About Alexander method in mapping class group

Question

The Alexander Method in Farb and Maraglit's "A Primer on Mapping Class Groups"

For a surface $S$ with marked points, we say that a collection $\left\{\gamma_{i}\right\}$ of curves and arcs fills $S$ if the surface obtained from $S$ by cutting along all $\gamma_{i}$ is a disjoint union of disks and once-marked disks.

Proposition 2.8 (Alexander method) Let $S$ be a compact surface, possibly with marked points, and let $\phi \in \operatorname{Homeo}^{+}(S, \partial S) .$ Let $\gamma_{1}, \ldots, \gamma_{n}$ be a collection of essential simple closed curves and simple proper arcs in $S$ with the following properties.

1. The $\gamma_{i}$ are pairwise in minimal position.
2. The $\gamma_{i}$ are pairwise nonisotopic.
3. For distinct $i, j, k$, at least one of $\gamma_{i} \cap \gamma_{j}, \gamma_{i} \cap \gamma_{k}$, or $\gamma_{j} \cap \gamma_{k}$ is empty.

(1) If there is a permutation $\sigma$ of $\{1, \ldots, n\}$ so that $\phi\left(\gamma_{i}\right)$ is isotopic to $\gamma_{\sigma(i)}$ relative to $\partial S$ for each $i$, then $\phi\left(\cup \gamma_{i}\right)$ is isotopic to $\cup \gamma_{i}$ relative to $\partial S$.

If we regard $\cup \gamma_{i}$ as a (possibly disconnected) graph $\Gamma$ in $S$, with vertices at the intersection points and at the endpoints of arcs, then the composition of $\phi$ with this isotopy gives an automorphism $\phi_{*}$ of $\Gamma$.

(2) Suppose now that $\left\{\gamma_{i}\right\}$ fills $S$. If $\phi_{*}$ fixes each vertex and each edge of $\Gamma$, with orientations, then $\phi$ is isotopic to the identity. Otherwise, $\phi$ has a nontrivial power that is isotopic to the identity.

The power of the Alexander method is that it converts the computation of a mapping class into a finite combinatorial problem.

(roughly) states that if $c_1,c_2$ are two filling curves in minimal position and $\phi$ is an orientation preserving homeomorphism of $S$ such that $\phi(c_i)$ is isotopic to either $c_1$ or $c_2$, then a power of $\phi$ is isotopic to the identity.

why Alexander method gives us a finite combinatorial problem ? in the sense of automorphism in graph theory ?

Answer 1

why Alexander method gives us a finite combinatorial problem ? in the sense of automorphism in graph theory ?

Yes, the "Alexander method" reduces a topological problem (deciding if a homeomorphism is isotopic to the identity) to a combinatorial one (understanding the automorphisms of a graph). It works because the complement of the graph is (assumed to be) a collection of disks, and also because the mapping class group of the disk, fixing the boundary pointwise, is trivial.