Past exam of the mathematics course of the University of Cambridge / 2022 / iii / Paper 156 / 4 / b

Fix punctures $p,q,r$. There is one isotopy class ${\ell }_p$ of essential arc from $p$ back to $p$: by the Jordan curve theorem, it separates $q$ from $r$, and any two such arcs are isotopic relative to the punctures.

Altogether $\mathcal{A}(S_{0,3})$ has six vertices:The joining arcs $a_{pq},a_{pr},a_{qr}$ span a triangle. For each puncture $p$, the three verticesspan another triangle. These are all the maximal simplices: ${\ell }_p$ must cross $a_{qr}$, and returning arcs based at distinct punctures cannot be made disjoint.

Every orientation-preserving permutation of the three punctures is realizable, while a homeomorphism acting trivially on the punctures is isotopic to the identity. ThusIt permutes the labels in the displayed description. There are two vertex orbits, the three joining arcs and the three returning arcs, as summarized by the arc complex of the three-punctured sphere.