Let $I, J$ be two bases of a matroid. For every $x$ in $I$, there is some $y$ in $J$ such that, if we exchange $x$ with $y$, then both resulting sets ($I \setminus x \cup y$ and $J \setminus y \cup x$) are bases (this is the [strong basis exchange property](https://blog.zilin.one/2019/11/03/on-the-basis-exchange-property/)).


Can we extend this property as follows: there exists a bijection $f$ between $I\setminus J$ and $J\setminus I$, such that for every $x$ in $I$, if we exchange $x$ with $f(x)$, then both resulting sets are bases?


The nearest result I found was in [lecture notes by Goemans](https://math.mit.edu/%7Egoemans/18438F09/lec11.pdf). In Lemma 5, he proves that there is a perfect matching between $I\setminus J$ and $J\setminus I$ in a bipartite graph that he denotes by $D\_M(I)$. This means that for every $x$ in $I$, if we exchange $x$ with $f(x)$, then $I \setminus x \cup f(x)$ is a base. But, it does not imply that $J \setminus f(x) \cup x$ is a base too.



Exchanges between independent sets of a matroid
**No**, not every matroid satisfies this property. For example, it is known to fail for the cycle matroid of $K\_4$. The matroids that satisfy your property are called *base orderable* matroids. There are important classes of matroids that are base orderable, such as transversal matroids. Moreover, base orderability is a minor-closed property, but Ingleton proved that there are actually an infinite number of excluded minors. See these [slides](https://blogs.gwu.edu/jbonin/files/2016/04/BaseOrderable-rjuvq2.pdf)
of Joseph Bonin for more information. For example, the slides include a proof that $M(K\_4) $ is not base orderable.


