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\title{On the connectivity function of a binary matroid\footnote{Research partially supported by Conselho Nacional de Desenvolvimento Cient\'\i fico e Tecnol\'ogico (CNPq, Brazil).}}

\author{Manoel Lemos\\
        Departamento de Matem\'atica\\
        Universidade Federal de Pernambuco\\
        Cidade Universit\'aria\\
        Recife, PE, 50740-540, Brazil}

\date{e-mail: manoel@dmat.ufpe.br}

\begin{document}

\maketitle

\begin{abstract}
In this paper, we shall consider the following problem: up to duality, is a connected matroid reconstructible from its connectivity function? Cunningham conjectured that this question has an affirmative answer, but Seymour gave a counter-example for it. In the same paper, Seymour proved that a connected binary matroid is always reconstructible from its connectivity function, up to duality, in the class of binary matroids. In this paper, we prove that when $M$ is a connected binary matroid and $N$ another matroid with the same connectivity function as $N$, then $M=N$ or $M=N^*$.  

Also, we shall discuss the implications of ours results on the following problem: when it exists, is the decomposition of the rank function of a 2-polimatroid as the sum of the rank functions of matroids unique?
\end{abstract}

\section{Introduction}

The matroid theory terminology used in this paper will follow Oxley [7]. The {\it connectivity function} of a matroid $M$ is defined as
$$\xi_M(X,Y)=r_M(X)+r_M(Y)-r(M)+1,$$
where $\{X,Y\}$ is a partition of $E(M)$. This function is invariant under duality, since it may be written as
$$\xi_M(X,Y)=r_M(X)+r_{M^*}(X)-|X|+1.$$
In particular, a matroid and its dual have the same connectivity function. When $M=M_1\oplus M_2$, then
$$\xi_M(X,Y)+1=\xi_{M_1}(X\cap E(M_1),Y\cap E(M_1))+\xi_{M_2}(X\cap E(M_2),Y\cap E(M_2)).$$
Thus, $M_1\oplus M_2$ and $M_1\oplus M_2^*$ have the same connectivity function.

In this paper, we shall study matroids with the same connectivity function. By the last identity, we may restrict our attention to connected matroids. In this case, Cunningham conjectured that, up to duality, a connected matroid is reconstructed from its connectivity function, that is:
\vspace{0.3cm}

\noindent {\bf Cunningham conjecture.} If $M$ and $N$ are connected matroids on the same ground set having the same connectivity function, then $M=N$ or $M=N^*$.
\vspace{0.3cm}

In [11], Seymour proved the Cunningham conjecture for the class of binary matroids, that is:
\vspace{0.3cm}

\noindent {\bf Theorem 1.} (3.3 of Seymour[11]) If $M$ and $N$ are connected binary matroids on the same ground set having the same connectivity function, then $N=M$ or $N=M^*$.\qeds
\vspace{0.3cm}

In the same paper, Seymour gave a counter-example for it in general. In [1], Lemos proved that this conjecture is true provided that $r(M)\not=r(M^*)$. In this paper, we shall prove that a connected binary matroid is reconstructible, not only in the class of binary matroids, but in the class of all matroids. Our main result is:
\vspace{0.3cm}

\noindent {\bf Theorem 2.} Suppose that $M$ is a connected binary matroid. If $N$ is a matroid having the same connectivity function as $M$ and $E(N)=E(M)$, then $N=M$ or $N=M^*$.
\vspace{0.3cm}

Note that 
$$r_M(X)+r_{M^*}(X)=r_N(X)+r_{N^*}(X),$$
for every $X\subseteq E$, when $M$ and $N$ are matroids over the same ground set $E$ having the same connectivity function. In general, we say that ${\bf M}=(M_1,M_2,M_3,$ $M_4)$ is a {\it quad} when $M_1,M_2,M_3$ and $M_4$ are matroids over the same ground set, which is denoted by $E({\bf M})$, such that
$$r_{M_1}(X)+r_{M_2}(X)=r_{M_3}(X)+r_{M_4}(X),$$
for every $X\subseteq E({\bf M})$. 

As one can guess from the definition of quads, this problem is related to the following problem which was proposed by Murty and Simon [6]: characterize polimatroid rank functions which are sums of rank functions of matroids? We also may ask: when this decomposition exists is it unique? It follows from ours results that a connected 2-polimatroid that is decomposed as the sum of rank functions of two binary matroids $M_1$ and $M_2$ cannot have another decomposition as the sum of rank functions of two other matroids $M_3$ and $M_4$, unless $(M_1,M_2,M_3,M_4)$ is a well described quad. Moreover, when the decomposition is not unique, then there is only two ways of doing it. In [1], Lemos proved that the decomposition of the rank function $\rho$ of a connected 2-polimatroid, defined on $E$, as the sum of the rank function of two matroids is unique, when it exists, provided that $\rho(E)$ is odd. We shall settle the problem of uniqueness of decomposition of the rank function of a connected 2-polimatroid in another paper, using the theory developed here.

We are setting a different notation from Lemos [1] to get a more compact description of the results. When ${\bf X}$ denotes a quad, $X_i$ denotes the $i$-th coordinate of this quad, for $i\in\{1,2,3,4\}$.

We say that two quads ${\bf M}$ and ${\bf N}$ are {\it equivalent}, and denote this by ${\bf M}\equiv {\bf N}$, when
$$\{\{M_1,M_2\},\{M_3,M_4\}\}=\{\{N_1,N_2\},\{N_3,N_4\}\}.$$
We say that two quads ${\bf M}$ and ${\bf N}$ are {\it isomorphic}, and denote this by ${\bf M}\simeq {\bf N}$, when there is a quad ${\bf H}$ such that ${\bf H}\equiv {\bf M}$ and a bijection $f:E({\bf H})\rightarrow E({\bf N})$ such that $f$ is an isomorphism between $H_i$ and $N_i$, for every $i\in\{1,2,3,4\}$.

For a quad ${\bf M}$ and disjoint subsets $X$ and $Y$ of $E({\bf M})$, we define a {\it minor} of ${\bf M}$ as  
$${\bf M}\backslash X/Y=(M_1\backslash X/Y,M_2\backslash X/Y,M_3\backslash X/Y,M_4\backslash X/Y)$$
and the {\it dual} of ${\bf M}$ as ${\bf M}^*=(M_1^*,M_2^*,M_3^*,M_4^*)$. We shall use ${\bf M}|X$ to denote ${\bf M}\backslash (E({\bf M})\setminus X)$. In [1], Lemos have proved that 

\bteo\rotulo{ia}
(1.1, 1.2 and 1.3 of [1]) ${\bf M}\backslash X/Y$ and ${\bf M}^*$ are quads.\qeds
\eteo

When ${\bf M}$ and ${\bf N}$ are quads such that $E({\bf M})\cap E({\bf N})=\emptyset$, we denote by ${\bf M}\oplus{\bf N}$ the quad
$$(M_1\oplus N_1,M_2\oplus N_2,M_3\oplus N_3,M_4\oplus N_4).$$
Case $N_1=N_2=N_3=N_4=N$, ${\bf M}\oplus {\bf N}$ is also denoted by ${\bf M}\oplus N$. For a matroid $N$ such that $E(N)\cap E({\bf M})=\{e\}$, ${\bf M}\oplus_2 N$ denotes
$$(M_1\oplus_2 N,M_2\oplus_2 N,M_3\oplus_2 N,M_4\oplus_2 N),$$ 
which was proved to be a quad by Lemos in [1].

In [1], Lemos proved the next result which characterizes a quad and it was proved by Seymour for a class of quads, in [11]:

\bteo\rotulo{ic}
(1.4 and 1.5 of [1]) Suppose that $M_i$ is a matroid having $E$ as its ground set, for every $i\in\{1,2,3,4\}$. $(M_1,M_2,M_3,M_4)$ is a quad if and only if 
$${\cal C}(M_1)\cup{\cal C}(M_2)={\cal C}(M_3)\cup{\cal C}(M_4)\hbox{ and }{\cal C}(M_1)\cap{\cal C}(M_2)={\cal C}(M_3)\cap{\cal C}(M_4).\hbox{\qeds}$$

Now, we outline the proof of theorem 2: when $M$ and $N$ satisfies the hypothesis of theorem 2 and its conclusion does not hold, then $(M,M^*,N,N^*)$ belongs to the class ${\cal S}$ of quads ${\bf M}$ satisfying: $M_1\not=M_3$, $M_1\not=M_4$ and both $M_1$ and $M_2$ binary matroids. The ``irreducible'' quads of ${\cal S}$ shall be constructed in section four and all of them have exactly one disconnected matroid as one of its coordinates. So, none of them can be equivalent to $(M,M^*,N,N^*)$. The construction of the ``irreducible'' quads of ${\cal S}$ shall be done by induction and the splitter-type theorem for quads is fundamental in this construction --- its statement and proof can be found in section three: essentially, it says that an ``irreducible'' quad can be obtained from a well described ``irreducible'' quad by a sequence of one element lift or extension without leaving the class of ``irreducible'' quads. 

\section{Connectivity results for quads}

In this section, we shall prove some connectivity results for quads which will be fundamental in the proof of the splitter-type theorem for quads which shall be presented in the next section. Here, the notions of connectivity and 3-connectivity for matroids are extended for quads.

Let ${\bf M}$ be a quad. For a partition $\{X,Y\}$ of $E({\bf M})$, we denote by
$$\xi_{\bf M}(X,Y)=\xi_{M_1}(X,Y)+\xi_{M_2}(X,Y)=\xi_{M_3}(X,Y)+\xi_{M_4}(X,Y).$$
When $\xi_{\bf M}(X,Y)=2$, then ${\bf M}=({\bf M}|X)\oplus ({\bf M}|Y)$. So, we can restrict our attention to quads satisfying $\xi_{\bf M}(X,Y)\ge 3$, for every partition $\{X,Y\}$ of $E({\bf M})$ such that $\min\{|X|,|Y|\}\ge 1$. Such quads are called {\it connected}. A quad ${\bf M}$ is said to be {\it non-trivial}, when $\{M_1,M_2\}\not=\{M_3,M_4\}$. In [1], Lemos proved the following result:

\bteo\rotulo{ib}
Suppose that ${\bf M}$ is a non-trivial connected quad. Then,
\begin{description}

\item{(i)} (2.6 of [1]) $\xi_{\bf M}(X,Y)\ge 4$, for every partition $\{X,Y\}$ of ${\bf M}$ such that $\min\{|X|,|Y|\}\ge 1$.

\item{(ii)} (2.7 of [1]) When $\xi_{M_1}(X,Y)=\xi_{M_2}(X,Y)=\xi_{M_3}(X,Y)=\xi_{M_4}(X,Y)=2$, for a partition $\{X,Y\}$ of $E({\bf M})$ such that $\min\{|X|,|Y|\}\ge 2$, then there is a matroid $N$ such that $E(N)=X\cup\{e\}$ or $E(N)=Y\cup\{e\}$, for an element $e\not\in E({\bf M})$, and a quad ${\bf N}$ such that $E({\bf N})=(E({\bf M})\setminus E(N))\cup\{e\}$ satisfying ${\bf M}={\bf N}\oplus_2 N$.\qeds
\end{description}
\eteo

As a consequence of \recorde{ib}(i), we have the following:

\bteo\rotulo{21}
If ${\bf M}$ is a non-trivial connected quad, then $M_i$ is both loopless and coloopless, for every $i\in\{1,2,3,4\}$.\qeds
\eteo

This is a particular instance of a more general result of Lemos [1]:

\bteo\rotulo{extra}
(2.3 of [1]) Let ${\bf M}$ be a connected quad. If $\{X,Y\}$ is a 1-separation for both $M_1$ and $M_3$ such that $M_1|X=M_3|X$, then ${\bf M}$ is trivial.\qeds
\eteo

\bteo\rotulo{id}
(theorem 2 of [1]) If ${\bf M}$ is a non-trivial connected quad, then 
$$r(M_1)=r(M_2)=r(M_3)=r(M_4).\hbox{\qeds}$$
\vspace{0.05cm}

By this theorem, for a non-trivial connected quad ${\bf M}$, we can define its {\it rank}, which is denoted by $r({\bf M})$, as the rank of  $M_i$, for $i\in\{1,2,3,4\}$.

We say that $\{X,Y\}$ is a $k$-{\it separation} for a quad ${\bf M}$, when $\{X,Y\}$ is an exact $k$-separation for $M_i$, for every $i\in\{1,2,3,4\}$. By \recorde{ib}(ii), when, $\{X,Y\}$ is a 2-separation for a non-trivial connected quad ${\bf M}$, then ${\bf M}={\bf N}\oplus_2 N$, where ${\bf N}$ is a non-trivial quad and $N$ is a matroid such that $E(N)=Z\cup\{e\}$ and $e\in E(N)\cap E({\bf N})$, for some $Z\in\{X,Y\}$. We say that $Z$ is the {\it dead} set associated with the 2-separation $\{X,Y\}$. A quad is said to be {\it irreducible}, when it does not have a 1-separation or a 2-separation. Observe that the deletion or the contraction of an element of an irreducible quad gives a connected quad.

\bteo\rotulo{lemma1}
Let ${\bf M}$ be a non-trivial connected quad. If $\{X,Y\}$ is a 2-separation for ${\bf M}$, then, for every $Z\in\{X,Y\}$, we have that
$$r_{M_1}(Z)=r_{M_2}(Z)=r_{M_3}(Z)=r_{M_4}(Z).$$
Moreover, when $|X|=2$, $X$ is contained in a series class of $M_i$, for every $i\in\{1,2,3,4\}$, or $X$ is contained in a parallel class of $M_i$, for every $i\in\{1,2,3,4\}$.
\eteo

\noindent {\bf Proof:} Without loss of generality, we may suppose that $Y$ is the dead set of this 2-separation. Thus $r_{M_1}(Y)=r_{M_2}(Y)=r_{M_3}(Y)=r_{M_4}(Y)$. Hence $r_{M_1}(X)=r_{M_2}(X)=r_{M_3}(X)=r_{M_4}(X)$, because $r_{M_i}(X)+r_{M_i}(Y)-r(M_i)=1$ and $r(M_i)=r({\bf M})$, for every $i\in\{1,2,3,4\}$. Moreover, when $|X|=2$, $X$ is independent in all these matroids or $X$ is dependent in all these matroids and the result follows. \qed

\bteo\rotulo{lemma2} 
If $\{X_1,Y_1\}$ and $\{X_2,Y_2\}$ are different 2-separations for a non-trivial connected quad ${\bf M}$ such that $X_1$ and $X_2$ are maximal dead sets, then $X_1\cap X_2=\emptyset$.
\eteo

\noindent {\bf Proof:} Suppose that $X_1\cap X_2\not=\emptyset$. Since both $X_1$ and $X_2$ are maximal dead sets, we have that $Y_1\cap X_2\not=\emptyset$ and $X_1\cap Y_2\not=\emptyset$. 

For $i\in\{1,2\}$, as $X_i$ is a dead set, it follows that ${\bf M}={\bf N}_i\oplus_2 N_i$, where ${\bf N}_i=(N_{i1},N_{i2},N_{i3},N_{i4})$ is a quad such that $E({\bf N}_i)=Y_i\cup\{e_i\}$ and $N_i$ is a matroid such that $E(N_i)=X_i\cup\{e_i\}$. 

First, we shall prove also that $Y_1\cap Y_2\not=\emptyset$. Suppose that $Y_1\cap Y_2=\emptyset$. In this case, we shall prove that $M_i=M_j$, for every $i$ and $j$ such that $\{i,j\}\subset\{1,2,3,4\}$, and we arrive at a contradiction. To demonstrate this, by symmetry, it is enough to prove that every circuit $C$ of $M_i$ is also a circuit of $M_j$. We may suppose that $C\cap Y_k\not=\emptyset$, for both $k\in\{1,2\}$, otherwise $C$ is a circuit of $N_k\backslash e_k$, and hence $C$ would be also a circuit of $M_j$. For $k\in\{1,2\}$, it follows that $(C\cap X_k)\cup\{e_k\}$ is a circuit of $N_k$. Hence, there is a circuit $C'$ of $M_j$ such that $C'\cap X_1=C\cap X_1$ and $C'\cap Y_1\not=\emptyset$. As $Y_2\subseteq X_1$ and $C\cap Y_2\not=\emptyset$, it follows that $(C'\cap Y_2)\cup\{e_2\}=(C\cap Y_2)\cup\{e_2\}$ is a circuit of $N_{2j}$. Since $(C\cap X_2)\cup\{e_2\}$ is a circuit of $N_2$, we have that $C=(C\cap Y_2)\cup(C\cap X_2)$ is a circuit of $M_j$. So, $Y_1\cap Y_2\not=\emptyset$.

For $i\in\{1,2\}$, we set $\{Z_i,W_i\}=\{X_i,Y_i\}$. Using submodularity, we arrive at the following:
$$\xi_{M_k}(Z_1\cap Z_2,W_1\cup W_2)+\xi_{M_k}(Z_1\cup Z_2,W_1\cap W_2)\le\xi_{M_k}(X_1,Y_1)+\xi_{M_K}(X_2,Y_2)=4,$$
for every $k\in\{1,2,3,4\}$. Thus, for ${\cal F}=\{X_1\cap X_2,X_1\cap Y_2,Y_1\cap X_2,Y_2\cap X_2\}$,
$$\sum_{Z\in{\cal F}}\xi_{M_k}(Z,E({\bf M})\setminus Z)\le 8.\eqno{\bf (1)}$$
By \recorde{ib}(i), for every $\{i,j\}$ equal to $\{1,2\}$ or $\{3,4\}$,
$$\xi_{M_i}(Z_1\cap Z_2,W_1\cup W_2)+\xi_{M_j}(Z_1\cap Z_2,W_1\cup W_2)\ge 4.$$
When, we take the sum over ${\cal F}$, we get
$$\sum_{Z\in{\cal F}}\xi_{M_i}(Z,E({\bf M})\setminus Z)+\sum_{Z\in{\cal F}}\xi_{M_j}(Z,E({\bf M})\setminus Z)\ge 16.\eqno{\bf (2)}$$
Hence, we have equalities in {\bf (1)} and {\bf (2)}. So, for every $k\in\{1,2,3,4\}$,
$$\xi_{M_k}(Z_1\cap Z_2,W_1\cup W_2)=1+s_k\hbox{ and }\xi_{M_k}(Z_1\cup Z_2,W_1\cap W_2)=3-s_k,\eqno{\bf (3)}$$
for some $s_k\in\{0,1,2\}$.

In this paragraph, we shall prove that $|Z|=1$ or $\{Z,E({\bf M})\setminus Z\}$ is a 2-separation for ${\bf M}$, for every set $Z\in{\cal F}$. Suppose that $|Z|>1$ and that $\{Z,E({\bf M})\setminus Z\}$ is not a 2-separation for $M_k$, say $k=1$, for some $Z\in{\cal F}$. Without loss of generality, by {\bf (3)}, we may suppose that $Z=Z_1\cap Z_2$ and that $\xi_{M_1}(Z_1\cap Z_2,W_1\cup W_2)=3$. Thus, by {\bf (3)}, $\xi_{M_1}(Z_1\cup Z_2,W_1\cap W_2)=1$. Hence $W_1\cap W_2=Y_1\cap Y_2$, otherwise $\{W_1\cap W_2,E(N_i)\setminus(W_1\cap W_2)\}$ is a 1-separation for $N_i$, where $i\in\{1,2\}$ satisfies $X_i=W_i$, and ${\bf M}$ is not connected. As ${\bf M}$ is non-trivial and connected, we have that $\xi_{M_2}(X_1\cup X_2,Y_1\cap Y_2)=3$, by \recorde{ib}(i). So, by {\bf (3)}, $\xi_{M_2}(X_1\cap X_2,Y_1\cup Y_2)=1$ and we arrive at a contradiction because $\{X_1\cap X_2,E(N_i)\setminus(X_1\cap X_2)\}$ is a 1-separation for $N_i$ for both $i\in\{1,2\}$. 

We can use all the 2-separations found in the previous paragraph to decompose each $M_i$. So, there is a set $E=\{e_Z:Z\in{\cal F}\}$ such that $|E|=4$ and $E\cap E({\bf M})=\emptyset$, and there are matroids $K_{iZ}$ and $J_i$ such that $E(K_{iZ})=Z\cup\{e_Z\}$, $E(J_i)=E$ and
$$M_i=J_i\oplus_2 K_{iX_1\cap X_2}\oplus_2 K_{iX_1\cap Y_2}\oplus_2 K_{iY_1\cap X_2}\oplus_2 K_{iY_1\cap Y_2},$$
for every $i\in\{1,2,3,4\}$ --- when $|Z|=1$, we take $K_{iZ}$ as the empty matroid and set $e_Z$ equal to the element belonging to $Z$. Observe that the 2-separations $\{X_1,Y_1\}$ and $\{X_2,Y_2\}$ induce 2-separations on $J_i$. Thus, $J_i$ is a circuit or cocircuit, for every $i\in\{1,2,3,4\}$.

For every $i\in\{1,2,3,4\}$, $J_i=H_i\oplus L_i$, where $H_i$ and $L_i$ are matroids such that $E(H_i)=\{e_1,e_{X_1\cap X_2},e_{X_1\cap Y_2}\}$ and $E(L_i)=\{e_1,e_{Y_1\cap X_2},e_{Y_1\cap Y_2}\}$. Observe that 
$$N_1=H_i\oplus_2 K_{iX_1\cap X_2}\oplus_2 K_{iX_1\cap Y_2},$$
for every $i\in\{1,2,3,4\}$. Thus,
$$K_{1X_1\cap Z_2}=K_{2X_1\cap Z_2}=K_{3X_1\cap Z_2}=K_{4X_1\cap Z_2},$$
for every $Z\in\{X,Y\}$, and $H_1=H_2=H_3=H_4$. From the last identity, we have that $J_1=J_2=J_3=J_4$. Similarly, we have that
$$K_{1Z_1\cap X_2}=K_{2Z_1\cap X_2}=K_{3Z_1\cap X_2}=K_{4Z_1\cap X_2},$$
for both $Z\in\{X,Y\}$. Hence ${\bf M}$ is equal to
$$(K_{1Y_1\cap Y_2},K_{2Y_1\cap Y_2},K_{3Y_1\cap Y_2},K_{4Y_1\cap Y_2})\oplus_2 N,$$
where $N=J_i\oplus_2 K_{iX_1\cap X_2}\oplus_2 K_{iX_1\cap Y_2}\oplus_2 K_{iY_1\cap X_2}$, for every $i\in\{1,2,3,4\}$. We arrive at a contradiction because $X_1\cup X_2$ cannot be a dead set. \qed 

By \recorde{lemma2}, we can decompose a non-trivial connected quad ${\bf M}$ as
$${\bf M}={\bf H}\oplus_2 N_1\oplus_2 N_2\oplus_2\cdots\oplus_2 N_k,$$
where ${\bf H}$ is a non-trivial irreducible quad and, for $i\in\{1,2,\dots,k\}$, $N_i$ is a connected matroid such that $E(N_i)\cap E({\bf H})=\{e_i\}$, for $e_i\not\in E({\bf M})$, $E(N_i)\cap E(N_j)=\emptyset$ and $e_i\not=e_j$, for every $i\not=j$. As this decomposition is unique, up to the labels of the base points of the 2-sums, by \recorde{lemma2}, we say that ${\bf H}$ is the {\it core} of ${\bf M}$ and that $N_1,N_2,\dots,N_k$ are the {\it leaves} of ${\bf M}$. We refer to it as {\it the canonical decomposition} for ${\bf M}$. We shall associate the following function with ${\bf M}$:
$$\gamma({\bf M})=\bigcup_{i=1}^{k}E(N_i)\setminus\{e_i\}.$$
Observe that $\gamma(M)=\emptyset$, when ${\bf M}$ is irreducible. 

During the rest of this section, we shall study the deletion and contraction of elements belonging to a non-trivial irreducible quad. First, we shall give some definitions. For an element $e$ of a non-trivial irreducible quad ${\bf M}$, we define $del(e)$ as the number of leaves of ${\bf M}\backslash e$, when ${\bf M}\backslash e$ is non-trivial, and as $\infty$ otherwise. Similarly, we define $con(e)$ as the number of leaves of ${\bf M}/e$, when ${\bf M}/e$ is non-trivial, and as $\infty$ otherwise. We also set
$$del({\bf M})=\min\{del(e):e\in E({\bf M})\}\hbox{ and }con({\bf M})=\min\{con(e):e\in E({\bf M})\}.$$
When $del({\bf M})=con({\bf M})=\infty$, we say that ${\bf M}$ is an {\it almost trivial} quad. Now, we shall present the characterization of the almost trivial quads. To do this, we need to set more notation: when $C_1,C_2,\dots,C_n$ are different circuit-hyperplanes of a matroid $M$, we denote by $M_{C_1,C_2,\dots,C_n}$ the matroid obtained from $M$ after the relaxation of all the circuit-hyperplanes $C_1,C_2,\dots,C_n$.

Now, we need the following corollary of a result of Mills, that appears in [5]:

\bteo\rotulo{corolario}
Let $M$ and $N$ be different matroids both loopless and coloopless defined in the same ground set. If, for every $e\in E(M)$, $M\backslash e=N\backslash e$ or $M/e=N/e$, then one of $M$ and $N$ is a relaxation of the other.\qeds
\eteo

\bteo\rotulo{lemma5}
If ${\bf M}$ is almost-trivial, then there is a matroid $M$ having a pair of circuit-hyperplanes $X$ and $Y$ such that $\{X,Y\}$ is a partition of $E({\bf M})$ and ${\bf M}\equiv(M_{X,Y},M,M_Y,M_X)$. This quad is denoted by ${\bf Q}(M,X,Y)$.
\eteo

\noindent {\bf Proof:} By hypothesis, for every $e\in E({\bf M})$, we have that 
$$\{M_1\backslash e,M_2\backslash e\}=\{M_3\backslash e,M_4\backslash e\}\hbox{ and }\{M_1/e,M_2/e\}=\{M_3/e,M_4/e\}.$$
For $i\in\{1,2\}$ and $j\in\{3,4\}$, when $M_i\backslash e=M_j\backslash e$, then $M_i/e=M_{7-j}/e$, otherwise $M_i/e=M_j/e$ and hence $M_i=M_j$ and ${\bf M}$ is trivial, a contradiction. Thus, there is a partition $\{X,Y\}$ of $E({\bf M})$ such that 
$$M_1\backslash x=M_3\backslash x,M_1/x=M_4/x,M_1/y=M_3/y\hbox{ and }M_1\backslash y=M_4\backslash y,$$
for every $x\in X$ and $y\in Y$. 

We denote by ${\cal C}_{ij}$ the set ${\cal C}(M_i)\bigtriangleup{\cal C}(M_j)$. By \recorde{ic} and \recorde{corolario}, we have that
$${\cal C}_{13}={\cal C}_{24}=\{X\}\cup\{X\cup\{y\}:y\in Y\}\hbox{ and}$$
$${\cal C}_{14}={\cal C}_{23}=\{Y\}\cup\{Y\cup\{x\}:x\in X\}.$$

To conclude the proof of this lemma, we observe that there is an $i\in\{1,2,3,4\}$, say $i=2$, such that $X$ and $Y$ are both circuit-hyperplanes of $M_2$, which is taken to be $M$. Thus, $M_1=M_{X,Y}$. Note that both $X$ and $Y$ cannot be a circuit of $M_j$, for $j\in\{3,4\}$, otherwise $M_2=M_j$ and ${\bf M}$ is trivial. Hence, by \recorde{ic}, we may suppose that $X$ is a circuit of $M_3$ and $Y$ is a circuit of $M_4$. So, $M_3=M_Y$ and $M_4=M_X$.\qeds 

\bteo\rotulo{corolario1}
If ${\bf M}$ is a non-trivial connected quad, then ${\bf M}$ has an almost-trivial quad as a minor. 
\eteo

\noindent {\bf Proof:} Suppose that this corollary is not true. Let ${\bf M}$ be a minimal counter-example for it. Observe that ${\bf M}$ is irreducible, otherwise, by the choice of ${\bf M}$, the core ${\bf N}$ of ${\bf M}$ has an almost-trivial quad as a minor which must be a minor of ${\bf M}$ as well. By its choice, ${\bf M}$ is not almost-trivial, and there is an element $e$ of $E({\bf M})$ such that ${\bf M}\backslash e$ or ${\bf M}/e$ is non-trivial, say ${\bf M}\backslash e$. As ${\bf M}\backslash e$ is connected because ${\bf M}$ is irreducible, it follows that ${\bf M}\backslash e$ has an almost-trivial quad as a minor. We arrive at a contradiction.\qed

We say that a quad ${\bf M}$ is {\it binary}, when, for every $i\in\{1,2,3,4\}$, $M_i$ is a binary matroid.

\bteo\rotulo{corolario2}
There exists no non-trivial connected binary quad.
\eteo

\noindent {\bf Proof:} By \recorde{corolario1}, every non-trivial connected binary quad must have an almost-trivial quad ${\bf N}$ as a minor. Thus, ${\bf N}$ must be binary. By \recorde{lemma5}, ${\bf N}\equiv{\bf Q}(N;X,Y)$, for some matroid $N$. As $N_X$ is a binary matroid, it follows that $|X|=|Y|=2$, otherwise $N_X\backslash Y'/X'$ is isomorphic to $U_{2,4}$, for some $X'\subset X$ and $Y'\subset Y$ such that $|X'|=|Y'|=r(N)-2$. Thus, $N=N|X\oplus N|Y$ and $N_{X,Y}$ must be isomorphic to $U_{2,4}$. So, ${\bf N}$ cannot exist and the result follows.\qed

Seymour's theorem that was stated in the introduction is a consequence of this result, since $(M,M^*,N,N^*)$ must be a non-trivial binary connected quad, when $M$ and $N$ are connected binary matroids having the same connectivity function such that $M\not=N$ and $M\not=N^*$. 

We say that a quad ${\bf M}$ is {\it semi-binary}, when $M_1$ and $M_2$ are binary matroids or $M_3$ and $M_4$ are binary matroids. Thus, the proof of \recorde{corolario2} can be easily adapted to prove the following result:

\bteo\rotulo{corolario3}
A semi-binary almost-trivial quad has four elements.\qeds
\eteo

Our goal is to construct all the semi-binary irreducible non-trivial quads, to conclude that they cannot be equivalent to the core of $(M,M^*,N,N^*)$, for a connected binary matroid $M$ having the same connectivity function as $N$. This shall be done in the last section of this paper.

Now, we shall prove a result that is not needed in this paper, but is fundamental to study the uniqueness of the decomposition of the rank function of a 2-polimatroid as the sum of rank functions of matroids. We shall state and prove it here for future reference.

\bteo\rotulo{poli}
If ${\bf M}$ is a non-trivial connected quad, then there is at most one $i\in\{1,2,3,4\}$ such that $M_i$ is not connected. Moreover, if $M_i$ is not connected, then it has two connected components.
\eteo

\noindent {\bf Proof:} Suppose that this result is not true and choose a minimal counter-example ${\bf M}$. Clearly ${\bf M}$ is irreducible. By \recorde{lemma5}, ${\bf M}$ cannot be an almost-trivial quad. So, ${\bf M}$ has an element $e$ such that ${\bf M}\backslash e$ or ${\bf M}/e$ is non-trivial, say ${\bf M}\backslash e$. As ${\bf M}$ is irreducible, it follows that ${\bf M}\backslash e$ is connected. Thus, by the choice of ${\bf M}$, there is at most one $i$ such that $M_i\backslash e$ is not connected. So, $e$ must be a loop or coloop of some $M_j$, for $j\not=i$. But this is contrary to \recorde{21}.\qeds 

\section{A splitter-type theorem for quads}

In this section, we shall prove a splitter-type theorem for irreducible quads. It says that it is possible to get a non-trivial irreducible quad from an almost-trivial quad or a wheel-whirl quad, which is going to be defined below, by a sequence of one elements lifts or extensions without leaving the class of irreducible quads. This theorem plays a central hole in the construction of all semi-binary irreducible quads, since permits an inductive proof. 

Let $N$ be a matroid having a circuit-hyperplane $C$ and let $U$ be a disconnected matroid defined on the same ground set as $N$ such that $U|C\simeq U_{|C|-1,|C|}$ and $U|(E(N)\setminus C)\simeq U_{1,|E(N)\setminus C|}$. By \recorde{ic}, $(N,U_C,N_C,U)$ is a quad, which is denoted by ${\bf Q}(N;C)$. When $N\simeq M(W_n)$, for $n\ge 2$, ${\bf Q}(N;C)$ is denoted by ${\bf W}_n$ and is called the {\it wheel-whirl} quad of rank $n$. Now, we state the main result of this section. 

\bteo\rotulo{ww}
If ${\bf M}$ is a non-trivial irreducible quad, then
\begin{description}

\item{(i)} there is $e\in E({\bf M})$ such that ${\bf M}\backslash e$ is non-trivial and irreducible; or

\item{(ii)} there is $e\in E({\bf M})$ such that ${\bf M}/e$ is non-trivial and irreducible; or

\item{(iii)} ${\bf M}$ is an almost-trivial quad; or

\item{(iv)} ${\bf M}$ is a wheel-whirl quad.

\end{description}
\eteo

Before the proof of \recorde{ww}, we shall give the definition of a chain of triangles and triads of a matroid with a small modification from the one which appears in Oxley [8]: for a matroid $M$ and integer $m\ge 4$, the sequence $z_1,z_2,\dots,z_m,z_{m+1},z_{m+2}$ of elements of $M$ is an {\it admissible sequence} of $M$, when (i) $z_1,z_2,\dots,z_m$ are all distinct; (ii) for $i\in\{1,2,\dots,m\}$, $Z_i=\{z_i,z_{i+1},z_{i+2}\}$ is a triangle when $i$ is odd and a triad when $i$ is even. We have the following theorem:

\bteo\rotulo{sow}
Suppose that $z_1,z_2,\dots,z_m,z_{m+1},z_{m+2}$ is an admissible sequence of a matroid $M$.
\begin{description}

\item{(i)} (6.1 of Seymour [10]) When $M$ is connected, $z_{m+1}=z_1$ and $z_{m+2}=z_2$, then $m=2n$ and $M$ is isomorphic to ${\cal W}_n$ or $M(W_n)$.
\vspace{0.1cm}

\item{(ii)} (see Tutte [12] or Oxley and Reid [9]) When $M$ is 3-connected and $z_{m+2}=z_1$, then $m=2n-1$ and $M$ is isomorphic to a ${\cal W}_n$ or $M(W_n)$.\qeds

\end{description}
\vspace{0.05cm}

We say that $Z_1,Z_2,\dots,Z_m$ is a {\it chain} of $M$. This chain is said to be {\it closed}, when $Z_1\cap Z_m\not=\emptyset$. In this case, we have (i) or (ii) of \recorde{sow}.
\vspace{0.3cm}

\noindent {\bf Proof of (3.1):} We may suppose that $0<\min\{del({\bf M}),con({\bf M})\}<\infty,$
otherwise the result follows: (a) when $del({\bf M})=0$, we have (i); (b) when $con({\bf M})=0$, we have (ii); (c) when $del({\bf M})=con({\bf M})=\infty$, then ${\bf M}$ is almost-trivial and we have (iii).

As the classes of almost-trivial quads and wheel-whirl quads are both closed under duality, we can replace ${\bf M}$ by ${\bf M}^*$, when necessary, and we may suppose that 
$$\gamma=\min_{e\in E({\bf M})}\{|\gamma({\bf M}\backslash e)|:del(e)<\infty\}\le\min_{e\in E({\bf M})}\{|\gamma({\bf M}/e)|:con(e)<\infty\}.$$ 
Choose an element $e$ belonging to $\{f\in E({\bf M}):\gamma=|\gamma({\bf M}\backslash f)|\}$ such that $del(e)$ is maximum. So, for $k=del(e)$,
$${\bf M}\backslash e={\bf H}\oplus_2 N_1\oplus_2 N_2\oplus_2\cdots\oplus_2 N_k,$$
where ${\bf H}$ is a non-trivial and irreducible quad and $N_i$ is a connected matroid, for each $i\in\{1,2,\dots,k\}$. Moreover, let $e_i$ be the element belonging to both $E({\bf H})$ and $E(N_i)$, for every $i\in\{1,2,\dots,k\}$. We also set $E_i=E(N_i)\setminus\{e_i\}$, for every $i\in\{1,2\dots,k\}$.

\bteo\rotulo{aaa}
For every $i\in\{1,2,\dots,k\}$ and $f\in E_i$, $N_i\backslash f$ is not connected.
\eteo

Suppose that $N_i\backslash f$ is connected, for some $f\in E_i$ and $i\in\{1,2,\dots,k\}$, say $i=k$. So, 
$${\bf M}\backslash\{e,f\}={\bf H}\oplus_2 N_1\oplus_2 N_2\oplus_2\cdots\oplus_2 N_{k-1}\oplus_2 N_k\backslash f$$
is the canonical decomposition of ${\bf M}\backslash\{e,f\}$. Hence ${\bf M}\backslash\{e,f\}$ is non-trivial and connected. Thus, ${\bf M}\backslash f$ is also non-trivial and connected. 

By the decomposition of ${\bf M}\backslash\{e,f\}$, we have that: if $\{X,Y\}$ is a 2-separation for ${\bf M}\backslash\{e,f\}$, then $X\subset E_k$ or $Y\subset E_k$ or $E_k\setminus\{f\}\subseteq X$ or $E_k\setminus\{f\}\subseteq Y$.

In this paragraph, we shall prove that the dead sets of ${\bf M}\backslash f$ are contained in $(E_k\setminus\{f\})\cup\{e\}$. Suppose that $\{X\cup\{e\},Y\}$ is a 2-separation for ${\bf M}\backslash f$. First, we shall prove that $|X|\ge 2$. Suppose that $|X|=1$, say $X=\{x\}$. By \recorde{lemma1}, we have two cases to deal with: (a) $\{e,x\}$ is contained in a parallel class of $M_i\backslash f$, for every $i\in\{1,2,3,4\}$. In this case, we arrive at a contradiction, because $\{\{e,x\},E({\bf M})\setminus\{e,x\}\}$ is a 2-separation for ${\bf M}$; or (b) $\{e,x\}$ is contained in a series class of $M_i\backslash f$, for every $i\in\{1,2,3,4\}$. Hence, $x$ is a coloop of $M_i\backslash\{e,f\}$ and we arrive at a contradiction, since ${\bf M}\backslash\{e,f\}$ is connected. Thus, $|X|\ge 2$ and $\{X,Y\}$ is a 2-separation for ${\bf M}\backslash\{e,f\}$. Note that $E_k\setminus\{f\}\not\subseteq Y$, otherwise $\{X\cup\{e\},Y\cup\{f\}\}$ is a 2-separation for ${\bf M}$ because $E_k\setminus\{f\}$ spans $f$ in $M_i$, for every $i\in\{1,2,3,4\}$. Similarly, we have that $E_k\setminus\{f\}\not\subseteq X$. So, $X\subset E_k$ or $Y\subset E_k$, by the previous paragraph. Hence the dead set of this 2-separation must be contained in $(E_k\setminus\{f\})\cup\{e\}$. 

So, by the choice of $e$, 
$$|\gamma({\bf M}\backslash e)|\le |\gamma({\bf M}\backslash f)|\le |E_k|=|\gamma({\bf M}\backslash e)|-\sum_{i=1}^{k-1}|E_i|.$$
Thus, $k=1$ and $|\gamma({\bf M}\backslash e)|=|\gamma({\bf M}\backslash f)|$. Hence, by the choice of $e$, ${\bf M}\backslash f$ has just one leave, which must have as its ground set $Z=(E_1\setminus\{f\})\cup\{e\}$. We arrive at a contradiction because $\{Z,E({\bf M})\setminus (Z\cup\{f\})\}$ is a 2-separation for ${\bf M}\setminus f$ and $Z$ spans $f$ in $M_i$, for every $i\in\{1,2,3,4\}$.

\bteo\rotulo{aab}
For every $i\in\{1,2,3,4\}$ and $j\in\{1,2,\dots,k\}$, $cl_{M_i}(E({\bf M})\setminus(E_j\cup\{e\}))\cap E_j=\emptyset$
\eteo

If $f\in cl_{M_i}(E({\bf M})\setminus(E_j\cup\{e\}))\cap E_j$, then $f$ must be in parallel with $e_j$ in $N_j$ and hence $N_j\backslash f$ is connected, which is contrary to \recorde{aaa}.

\bteo\rotulo{lemma4}
$k=1$ and $|E_1|=2$.
\eteo

Without loss of generality, we may suppose that
$$|E_1|\le |E_2|\le\cdots\le |E_k|.$$
If $f\in E(N_k)\setminus\{e_k\}$, then, by \recorde{aaa}, $N_k/f$ is connected and the canonical decomposition of ${\bf M}\backslash e/f$ is the following
$${\bf M}\backslash e/f={\bf H}\oplus_2 N_1\oplus_2 N_2\oplus_2\cdots\oplus_2 N_{k-1}\oplus_2 N_k/f.$$

Now, we shall study the maximal dead sets of ${\bf M}/f$. Suppose that $\{X\cup\{e\},Y\}$ is a 2-separation for ${\bf M}/f$. We shall prove that its dead set is equal to: (a) $X\cup\{e\}$ and $|X|=1$; or (b) $X\cup\{e\}$ and $X\subseteq E_i\setminus\{f\}$, for some $i\in\{1,2,\dots,k\}$.

If $|X|=1$, then $X\cup\{e\}$ is its dead set. Thus, we may suppose that $|X|\ge 2$. So, $\{X,Y\}$ is a 2-separation for ${\bf M}\backslash e/f$. Now, we have two cases to deal with: (c) $Y\subseteq E_i\setminus\{f\}$, for some $i\in\{1,2,\dots,k\}$ or (d) $X\subseteq E_i\setminus\{f\}$, for some $i\in\{1,2,\dots,k\}$. When (d) happens, we have (b). So, we may suppose that (c) happens. Thus, $Y$ must span $f$ in $N_i$, for some $i\in\{1,2,3,4\}$, otherwise $\{X\cup\{e,f\},Y\}$ is a 2-separation for ${\bf M}$. By \recorde{aab}, this cannot happens when $i<k$. So, $i=k$ and let $C$ be a circuit of $N_k$ such that $f\in C\subseteq Y\cup\{f\}$. As $N_k\backslash g$ is not connected, for every $g\in C$, by \recorde{aaa}, it follows that $C\setminus\{f\}$ contains a 2-cocircuit $D$ of $N_k$, by a result of Oxley (lemma 10.2.1 of [7]). Observe that $D$ is a 2-cocircuit of $M_i/f$, for every $i\in\{1,2,3,4\}$, since $X$ spans $e$ in $M_i/f$. Hence $D$ is a 2-cocircuit of $M_i$, for every $i\in\{1,2,3,4\}$, and we arrive at a contradiction. 

%Observe that
%$$\gamma({\bf M}/f)\subseteq E_1\cup E_2\cup\cdots\cup E_{k-1}\cup E_k',$$
%where $E_k'=(E_k\setminus\{f\})\cup\{e\}$ or $E_k'=Z$, when there is a dead set $Z$ such that $|Z|=2$ and $e\in Z$. Hence $|\gamma({\bf M}/f)|\le|\gamma({\bf M}\backslash e)|$ and, by the choice of $e$, we must have $|\gamma({\bf M}/f)|=|\gamma({\bf M}\backslash e)|$.
%Thus, the dead sets of ${\bf M}/f$ are: (e) $E_i\cup\{e\}$, for every $i\in\{1,2,\dots,k-1\}$; and (f) $(E_k\cup\{f\})\setminus\{e\}$ or $\{e,x\}$, for some $x\in E({\bf M})$. 

By \recorde{lemma2}, the maximal dead sets are disjoint. So, ${\bf M}/f$ has just one dead set. Hence
$$2(k-1)+|E_k|\le\sum_{i=1}^k|E_i|=|\gamma({\bf M}\backslash e)|\le|\gamma({\bf M}\backslash f)|\le |E_k|+1.$$
Thus, $k=1$.

%Hence (g) $k=1$; or (h) $k=2$ and $|E_1|=|E_2|=2$. When case (h) happens, we must have $E_1\cup\{e\}$ as a dead set of ${\bf M}/f$.

%In this paragraph, we prove that $k=1$. Suppose that $k\ge 2$. First, we shall establish that 
%$$\hbox{$r_{M_l}(E_1\cup E_2)=3$, for every $l\in\{1,2,3,4\}$.}$$
%We have two cases to deal with, since $E_1\cup\{e\}$ is a dead set of ${\bf M}/f$: (a) $E_1$ does not span $e$ in $M_l$. So, $E_1\cup\{e\}$ spans $f$ in $M_l$ and hence there is a circuit $C_f$ of $M_l$ such that $f\in C_f\subseteq E_1\cup\{e,f\}$. As $f$ can be any element of $T_2\setminus\{e\}$, it follows that $E_1\cup\{e\}$ spans $E_2$ in $M_l$. Thus, $r_{M_l}(E_1\cup E_2)=3$; (b) $E_1$ spans $e$ in $M_l$. As $e$ cannot be in parallel with an element of $E_1$ in $M_l$, it follows that $E_1\cup\{e\}$ is a triangle of $M_l$. By symmetry, we have that $E_2\cup\{e\}$ is a triangle of $M_l$, otherwise $r_{M_l}(E_1\cup E_2)=3$, by (a). Thus, $E_1\cup E_2$ contains a circuit of $M_l$ and $r_{M_l}(E_1\cup E_2)=3$. We arrive at a contradiction, since
%$$r_{M_l}(E({\bf M}\backslash e)\setminus (E_1\cup E_2))+r_{M_l}(E_1\cup E_2)\le (r(M_l)-2)+3=1$$
%and $\{E_1\cup E_2,E({\bf M}\backslash e)\setminus (E_1\cup E_2)\}$ is a 2-separation for $M_l$, for every $i\in\{1,2,3,4\}$. Thus, $E_1\cup E_2$ is a dead set for ${\bf M}\backslash e$ and we arrive at a contradiction. So, $k=1$.

Now, we shall prove that $|\gamma({\bf M}\backslash e)|=2$. When $|\gamma({\bf M}\backslash e)|>2$, we must have $|E_1|>2$ and $(E_1\setminus\{f\})\cup\{e\}$ as a dead set of ${\bf M}/f$, otherwise $\{e,x\}$ is a dead set for ${\bf M}/f$ and we arrive at a contradiction by the choice of $e$. As $\{(E_1\setminus\{f\})\cup\{e\}, E({\bf M})\setminus (E_1\cup \{e\})\}$ cannot be a 2-separation for ${\bf M}/f$, since $ E({\bf M})\setminus (E_1\cup \{e\})$ does not span $f$ in $M_l$, for every $l\in\{1,2,3,4\}$, by \recorde{aab}, we arrive at a contradiction and $|E_1|=2$. 
\vspace{0.3cm}

Consider the following two sets:
$$Z=\{e\in E({\bf M}):|\gamma({\bf M}\backslash e)|=2\}\hbox{ and }W=\{e\in E({\bf M}):|\gamma({\bf M}/e)|=2\}.$$
By \recorde{lemma4}, $Z\cup W\not=\emptyset$. Thus, for each $e\in Z$ and $f\in W$, we have that
$${\bf M}\backslash e={\bf H}_e\oplus_2 N_e\hbox{ and }{\bf M}/f={\bf H}_f\oplus_2 N_f,$$
where ${\bf H}_e$ and ${\bf H}_f$ are  non-trivial irreducible quads and $N_e$ is a matroid isomorphic to $U_{2,3}$ and $N_f$ is a matroid isomorphic to $U_{1,3}$. We set $W_e=E({\bf M})\cap E(N_e)$ and $Z_f=E({\bf M})\cap E(N_f)$. Observe that $W_e\subseteq W,Z_f\subseteq Z$ and $|W_e|=|Z_f|=2$. Moreover, 

\bteo\rotulo{lemmaE}
For each $i\in\{1,2,3,4\},e\in Z$ and $f\in W$, $W_e$ or $W_e\cup\{e\}$ is a cocircuit of $M_i$ and $Z_f$ or $Z_f\cup\{f\}$ is a circuit of $M_i$.
\eteo

\bteo\rotulo{lemmaF}
$f\in W_e$ if and only if $e\in Z_f$.
\eteo

\bteo\rotulo{lemmaG}
When $W_e\cup\{e\}$ is a triad of $M_i$, for $e\in Z$ and $i\in\{1,2,3,4\}$, then $Z_{f_1}\cup\{f_1\},W_e\cup\{e\},Z_{f_2}\cup\{f_2\}$ is a chain of $M_i$, where $W_e=\{f_1,f_2\}$.
\eteo

First, we shall prove that $Z_f\cup\{f\}$ is a triangle of $M_i$, for every $f\in W_e$. When $Z_f\cup\{f\}$ is not a triangle of $M_i$, then $Z_f$ is a circuit of $M_i$, by \recorde{lemmaE}. Observe that $e\in Z_f\cap (W_e\cup\{e\})$, by \recorde{lemmaF}. Thus, by orthogonality, we have that $Z_f=\{e,g\}$, where $g\in W_e\setminus\{f\}$. As $\{e\}$ is a loop of $M_i/g$ and ${\bf M}/g$ is connected, by \recorde{21}, it follows that ${\bf M}/g$ is trivial, a contradiction. Hence $Z_f\cup\{f\}$ is a triangle of $M_i$. 

Now, we shall prove that $|(Z_f\cup\{f\})\cap(W_e\cup\{e\})|=2$. When $X=Z_f\cup\{f\}=W_e\cup\{e\}$, it follows that $\{X,E({\bf M})\setminus X\}$ is a 2-separation for $M_i$. As $g\in Z_f\cap W_e$, it follows that $X$ is a triad of $M_j$, for every $j\in\{1,2,3,4\}$, otherwise, by \recorde{lemmaE}, $W_e$ is a cocircuit of $M_j$, for some $j\in\{1,2,3,4\}$, and $f$ is a coloop of $M_j\backslash g$, which is contrary to the fact that ${\bf M}\backslash g$ is non-trivial and connected, by \recorde{21}. Similarly, we have that $X$ is a triangle of every $M_j$, for every $j\in\{1,2,3,4\}$. Hence $\{X,E({\bf M})\setminus X\}$ is a 2-separation for ${\bf M}$.
\vspace{0.3cm}

Now, we shall state the dual of this result:

\bteo\rotulo{lemmaH}
When $Z_f\cup\{f\}$ is a triangle of $M_i$, for $f\in W$ and $i\in\{1,2,3,4\}$, then $W_{e_1}\cup\{e_1\},Z_f\cup\{f\},W_{e_2}\cup\{e_2\}$ is a chain of $M_i$, where $Z_f=\{e_1,e_2\}$.
\eteo

By \recorde{lemma4} and \recorde{lemmaE}, for some $e\in E({\bf M})$, it follows that $W_e\cup\{e\}$ is a triad of $M_1$ or $M_2$, say $M_1$, since ${\bf M}$ is irreducible. By \recorde{ic}, it must be a triad of $M_3$ or $M_4$, say $M_3$. Thus, by \recorde{lemmaG} and \recorde{lemmaH}, this triad is contained in a chain ${\cal C}$ of both $M_1$ and $M_3$ which must close as in (i) of \recorde{sow}. Let $R$ be the set of elements of $E({\bf M})$ belonging to two triangles of ${\cal C}$ and let $S$ be the set of elements of $E({\bf M})$ belonging to two triads of ${\cal C}$. We define $X=R\cup S$. Thus, by \recorde{sow}(i), $\{X,E({\bf M})\setminus X\}$ is a 1-separation for both $M_1$ and $M_3$ and $M_1|X$ and $M_3|X$ are equal to a wheel or a whirl having rim $R\subseteq Z$ and set of spokes $S\subseteq W$. Thus, one is a whirl and the other to a wheel, otherwise ${\bf M}$ is trivial, by \recorde{extra}. As $M_2|X$ and $M_4|X$ cannot be connected components of $M_2$ and $M_4$, respectively, equal to the same wheel or whirl, by \recorde{extra}, it follows that ${\cal C}$ is not a chain for both $M_2$ and $M_4$ and $W_e$ and $Z_f$ is a cocircuit and a circuit, respectively, for every $e\in S$ and $f\in R$, by \recorde{lemmaE}. As $S$ is a circuit of $M_2$ or $M_4$, by \recorde{ic}, it follows that $S$ is circuit which is a connected component of one of these matroids, since $S$ is contained in one of its series classes. Similarly, $R$ is a cocircuit which is a connected component of $M_2$ or $M_4$. As ${\bf M}|X$ is a non-trivial connected quad, it follows that $r_{M_l}(X)=r({\bf M}|X)$ is independent of $l$. So, $R$ and $S$ are connected components of the same matroid $M_i$, for $i\in\{2,4\}$, since $S$ is a basis for $X$ in the other. Thus, $\{X,E({\bf M})\setminus X\}$ is a 1-separation for ${\bf M}$. and it follows that $X=E({\bf M})$. We have (iv) of theorem 1.\qed

\section{Irreducible semi-binary quads}

The main result of this section gives a description of all the semi-binary irreducible quads and from it we derive theorem 2. We need the following lemma which is similar to ones of Lemos [3] or Oxley [6]:

\bteo\rotulo{ch}
Let $N$ be a matroid having a circuit-hyperplane $C$ without parallel elements. 
\begin{description}

\item{(i)} If $\{X,Y\}$ is a $1$-separation for $N$, then $\{X,Y\}=\{C,E(N)\setminus C\}$.

\item{(ii)} If $\{X,Y\}$ is a $2$-separation for $N$, then $\{X,Y\}=\{C,E(N)\setminus C\}$ or $Z\subset C$ is contained in a series class of $N$, for some $Z\in\{X,Y\}$. 

\item{(iii)} If $|C|\ge 3$ and $N$ is binary and not 3-connected, then $C$ contains a non-trivial series class of $N$.

\end{description}
\vspace{0.05cm}

\noindent {\bf Proof:} (i) Observe that $\{X,Y\}=\{C,E(N)\setminus C\}$ because $C$ is a circuit of $N$ and $E(N)\setminus C$ is a cocircuit of $N$. 
\vspace{0.15cm}

\noindent (ii) For $Z\in\{X,Y\}$, we have that
$$r(Z)\ge\min\{|C|-1,|Z\cap C|\}+\min\{1,|Z\setminus C|\}.$$
When $C\cap Z\not=\emptyset$, for both $Z\in\{X,Y\}$, we have that
$$|C|+1=r(N)+1=r(X)+r(Y)\ge |C|+\min\{1,|X\setminus C|\}+\min\{1,|Y\setminus C|\}.$$
So, $Z\setminus C=\emptyset$, for some $Z\in\{X,Y\}$ and the result follows. When $C\cap Z=\emptyset$, for some $Z\in\{X,Y\}$, say $Z=X$, we have that $Z=C$, otherwise $r(Y)=r(N)$ and $X$ is a parallel class of $N$.
\vspace{0.15cm}

\noindent (iii) From (i) and (ii), we have (iii) unless $\{C,E(N)\setminus C\}$ is the only 2-separation for $N$ and it is exact. In this case, we can write $N=N_1\oplus_2 N_2$, where $N_1$ and $N_2$ are binary matroids such that $E(N_2)=C\cup\{e\}$, for a new element $e$. As $r^*(N_2)=2$ and $|E(N_2)|=|C|+1\ge 4$, it follows that $N_2$ has a non-trivial series class which must be contained in $C$.\qeds

\bteo\rotulo{teorema3}
${\bf M}$ is a non-trivial semi-binary irreducible quad if and only if there is a 3-connected binary matroid $N$ having a circuit-hyperplane $C$ such that ${\bf M}\equiv{\bf Q}(N;C)$ or ${\bf M}\simeq{\bf W}_2$.
\eteo

\noindent {\bf Proof:} We shall prove the ``only if'' part of the theorem, since the ``if'' is trivial. Suppose that it is not true and choose a counter-example ${\bf M}$ which is minimal. By \recorde{corolario3}, an almost-trivial matroid cannot be semi-binary, unless it is isomorphic to ${\bf W}_2$. So, by \recorde{ww}, (i) ${\bf M}$ is isomorphic to a wheel-whirl quad. In this case, we may take $N$ equal to the wheel and $C$ to be its rim and the result follows; or (ii) ${\bf M}$ has an element $e$ such that ${\bf M}\backslash e$ or ${\bf M}/e$ is irreducible. As this theorem is invariant under duality, we may suppose that ${\bf M}$ has an element $e$ such that ${\bf M}\backslash e$ is irreducible. So, by the choice of ${\bf M}$, ${\bf M}\backslash e\equiv {\bf Q}(N;C)$, for a binary 3-connected matroid $N$, or ${\bf M}\backslash e\simeq {\bf W}_2$. When the first possibility occurs, we suppose that $M_1\backslash e=N$ and $M_3\backslash e=N_C$. We divide the proof of this theorem in some lemmas. At the end, we shall arrive at a contradiction.

\bteo\rotulo{step1}
${\bf M}\backslash e\equiv {\bf Q}(N;C)$ and $\min\{r(N),r(N^*)\}\ge 3$.
\eteo

By the next lemma, up to isomorphism, there is just one non-trivial semi-binary irreducible quad having rank equal to 2: ${\bf W}_2$. So, ${\bf M}\backslash e$ cannot be isomorphic to it, otherwise, by {\bf (4.4)}, ${\bf M}$ must also be isomorphic to ${\bf W}_2$. Thus, ${\bf M}\backslash e\equiv {\bf Q}(N;C)$ and both $r(N)$ and $r(N^*)$ are at least three.

\bteo\rotulo{lemma}
If ${\bf H}$ is a non-trivial semi-binary irreducible quad such that $r({\bf H})=2$, then ${\bf H}\simeq{\bf W}_2$.
\eteo

\noindent {\bf Proof:} It is enough to prove that ${\bf W}_2$ does not have an irreducible semi-binary extension. So, suppose that ${\bf H}\backslash e$ is isomorphic to ${\bf W}_2$ and that $E({\bf H})=\{a,b,c,d,e\}$. By \recorde{ic}, we may suppose that $\{a,b\}$ and $\{c,d\}$ are circuits of $H_2$, $\{a,b\}$ is a circuit of $H_3$ and $\{c,d\}$ is a circuit of $H_4$. As $H_3$ and $H_4$ do not have a minor isomorphic to $U_{2,4}$, since ${\bf H}$ is semi-binary, it follows that $\{e,e_i\}$ is a circuit of $H_i$, for $i\in\{3,4\}$. Observe that $e_3\not=e_4$, otherwise $\{\{e,e_3\},E({\bf H})\setminus\{e,e_3\}\}$ is a 2-separation for ${\bf H}$, by \recorde{ic}. Thus, by \recorde{ic}, $\{e,e_3\}$ or $\{e,e_4\}$ is a circuit of $H_2$, otherwise $\{e_3,e_4\}$ is a circuit of $H_1$. So, $H_2$ has a parallel class $P$ with three elements. $P$ must be a parallel class of $H_3$ or $H_4$ because ${\bf H}/e$ is connected and trivial, since it has loops, by \recorde{21}. So, $H_2/e=H_i/e$, for some $i\in\{3,4\}$, and $P$ is a parallel class of $M_i$. This cannot occurs, since $e_3\in\{a,b\}$ if and only if $e_4\in\{c,d\}$. That is, both $M_3$ and $M_4$ have a three element parallel class or both does not have it.\qeds

\bteo\rotulo{step1'}
$C$ does not span $e$ in $M_1$.
\eteo

Suppose that $C$ spans $e$ in $M_1$. Choose a circuit $C'$ of $M_i$, for $i\in\{1,2\}$, such that $e\in C'\subseteq C\cup\{e\}$ and $|C'\cap C|$ is minimum. First, we shall prove that $|C\setminus C'|\ge 2$. As $M_1$ is binary and $C$ spans $e$ in $M_1$, there are circuits $C_1$ and $C_2$ of $M_1$ such that $C_1\cap C_2=\{e\}$ and $C_1\bigtriangleup C_2=C$. We may suppose that $|C_1|\le |C_2|$. Thus, $|C'|\le |C_1|$ and $|C\setminus C'|\ge |C\setminus C_1|=|C\cap C_2|\ge 2$, since $|C|\ge 3$, by \recorde{step1}. 

In this paragraph, we prove that $C'$ is a circuit of $M_i$, for every $i\in\{1,2,3,4\}$. Observe that ${\bf M}/(C'\cap C)$ is non-trivial, since $|C\setminus C'|\ge 2$ and hence 
$$M_3\backslash e/(C'\cap C)=N_C/(C'\cap C)=[N^*_{E(N)\setminus C}\backslash (C'\cap C)]^*$$ 
is non-binary. So, ${\bf M}/(C'\cap C)$ is not connected, by \recorde{21}, because $M_i/(C'\cap C)$ has a loop, for some $i\in\{1,2\}$. By the dual of \recorde{ch}(i), $M_1\backslash e/(C'\cap C)=N/(C'\cap C)$ is connected. So, the only 1-separation for ${\bf M}/(C'\cap C)$ has $\{e\}$ as one of its sets. Thus, $e$ is a loop of $M_i/(C'\cap C)$, for every $i\in\{1,2,3,4\}$, otherwise it would be a coloop of $M_i$, for some $i\in\{1,2,3,4\}$, which is contrary to \recorde{21}. Thus, by \recorde{ic}, $C'$ is a circuit of $M_i$, for every $i\in\{1,2,3,4\}$.

Observe that $|C'|\ge 3$, otherwise $C'$ is a parallel class of $M_i$, for every $i\in\{1,2,3,4\}$ and ${\bf M}$ is not irreducible. Now, $C''=(C\setminus C')\cup\{e\}$ is a circuit of $M_1$. Similarly to the previous paragraph, we have that ${\bf M}/(C''\cap C)$ is a non-trivial quad, since $|C\setminus C''|=|C\cap C'|\ge 2$ and $M_3/(C''\cap C)$ has $U_{2,4}$ as a minor. Observe that the only 1-separation for $M_1/(C''\cap C)$ has $\{e\}$ as one of its sets, by the dual of \recorde{ch}(i) applied to $N/(C''\cap C)=M_1/(C''\cap C)\backslash e$. But $M_2/(C''\cap C)$ does not have a 1-separation having $\{e\}$ as one of its sets, because there is just one circuit of $M_2$ contained in $C\cup\{e\}$, namely $C'$. We arrive at a contradiction, by \recorde{21}, since ${\bf M}/(C''\cap C)$ is a non-trivial connected quad having a matroid with a loop.

\bteo\rotulo{step2}
$M_1$ is a 3-connected matroid.
\eteo

By \recorde{21}, it is enough to prove that $\{e,f\}$ is not a circuit of $M_1$ for every $f\in E({\bf M})\setminus\{e\}$. Suppose that $\{e,f\}$ is a circuit of $M_1$. By \recorde{step1'}, $f\not\in C$. By \recorde{ic}, we have two cases to consider: (i) $\{e,f\}$ is a circuit of $M_3$; or (ii) $\{e,f\}$ is a circuit of $M_4$. In both cases, we shall arrive at a contradiction proving that $\{e,f\}$ is a circuit of $M_i$, for every $i\in\{1,2,3,4\}$. 

First, we shall prove that $\{e,f\}$ cannot be a circuit of $M_4$. If $\{e,f\}$ is a circuit of $M_4$, then $E({\bf M})\setminus C$ is a parallel class of $M_4$. As $|E({\bf M})\backslash C|\ge 3$, by \recorde{step1}, we can choose different elements $a$ and $b$ of $E(M)\setminus (C\cup\{e\})$. By \recorde{ic}, it follows that $\{e,a\}$ or $\{e,b\}$ is a circuit of $M_2$, otherwise $\{a,b\}$ is a parallel class of $M_1$. Hence $e$ is in parallel with $f$ in $M_2$, since both $a$ and $b$ are parallel to $f$ in this matroid. By \recorde{ic}, $\{e,f\}$ is also a circuit of $M_3$. We arrive at a contradiction.

When $\{e,f\}$ is a circuit of $M_3$, $C\cup\{e\}$ is also a circuit of $M_3$, since $e$ and $f$ are parallel in this matroid. As $C\cup\{e\}$ is not a circuit of $M_1$, by \recorde{step1'}, it follows that $C\cup\{e\}$ is a circuit of $M_2$, by \recorde{ic}. Hence $(C\cup\{e\})\bigtriangleup (C\cup\{f\})=\{e,f\}$ is also a circuit of $M_2$, since $M_2$ is binary. Again, we arrive at a contradiction using \recorde{ic}. 

\bteo\rotulo{step3}
For every element $f\in E({\bf M})\setminus (C\cup\{e\})$, there is a triad $T_f$ of $M_1$ such that $f\in T_f$ and $T_f\setminus\{f\}\subseteq C$. 
\eteo

First, we shall prove that $M_1\backslash f$ is not 3-connected. When $M_1\backslash f$ is 3-connected, it follows that ${\bf M}\backslash f$ is an irreducible non-trivial semi-binary quad. Thus, by the choice of ${\bf M}$, ${\bf M}\equiv{\bf Q}(N';C')$, for some 3-connected binary matroid $N'$. So, $C=C'$. Hence $C\cup\{e\}$ is also a circuit of $M_3$ and $e$ is not spanned by $M_1$. That is, $C$ is a circuit-hyperplane of $M_1$ and $M_3=(M_1)_C$. We arrive at a contradiction and $M_1\backslash f$ is not 3-connected.

Let $\{X\cup\{e\},Y\}$ be a 2-separation for $M_1\backslash f$. Now, we shall prove that $|X|\ge 2$. If $|X|=1$, say $X=\{x\}$, then $\{x,e\}$ is a series class of $M_1\backslash f$, by \recorde{step2}. So, $\{e,f,x\}$ is a triad of $M_1$ and $M_1\backslash e=N$ is not 3-connected, a contradiction. Hence $|X|\ge 2$ and $\{X,Y\}$ is a 2-separation for $M_1\backslash\{e,f\}=N\backslash f$. By \recorde{ch}(iii), it follows that there is $Z'\subseteq Z\subseteq C$, for some $Z\in\{X,Y\}$, such that $|Z'|=2$ and $Z'$ is contained in a series class of $M_1\backslash\{e,f\}=N\backslash f$. As $X$ spans $e$ in $M_1$, it follows that $Z=Y$, by \recorde{step1'}, and $Y'$ is contained in a series class of $M\backslash f$. Thus, we can take $T_f=\{f\}\cup Y'$.

\bteo\rotulo{step4}
If $g\in T_f\cap C$, for some $f\in E({\bf M})\setminus (C\cup\{e\})$, then there is a triangle $T_g$ of $M_1$ such that $\{f,g\}\subseteq T_g$ and $T_g\cap C=\{g\}$.
\eteo

First, we shall prove that $M_1/g$ is not 3-connected. If $M_1/g$ is 3-connected, then ${\bf M}/g$ is a non-trivial quad because $M_1\backslash e/g\not=M_3\backslash e/g$, by \recorde{step1}. So, ${\bf M}/g$ is an irreducible non-trivial semi-binary quad and, by the choice of ${\bf M}$, ${\bf M}/g\equiv{\bf Q}(N',C')$ for a 3-connected binary matroid $N'$. Thus, $C'=C\setminus\{g\}$ and $C\cup\{e\}$ or $(C\setminus\{g\})\cup\{e\}$ is a circuit of $M_3$. Hence $C''=(C\setminus\{g\})\cup\{e\}$ is a circuit of $M_3$, otherwise $M_3=(M_1)_C$.  As $C''\cap T_f=\{g'\}$, where $T_f\cap C=\{g,g'\}$, it follows that $T_f$ is not a triad of $M_3$. Now, we shall prove that $T_f$ is a triad of $M_3\backslash e$. Observe that $T_f$ is a triad of $M_1\backslash e=N$, since this matroid is 3-connected. Thus, by the dual of \recorde{ic}, $T_f$ is a triad of $M_3\backslash e$ or $M_4\backslash e$. As, by orthogonality, $T_f$ cannot be a triad of $M_4\backslash e$, it follows that $T_f$ is a triad of $M_3\backslash e$. Hence $T_f\cup\{e\}$ is a cocircuit of $M_3$. So, $T_f\cup\{e\}$ is a cocircuit of $M_2$, by the dual of \recorde{ic}. We arrive at a contradiction, by orthogonality, because $P=E({\bf M})\setminus(C\cup\{e\})$ is contained in a parallel class of $M_2$ and $P\cap (T_f\cup\{e\})=\{f\}$. Thus, $M_1/g$ is not 3-connected.

Let $\{X\cup\{e\},Y\}$ be a 2-separation for $M_1/g$. When $|X|=1$, say $X=\{x\}$, then $\{x,e\}$ is a parallel class of $M_1/g$, by \recorde{step2}. So, we can take $T_g=\{e,g,x\}$, since $M_1$ is 3-connected and $x\not\in C$, otherwise $C$ spans $e$ in $M_1$, which is contrary to \recorde{step1'}. Hence, we may suppose that $|X|\ge 2$. Thus, $\{X,Y\}$ is a 2-separation for $M_1\backslash e/g=N/g$. By the dual of \recorde{ch}(iii), it follows that there is $Z\in\{X,Y\}$ and $Z'\subseteq Z$ such that $Z'\cap C=\emptyset$, $|Z'|=2$ and $Z'$ is contained in a parallel class of $M_1\backslash e/g=N/g$. Thus, we can take $T_g=\{g\}\cup Z'$. 
\vspace{0.3cm}
%We have two cases to consider: (i) when $Z=Y$, it follows that $Y$ is contained in a parallel class of $M_1/g$. Thus, we can take $T_g=\{g\}\cup Y'$, where $|Y'|=2$ and $Y'\subseteq Y$; or (ii) when $Z=X$, it follows that $X$ spans $e$ in $M_1/g$ and $X\cup\{e\}$ is a parallel class of $M_1/g$. So, we can take $T_g=\{g\}\cup X'$, where $X'\subseteq X\cup\{e\}$ and $|X'|=2$.
%\vspace{0.3cm}

By \recorde{step3} and \recorde{step4}, we can construct a chain of $M_1$ having its rim contained in $C$ and spokes in $E({\bf M})\setminus C$. This chain must close, since the only element of $E({\bf M})\setminus C$ which may not belong to a triad of $M_1$ is $e$. It may happens that this chain starts and ends with triangles that contain $e$. But, by \recorde{step2}, $M_1$ is 3-connected, and by \recorde{sow}(ii), it follows that $M_1$ must be a wheel or whirl. We arrived at the final contradiction.\qed

Now, we shall prove the main result of this paper:
\vspace{0.3cm}

\noindent {\bf Proof of theorem 2:} Suppose that theorem 2 fails. Choose a pair of matroids $M$ and $N$ which contradict it. Observe that ${\bf M}=(M,M^*,N,N^*)$ is a semi-binary non-trivial connected quad having only connected matroids as coordinates. Thus, the core ${\bf N}$ of ${\bf M}$ must be semi-binary. As ${\bf N}$ is irreducible, by \recorde{teorema3}, it must be equivalent to some ${\bf Q}(H;C)$, for a 3-connected binary matroid $H$ or it is isomorphic to ${\bf W}_2$. We arrive at a contradiction because these quads has exactly one matroid which is not connected as a coordinate. \qeds

\section{Acknowledgements}

The author would like to thank Professor James Oxley for some discussions on the topic of this paper and all the staff of the Department of Mathematics of Louisiana State University for their hospitality.

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