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Let $X$ be a finite graph. Its fundamental group is the free group $F_n$ on (say) $n$ generators. Let further an automorphism $\phi$ of $F_n$ be given. It is not true in general that this automorphism is induced by a homeomorphism of the graph. So my question is:

Can we always thicken the graph up such that this automorphism is induced by a homeomorphism of the thickening?

A motivating example is the case of the automorphism $a,b\mapsto a,ab$ in which case we can choose the thickening to be a torus with a disc cut out and the homeomorphism is given by a Dehn twist along one of the generators. More generally those automorphisms for which the thickening can be chosen as a surface with boundary are called geometric.

However I would also allow higher dimensional thickenings.

Actually I am not sure what I mean by thickening and how far the answer depends on that choice. My first guess would be a compact CW-complex that deformation retracts onto the graph.

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    $\begingroup$ FYI there is a highly developed theory, with many applications, of the "best" graph homotopy equivalences representing an (outer) automorphism of $F_n$, known as "relative train track representatives". $\endgroup$
    – Lee Mosher
    Commented May 8, 2015 at 18:35
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    $\begingroup$ Regarding your motivating example, perhaps it's worth pointing out that all automotphisms of $F_2$ are geometric, and can be realized on the punctured torus. $\endgroup$
    – HJRW
    Commented May 8, 2015 at 20:07

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Every automorphism of $F_n$ can be realized by an automorphism of the 3-dimensional handlebody of genus $n$, obtained by attaching $n$ 1-handles to a 3-ball.

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