by a pseudo-Anosov if and only if $\psi$ is induced by a pseudo-Anosov.

Let $H \le F*_\phi$ and $H' \le F'*_\psi$ be isomorphic finite index subgroups. Without loss of generality, suppose $\phi$ is induced by a pseudo-Anosov homeomorphism on a punctured surface. Then $\phi$ is an automorphism and $F*_\phi = F \rtimes_\phi \mathbb Z$. By Thurston’s hyperbolization theorem, $F \rtimes_\phi \mathbb Z$, and hence $H' \cong H$, is a fundamental group of a complete finite-volume hyperbolic 3-manifold $M$. Assume $H' \trianglelefteq F' *_\psi$, then Mostow’s rigidity implies the finite group $Q = (F' *_\psi)/H'$ acts on $M$ by isometries. The action is free since $F' *_\psi$ is torsion-free. Thus $F' *_\psi$ is the fundamental group of the hyperbolic 3-manifold $M/Q$. By symmetry of the BNS-invariant and Stallings’ fibration theorem, $\psi$ is induced by a surface homeomorphism. As $\psi$ has a hyperbolic mapping torus, it is induced by a pseudo-Anosov.

As mentioned before the proposition, the proof can be replaced by an argument using the theory of train tracks. However, the next theorem is a geometric statement and there is no apparent way around Thurston’s hyperbolization theorem. In fact, we will also need Schwartz’ rigidity [@Sch95] to reduce the theorem to 