Let $F$ be a fixed free group of finite rank. If $H$ is a finitely generated subgroup of $F$ and $A$ is a basis for $F$, then we can form the Stallings graph $\Gamma\_A(H)$ for $H$. It is the unique smallest ($|A|$-labelled) subgraph of the covering space of a bouquet of $A$-circles corresponding to $H$ that contains the base point and carries the fundamental group. 


It is easy to see that if $\Gamma\_A(H)$ embeds in $\Gamma\_A(K)$, then $H$ is a free factor in $K$. If one fixes the basis $A$ then there are free factors of $F$ itself that are not induced by graph embeddings. 


Let us say that a finitely generated subgroup $H$ of $F$ is a *geometric free factor* of a subgroup $K$ if there exists some basis $B$ for $F$ such that $\Gamma\_B(H)$ embeds as a labelled graph in $\Gamma\_B(K)$. 



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> > **Question.** Is every free factor $H$ of a finitely generated subgroup $K$ of a free group $F$ a geometric free factor?
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**History.** I came upon this question in some joint work with Auinger on finite semigroups. To solve some problem we needed all geometric free factors of open subgroups to be closed in the pro-C topology in order for nice things to happen. We would have liked to have been able to drop the word "geometric." Enric Ventura has independently considered this question. 



Are all free factors of finitely generated subgroups of free groups geometric?
If you take a free group on two generators to be $F=\langle a,b\rangle$, and take an index 2 subgroup to be $K$, then the subgroup $H$ generated by $[a,b]$ is not a geometric free factor in $K$ for any choice of generators for $F$ (but is a free factor). One may verify this for all three 2-generator subgroups of $F$ with respect to the generators $a,b$. Changing to another set of generators (applying an automorphism of $F$) sends the commutator to a conjugate of commutator of generators, which is therefore not geometric with respect to any generating set. 


