Pseudo-Anosov diffeomorphisms vs reducible diffeomorphisms I was wondering if anyone knew a 'simple' proof of the fact that a pseudo-Anosov diffeomorphism of a closed surface $\Sigma$ is not reducible, in the sense that it does not fix the free homotopy class of a multi-curve on $\Sigma$? 
Thanks for your attention!
 A: Here are some of the details for the answers given by Ian and Lee in the comments.

Suppose that $S$ is the given (closed, connected, oriented) surface.  Suppose that $f : S \to S$ is the given pseudo-Anosov map.  By the definition of pseudo-Anosov, there is a pair of transverse, transversely measured foliations $F^+$ and $F^-$ in $S$ which are both preserved by $f$.  Also, $f$ expands the transverse measure of one and contracts the transverse measure of the other, both by the same factor $\lambda > 1$.  There are different conventions about which foliation gets called "stable" and which gets expanded...  I'll call $F^+$ the unstable foliation and "draw" it as horizontal.
Suppose that $\gamma$ is a small arc in $S$, transverse to both foliations. Then we can "integrate" $\gamma$ against $F^\pm$ and get a pair of real numbers.  I'll call these the "imaginary" and "real" periods of $\gamma$.  This gives $S$ a locally euclidean metric, away from the common singularities of $F^\pm$.  It is an exercise to show that the cone angle at a $k$-prong singularity equals $k \cdot \pi$.  (Recall that $S$ is closed.)
Thus the transversely measured foliations $F^\pm$ gives us a "singular euclidean metric" on $S$.  Let's call this metric $q = q_f$.  This is the "induced geometric structure" we need.  Note that $f$ does not preserve $q$.  Instead (by our convention above) $f$ stretches the horizontal foliation (that is, $F^+$) and shrinks the vertical.  That is, $f$ acts on $(S, q)$ as a piecewise affine map, non-affine only at the singularities, preserving area.
Suppose that $\alpha \subset S$ is a simple closed essential curve. We can "pull $\alpha$ tight" and obtain a $q$-geodesic $\alpha^*$.  Note that $\alpha^*$ is typically unique - when it is not unique the union of all geodesic representatives of $\alpha$ foliate a "flat cylinder" $A(\alpha)$.  Thus either $\alpha^*$ or $A(\alpha)$ is an invariant of the homotopy class of $\alpha$.
Finally, consider any curve $\alpha$.  Set $\alpha_k = f^k(\alpha)$.  We now compute the $q$-geodesic representatives $\alpha_k^*$.  As  Ian and Lee point out these tend (exponentially quickly in $k$) towards horizontal.  Since $f$ does not preserve (any!) geodesic representatives, it does not preserve any homotopy classes.
