= 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.
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