Past exam of the mathematics course of the University of Cambridge / 2022 / iii / Paper 156 / 2 / c / Solution

If two arcs bound a bigon, pushing one side across its disc is an isotopy relative to the ends that removes the two corners. Such representatives cannot be in minimal position.

Conversely, compare $\alpha$ with an isotopic representative having the fewest intersections with $\beta$, and lift the isotopy to $\overline{{\mathbb{H}}^2}$. At the first stage where the original excess intersections disappear, two lifted arcs enclose an innermost disc. Part b rules out escape through a puncture or repeated intersections at the ideal boundary, so this disc projects injectively to a bigon on $S$. Thus absence of a bigon implies minimal position. This proves the bigon criterion for essential simple proper arcs.