curve. Since this holds for any boundary component $\gamma$ of the core of $\hat \Sigma$, it must be that $A_j$ is a proper free factor corresponding to a boundary component of $\Sigma$. But $f$ acts transitively on the boundary components, hence there is a one-to-one correpondence between the proper free factors $A_1, \ldots, A_k$ and the boundary components of $\Sigma$. This is a contradiction, as all boundary components of a surface can not be simultaneously realized as proper free factors of the surface group. Therefore, $\phi$ is irreducible.

If $b=1$, then all powers of $f$ are pseudo-Anosovs that act transitively on the boundary component. Thus, all powers of $\phi$ are irreducible. If $b \ge 2$, then any boundary component of $\Sigma$ determines a periodic proper free factor.

For the rest of the appendix, we will prove the converse. Bestvina-Handel use the following proposition to prove that fully irreducible automorphisms with periodic nontrivial conjugacy classes are induced by pseudo-Anosov homeomorphisms of surfaces with one boundary component [@BH92 Proposition 4.5].

Suppose $\phi:F \to F$ is fully irreducible and there exists $k \ge 1$ and nontrivial conjugacy class $[c]$ of $F$ such that $[\phi^k(c)] = [c]$.

Then some iterate of $\phi$ has an 