adjacent.Likewise, vertices v and u 2 are not adjacent.By the maximality of F , vertices u and v are not adjacent.We deduce that either u is adjacent to u 3 or v is adjacent to v 3 .Adjust notation so that u is adjacent to u 3 .Then, k = 3 (Figure 21).Proof.Assume, to the contrary, that N(u This implies that G has only eight vertices.As v is not adjacent to u 2 , it follows that N(v) = {u, v 1 , v 3 }.In particular, u and v are adjacent, a contradiction.We conclude that u 3 has degree four or more in G.This conclusion implies that u 2 u 3 has index one.Thus, G − u 2 u 3 has six or more vertices.But in G − u 2 u 3 , vertex v 1 is adjacent only to v and to the contraction vertex of G − u 2 u 3 .Thus, G − u 2 u 3 is not a brace.We deduce that u 2 u 3 is not thin in G.
By Lemma 40, with u 1 v 1 playing the role of e, u 2 u 3 the role of e 1 and uu 3 the role of e 2 , we have that uu 3 lies in T .Edge u 3 v 3 also lies in T .Vertex u 3 has degree four or more, therefore both uu 3 and u 3 v 3 have index less than two.By the hypothesis of the case, they are both strictly thin.We conclude that u 3 u and u 3 v 3 are strictly thin edges in T ⋆ .Case 2.3.Edges e 1 and e 2 share a common w end not in X ∪ Y (Figure 15(c)).
In this case, we shall also prove that at least one of e 1 and e 2 lies in T and is strictly thin.

Vertex w has degree four or more in G.
Proof.Assume that w has degree three.In the retract H of G − e, vertex w is adjacent only to two vertices.Then, H has only four vertices.We conclude that G has only eight vertices (Figure 22).The case in which x 1 is adjacent to both y 1 and y 2 has already been considered in Case 2.1, therefore x 1 has degree three in G.Likewise, x 2 has degree three in G. Thus, the four vertices of the part of G that contains vertex x 0 have degree three in G. Thus, G is cubic.The only cubic brace on eight vertices is the cube.This is a contradiction, as the cube is a prism.Consider first the case in which both e 1 and e 2 are in T .Vertex w has degree four or more.Thus, e 1 has index zero or one.If e 1 is not strictly thin then it has index one, the electronic journal of combinatorics 22(4) (2015), #P4.14 a case already considered (Case 1).Thus, e 1 is strictly thin.Likewise e 2 is strictly thin.The assertion holds.
We may thus assume that one of e 1 and e 2 does not belong to T .Adjust notation so that e 1 is not in T .By Lemma 40, edge e 2 lies in T and has index zero or one.Edges in T \ T * of index one have been considered in Case 1.We may thus assume that e 2 is strictly thin.
To complete the proof, we must now prove that G has an edge distinct from e 2 that also lies in T and is strictly thin.Edge e 1 does not lie in T .By Lemma 40, G has an edge f in T distinct from e 2 and such that e 2 is not a multiple edge in the retract of G − f .
If f is strictly thin then the assertion holds, because f and e 2 are distinct.We may thus assume that f is not strictly thin.The case in which f has index one has already been considered (Case 1).We may thus assume that f has index two.Let f 1 and f 2 denote two parallel edges of the retract of G − f .The case in which f 1 and f 2 share an end adjacent with an end of f has already been considered (Case 2.1).The case in which f 1 and f 2 are not adjacent has already been considered (Case 2.2).We may thus assume that f 1 and f 2 share a common end not adjacent to an end of f .We have seen that in this case at least one of f 1 and f 2 lies in T and is strictly thin.Adjust notation so that f 2 lies in T and is strictly thin.Edge e 2 is not a parallel edge in the retract of G − f .Thus e 2 and f 2 are distinct strictly thin edges in T .The assertion holds.
With the aid of the above lemma and Theorems 31 and 32, it is now straightforward to deduce the validity of Theorem 34.
Proof of the Main Theorem.Let us first prove the validity of item (i) of the Main Theorem.By Theorem 31, G has a thin edge e such that J is a matching minor of G − e.If e is strictly thin, then item (i) holds.We may thus assume that e is not strictly thin.By Lemma 41, G has two strictly thin edges f and g such that J is a matching minor of both G − f and G − g.In both alternatives, item (i) holds.
Let us now prove that G has at least two strictly thin edges.By Theorem 32, G has two thin edges, e and f .If both e and f are strictly thin then the Theorem is proved.Adjust notation so that e is not strictly thin.Choose any simple brace J that is a matching minor of G − e.For instance, let J := C 4 .By Lemma 41, G has two strictly thin edges e 1 and e 2 such that C 4 is a matching minor of both G − e 1 and G − e 2 .In both alternatives, G has two strictly thin edges.In sum, the Main Theorem is reduced to the Key Lemma.

Braces with just Two Strictly Thin Edges
In this section we give examples of simple braces which have just two strictly thin edges thereby showing that our Main Theorem 34 provides the best possible lower bound on the number of strictly thin edges in a brace.Our constructions are based on the operation of 4-sum which appears in the works of Robertson, Seymour, and Thomas [14] and of McCuaig [12].
Let G 1 , G 2 , . . ., G r be r distinct graphs, and let Q be a cycle of length four such that G i ∩ G j = Q, for 1 i < j r.Then, for any fixed subset R (possibly empty) of the The graphs G − uv and G − xy are simple braces which are also 4-sums of three copies of K 3,3 (with different proper subsets of E(Q) designated as the set R). Therefore both uv and xy are strictly thin edges (of index zero) of the brace G.However, no edge in E(G) − {uv, xy} is strictly thin.To see this, using the symmetries of G, it suffices to check that the two edges au and ab are not strictly thin.The edge au is not strictly thin because, in the retract of G − au, the edges xy and by are parallel.And, the edge ab is not strictly thin because, in the retract of G − ab, the edges uv and xy are parallel.
In exactly the same manner as above, it can be shown that the 4-sum of any r copies of K 3,3 , with r 3, where R consists of two nonadjacent edges of their shared 4-cycle, is a brace with exactly two strictly thin edges.

Thin Edges in Bricks
The notions of thin edges and strictly thin edges extend in an obvious manner to bricks.A removable edge e of a brick G is thin if the retract G − e of G − e is a brick; and if G − e is a simple brick, then e is a strictly thin edge of G.In this concluding section, we briefly review some results related to thin and strictly thin edges in bricks which are analogous to those concerning braces discussed in this article.
A removable edge e of a brick G is b-invariant if G−e has at most one brick.Confirming a conjecture made by Lovász in 1987, we showed in [3] that every brick different from K 4 , C 6 , R 8 (which is obtained by splicing K 4 and C 6 ), and the Petersen graph has at least two the electronic journal of combinatorics 22(4) (2015), #P4.14 b-invariant removable edges.From this result, we deduced in [5] that every brick distinct from K 4 , C 6 and the Petersen graph has a thin edge, and used this conclusion to describe a generation procedure for bricks.
Theorem 43.Given any brick G, there exists a sequence (G 1 , G 2 , . . ., G r ) of bricks such that: (i) G 1 ∈ {K 4 , C 6 , Petersen graph}, (ii) G r ∼ = G, and (iii) for 2 i r, the brick G i has a thin edge e i such that G i−1 ∼ = G i − e i implying that G i can be obtained from G i−1 by one of four types of elementary expansion operations.
Just as there are families of braces (prisms, Möbius ladders and biwheels) which do not have strictly thin edges, there are families of bricks which do not have strictly thin edges.Norine and Thomas [13] discovered that, apart from the Petersen graph, there are five infinite families of such bricks.They include prisms and Möbius ladders whose orders are divisible by four, odd wheels, and two other families which Norine and Thomas refer to as prismoids and staircases.For convenience, let us denote by N T the class of bricks consisting of the Petersen graph and members of these five infinite families.In the same paper cited above, Norine Thomas proved that any simple brick which does not belong to N T has a strictly thin edge.This significant work was quite independent of our work, and used methods entirely different from ours.They state their result as a generation procedure for simple brick which is analogous to the procedure for generating simple braces due to McCuaig [11].(The interpretation in terms of strictly thin edges is ours.)Subsequently, we were able to show that their result can be deduced from our theorem on thin edges [5] and described this in [6].We submitted this paper to a leading journal; it was rejected on the grounds that it, in their opinion, merely presents a new proof of a known result.We urge the interested reader to take a look at this unpublished article for an alternative perspective on the important works of McCuaig [11] on braces and Norine and Thomas [13] on bricks.We do not know whether or not every brick (other than a few exceptions) has at least two thin edges.However, Nishad Kothari (a graduate student at the University of Waterloo) has found, by means of extensive computations, a number of bricks with just one strictly thin edge.One of the bricks he discovered is shown in Figure 24.This brick has five thin edges, e i , 1 i 5, of which only e 5 is strictly thin.Thus the natural analogue of Theorem 34 to bricks does not hold.

Proposition 6 (
Modularity).Let G be a matching covered graph, C := ∂(X) andD := ∂(Y ) two tight cuts of G.If |X ∩ Y | is odd then each of ∂(X ∩ Y ) and ∂(X ∩ Y ) is tight and no edge of G joins a vertex of X ∩ Y to a vertex of X ∩ Y .
matching, for any two vertices a 1 and a 2 in A and any two vertices b 1 and b 2 in B; (c) |N(X)| > |X| + 1, for every subset X of A such that 0 < |X| < |A| − 1.

Lemma 12 .
Let G := G[A, B] be a brace.Let e and f be two (not necessarily distinct) edges of G such that each of the graphs G − e, G − f and G − e − f is matching covered.Let C := ∂(X) and D := ∂(Y ) be two crossing cuts of G such that C − e is tight in G − e the electronic journal of combinatorics 22(4) (2015), #P4.14 and D

Figure 10 :
Figure 10: The graph G − e has two braces of order six

Proof.
By the definition of r(e), there is a brace of G − e of order r 0 (e) which has J as a matching minor.As e is not thin by hypothesis, G − e has at least two braces, and r(e) < |V (G)|.For every tight cut decomposition of G − e, every brace has at most two contraction vertices, by Lemma 17.Consider all the tight cut decompositions of G − e.Let G ⋆ denote the set of those braces G ⋆ of G − e that satisfy the following properties:

Figure 12 :
Figure 12: The four quadrants in Case 2 up to multiple edges, by Corollary 7. By definition, the graph (G − e)/X has G ⋆ as a brace.It follows then that the graph (G − e − f )/Y /(X ∩ Y ) has a brace isomorphic to G ⋆ , up to multiple edges.Let S denote the subset of X ∩ Y , and H := (G − f )/Y /S denote the brace as defined in Corollary 28.Then ∂(X ∩ Y ) − e − f is a tight cut of H − e, implying that (G − e − f )/Y /X ∩ Y is a matching minor of H − e.Thus, the underlying simple graph of G ⋆ is a matching minor of H.As J is simple, it follows that J is a matching minor of H. Thus, f lies in R. Now, as H is a brace of G − f , we have: r 0 (f ) |V (H)|, by the definition of r 0 (f ) |X ∩ Y | + 2, by Corollary 28 = |X| + 1, because, by assumption, |X ∩ Y | = 1 r(e), by (2).

Case 2 . 2 .
The cut ∂(X ∩ Y ) − e is nontrivial and tight in G − e.the electronic journal of combinatorics 22(4) (2015), #P4.14 Furthermore, the graphs (G − f )/Y /(X ∩ Y ) and (G − e)/X/(X ∩ Y ) are braces, which are isomorphic, up to multiple edges.Proof.Since ∂(X ∩ Y ) − e is nontrivial and tight in G − e, the end v of e in X belongs to Y .But, as v ∈ B, it follows that v belongs, in fact, to Y − .This implies that all neighbours of v in G − f belong to Y + because ∂(Y ) − f is a tight cut of G − f .In particular, vertex u, which is joined to v by e, belongs to Y + .Thus, both ends of e lie in Y , whence e does not lie in ∂

1 and a 2
are two vertices in A ∩ Y ∩ Z, whereas b 1 and b 2 are two vertices in B ∩ Y ∩ Z. Graph G is a brace.By Theorem 10(b), G − a 1 − a 2 − b 1 − b 2 has a perfect matching, M. Vertices b 1 and b 2 lie in B ∩ Y = Y + , whereas vertices a 1 and a 2 lie in Y .As ∂
, and Y := {y 0 , y 1 , y 2 }.Finally, for brevity, we shall use the following notation for the retract of G − e and its underlying simple graph: H: the retract G − e of G − e, and H 0 : the underlying simple graph of H.

4 Figure 16 :
Figure 16: Index of e is two, e 1 is not thin

(i) edges e 1
and e 2 share an end w that is not adjacent to any end of e, as inFigure  15(c);

Proof.
Since e is a thin edge of G of index two, the graph G − e has precisely three braces, one is H, the other two have order four.Moreover, H has two contraction vertices.Thus, G − e − e 1 also has three braces, one is H − e 1 , the other two have order four.Brace H has order |V (G)| − 4 and H − e 1 is a brace of G − e − e 1 .By hypothesis, edge e 1 is not thin.Thus G has a cut C := ∂(Z) such that C − e 1 is tight in G − e 1 and both Z and Z have five or more vertices.Consequently, G has 10 or more vertices, and H has six or more vertices.Moreover, one of the (C − e − e 1 )-contractions of G − e − e 1 is isomorphic to H − e 1 , the other (C − e − e 1 )-contraction has two braces, both of order four.Adjust notation so that (G − e − e 1 )/(Z → z) ≃ H − e 1 , whereupon |Z| = 5.Because H − e 1 is a brace of order at least six, no vertex of Z has degree two in G − e − e 1 .Thus, edge e has both ends in Z.Likewise, the end w of e 1 in Z has degree four or more in G.As e has both ends in Z, (G − e 1 )/(Z → z) ≃ H − e 1 (See Figure 17).

Figure 17 :
Figure 17: The cut C

Case 2 . 1 .
Edges e 1 and e 2 share a common end in X ∪ Y .Adjust notation so that e i = x 1 y i , for i = 1, 2 (Figure 15(a)).By Lemma 40, edges e 1 and e 2 are both thin, whence, by Corollary 38, they belong to T .

2 w
Figure 22: Brace G is the cube

5 Figure 24 :
Figure24: A brick with a unique strictly thin edge (e 5 )