A matroid is said to be *strongly base-orderable* if for any two bases $B\_1,B\_2$ there is a bijection $f:B\_1 \to B\_2$ such that for any $S\subseteq B\_1$ set $(B\_1 \setminus S) \cup f(S)$ is also a base.


Are there strongly base-orderable matroids $M\_k = (E, \mathcal{I}\_k)$ for $k=1,2$ on the same ground set $E$, such that the union of $M\_1$ and $M\_2$ is **not** strongly base-orderable?



Is the union of strongly base-orderable matroids strongly base-orderable?
Theorem 42.11 in A. Schrijver, *Combinatorial Optimization: Polyhedra and Efficiency*, 2003 (he references Brualdi, *Common Transversals and Strong Exchange Systems*, 1970):



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> Any truncation of a strongly base orderable matroid is strongly base orderable again.
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Given that, we may assume (w.l.o.g.) that $r(M\_1 \vee M\_2) = r(M\_1) + r(M\_2)$ since if not, then we may take an appropriate truncation $M\_1'$ of $M\_1$ such that $r(M) = r(M\_1') + r(M\_2)$. And $r(M\_1 \vee M\_2) = r(M\_1) + r(M\_2)$ gives us that the bases of $M\_1 \vee M\_2$ are given by $B\_1 \cup B\_2$ such that $B\_1 \cap B\_2 = \emptyset$ (where $B\_i$ is a base of $M\_i$). And in this case it is easy to see that we get the desired bijection $f: B\_1 \cup B\_2 \to B\_1' \cup B\_2'$ from the bijections $B\_1 \to B\_2$ and $B\_1' \to B\_2'$.


