outer automorphism classification I am trying to understand Bestvina's "A Bers-like proof of the existence of train tracks for free group automorphisms". I'm going to ask a probably trivial question ... Here we go:
The automorphism $\Phi\in Out(F_n)$ is called reducible if there is some $\phi:\Gamma\to\Gamma$ that represents $\Phi$ and leaves a homotopically nontrivial proper subgraph invariant. Otherwise $\Phi$ is irreducible.
The aim of the paper is to show that

Every irreducible automorphism $\Phi$ is represented by a train track map $\phi:\Gamma\to\Gamma$.

Maybe I'm wrong but I think that the aim is to show that only parabolic automorphisms are reducible. $(*)$
For hyperbolic $\Phi$ Proposition 6 shows that there is an optimal map $\phi:\Gamma\to \Gamma\Phi$ st $\phi(\Delta)\subset\Delta$. Also $\Delta$ is a core graph hence homotopically nontrivial. Doesn't this contradict $(*)$? What am I missing?
 A: If I've understood correctly, your concern is that for an optimal representative of a hyperbolic automorphism, the tension subgraph is a homotopically non-trivial subgraph left invariant by $\phi$, showing that the automorphism is reducible.
In fact, hyperbolic (meaning that the infimum of the displacement is positive and realised) automorphisms of the free group are allowed to be reducible! For an easy example, add an extra loop to the hyperbolic automorphism represented by Figure 2 in the paper and extend the automorphism by the identity on this loop. Then this topological representative minimises displacement for the resulting automorphism of $F_3$ (for example, by Remark 3.3), but it clearly has a homotopically non-trivial subgraph.
This situation is in contrast with the case of surfaces, where the map is pseudo-Anosov whenever a positive infimal displacement in the Teichmuller metric is realised.
The proof of the main theorem, that irreducible automorphisms admit train tracks, still stands: the point is that the hyperbolic automorphism is irreducible if and only if the tension subgraph in Proposition  is all of $\Gamma$.
