# Is this condition equivalent to being a matroid quotient?

Here is the condition, which arose in contemplating polytopes associated to matroid quotients:

Let $M$ and $N$ be matroids on $E$. If $X \subseteq Y \subseteq E$ such that $X$ is indepedent in $N$ and $Y$ is spanning in $M$, then $X$ is independent in $M$ and $Y$ is spanning in $N$.

If $M \to N$ is a matroid quotient, then this condition does hold.

Proof: First, recall that $M \to N$ is a matroid quotient if and only if the following exchange condition holds: if $B_N$ is a basis of $N$, $B_M$ is a basis of $M$, and $i \in B_N - B_M$, then there exists $j \in B_M - B_N$ such that $B_N-i+j$ and $B_M-j+i$ are both bases. This follows since the fundamental cocircuit of $(B_N,i)$ and fundamental circuit of $(B_M,i)$ must intersect in more than one element.

If $X\subseteq Y \subseteq E$ such that $X$ is independent in $N$ and $Y$ is spanning in $M$, then there is a basis $B_N \supseteq X$ of $N$ and a basis $B_M \subseteq Y$ of $M$, respectively. To show $Y$ is spanning in $N$: Let $B \supset X$ be a basis of $N$ such that $|B-Y|$ is minimized. Then if $x \in B - Y$, $x$ is not in $B_M$, and the above exchange condition gives a $y \in B_M - B$ such that $B - x + y$ is a basis of $N$, which contradicts our construction of $B$. Hence, $B \subseteq Y$ and so $Y$ is spanning in $N$. The argument for $X$ is dual.

The proof of this condition only uses one half of the quotient basis exchange condition at a time. However, for a single matroid, the strong and weak basis exchange conditions are equivalent, which gives me hope (although the proof of this equivalence does not seem to generalize to this problem).