Irreducible outer automorphism and pseudo-Anosov homeomorphisms In Bestvina and Handel's paper on train track maps, they state that a pseudo-Anosov map on a punctured surface with one orbit of punctures induces an irreducible automorphism, which is not fully irreducible if there's more than one puncture (Example 1.4 on Page 6). They don't give a proof of this and I haven't been able to find a proof (nor a sketch of one) anywhere.
Suppose $p:\Sigma \to \Sigma$ was the homeomorphism, $\phi = [p_*]$ was the induced outer automorphism in $Out(F_n) \cong Out(\pi_1(\Sigma))$, and $f:\Gamma \to \Gamma$ was map of graphs that was a reduction for $\phi$, i.e., $\phi = [f_*]$ in $Out(F_n) \cong Out(\pi_1(\Gamma))$ and it has a non-contractible invariant proper subgraph. How am I supposed to relate $\Sigma$ and $\Gamma$ to get a contradiction? The two are homotopy equivalent but there's no reason for $\Gamma \to \Sigma$ to be an embedding.
 A: In that paper, the definition of an irreducible automorphism is one that does not fix (the conjugacy class of) any proper free factor.  This is equivalent to the invariant subgraph definition, but for this example it's easier to think about free factors.  Suppose $\Sigma$ has $k$ punctures and choose a basis for $\pi_1(\Sigma)$ that contains a peripheral curve (a curve $c$ such that one component of $\Sigma-c$ is a once-punctured disk).  Then $\phi(c)\neq c$ since $f$ induces a non-trivial permutation of the punctures, but $\phi^{k!}(c)=c$ as the permutation of the punctures has order dividing $k!$.  The point is that $[c]$ is a free factor, since $c$ is primitive.  Since $f$ is pseudo-Anosov, the only (homotopy classes of closed) curves on the surface that are fixed by $\phi$ are peripheral ones.
A: I thought I would give an answer to this old question.
To get a good answer, one should use some later technology than in that paper you were looking at. Instead one should use the technology from the Bestvina, Feighn, Handel paper on the Tits Alternative for $\text{Out}(F_n)$, part I. That technology includes: free factor systems; free factor supports; attracting laminations; weak attraction; etc. 
That paper defines the set of attracting laminations for $\phi \in \text{Out}(F_n)$. Furthermore, for $\phi$ represented by a pseudo-Anosov surface homeomorphism $p : \Sigma \to \Sigma$, the attracting lamination $\Lambda_\phi$ of $\phi$ is identified with the unstable measured geodesic lamination $\Lambda^u \subset S$ of $p$.
That paper also considers the question: Which conjugacy classes of $F_n$ are weakly attracted to $\Lambda_\phi$ under iteration by $\phi$? Transferring the question over to the surface $S$, and assuming the boundary of $S$ to be totally geodesic, this is equivalent to asking: Which closed geodesics (not necessarily simple) are weakly attracted to $\Lambda^u$ under iteration of $p$? To say that a closed geodesic $c$ is weakly attracted to $p$ means that, for each $i \ge 0$, letting $c_i$ be the closed geodesic homotopic to $p^i(c)$, and fixing an $\epsilon > 0$, the maximal length of a subsegment of $c_i$ staying within $\epsilon$ of a leaf segment of $\Lambda^u$ goes to $\infty$ as $i \to \infty$. And the answer to this question is: only the boundary geodesics, and their iterates, are not weakly attracted to $\Lambda^u$; every closed geodesic not contained in the boundary is weakly attracted to $\Lambda^u$.
Now, transferring the question back over to the $F_n$ setting, suppose that there is a map of graphs $f : \Gamma \to \Gamma$ representing $\phi$, and a proper subgraph $H \subset \Gamma$ with noncontractible components such that $f(H) \subset H$. When leaves of $\Lambda_\phi$ are realized in $\Gamma$, the maximal length of a leaf segment in $H$ is uniformly bounded. It follows that if a conjugacy class $c$ is realized by a circuit in $H$ then $c$ is not weakly attracted to $\Gamma$, because when $f^i(c)$ is straightened to yield a circuit $C_i$ in $H$, the maximal length subsegment that $C_i$ shares with a leaf segment of $\Lambda_\phi$ is bounded. 
So, the only conjugacy classes supported in $H$ are those representing the boundary components of $S$. And since those boundary components are transitively permuted, each of them is supported in $H$. But the contradiction is that the boundary components of a closed surface fill the free group $F_n$: in any core graph $\Gamma$ with fundamental group marked by an isomorphism with $F_n$, the union of the circuits in $\Gamma$ representing the components of $\partial S$ are surjective in $\Gamma$.
A: This geometric solution is a slight modification of the solution I mention under Lee Mosher's answer. Let $p:\Sigma \to \Sigma$ be the pseudo-Anosov homeomorphism inducing the outer automorphism $\phi = [p_*]$. Any finitely generated subgroup of $F \cong \pi_1(\Sigma)$ can be realized as a connected subsurface of a finite cover of $\Sigma$. This fact is Theorem 2.1 in Scott's paper "Subgroups of surface groups are almost geometric" (https://doi.org/10.1112/jlms/s2-17.3.555) the proof of which is essentially the same as Hempel's or Stallings' proofs of Marshall Hall's theorem.
Suppose $A_1 * \cdots * A_k$ is a $\phi$-fixed proper free factor system of $F$ whose free factors $A_i$ are cyclically permuted by $\phi$. Any $A_i$ has infinite index and is $\phi$-periodic. Realize $A_i$ as a proper subsurface $S_i$ of a finite cover $\bar \Sigma_i \to \Sigma$ and, after passing to a power, $p$ lifts to a pseudo-Anosov $\bar p_i: \bar \Sigma_i \to \bar \Sigma_i$ such that $S_i$ is $\bar p_i$-fixed. But the only fixed proper subsurfaces of a pseudo-Anosov are the boundary components. So $S_i$ is a boundary component of $\bar \Sigma_i$ and $A_i$ is a cyclic free factor corresponding to a boundary component of $\Sigma$. Since the $A_i$ are cyclically permuted by $\phi$ and $p$ acts transitively on the boundary components, there is a one-to-one correspondence between the free factors $A_i$ and the boundary components of $\Sigma$. But this is a contradiction since the boundary components cannot be simultaneously realized as a free factor system.
