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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?

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  • $\begingroup$ May I ask, what are the reasons to expect this to be true? $\endgroup$ – Fedor Petrov Mar 26 '15 at 7:33
  • $\begingroup$ I don't expect it to be true, but I didn't find a counterexample.. $\endgroup$ – Dominic van der Zypen Mar 26 '15 at 7:35
  • $\begingroup$ Well, this is a reason. Have you tried unions of gammoids? $\endgroup$ – Fedor Petrov Mar 27 '15 at 9:08
  • $\begingroup$ I haven't ... Can you construct an example? (I just formulated the question a bit differently.) $\endgroup$ – Dominic van der Zypen Mar 27 '15 at 9:48
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Theorem 42.11 in A. Schrijver, Combinatorial Optimization: Polyhedra and Efficiency, 2003 (he references Brualdi, Common Transversals and Strong Exchange Systems, 1970):

Any truncation of a strongly base orderable matroid is strongly base orderable again.

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 $B_1' \to B_2'$.

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