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A sufficient condition is that both collections are Alexander systems: within each collection the essential simple closed curves are pairwise nonisotopic, are in pairwise minimal position of curves or arcs, have no triple intersection points, and no three curves intersect pairwise. If ${\alpha }_i$ is isotopic to ${\beta }_i$ for every $i$, the simultaneous-isotopy lemma for Alexander systems gives an ambient isotopy $H_t$ with $H_1({\alpha }_i)={\beta }_i$ for all $i$. Successive applications of the bigon criterion prove the lemma while preserving the curves already matched.

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.

Whenever $\alpha$ and $\beta$ bound a bigon, isotope one side across the bigon. The two removed crossings have opposite local signs: the induced directions around the two corners of an oriented disc are opposite. Thus the move reduces the geometric intersection number by two while leaving the algebraic intersection number of curves on an oriented surface unchanged.

Repeatedly remove bigons. The process terminates because the intersection count is a nonnegative integer, and the bigon criterion says that the resulting curves ${\alpha }_0$ and ${\beta }_0$ are in minimal position. Since every removal cancelled one positive and one negative crossing,

On a genus-two surface, let $\alpha$ be a separating simple closed curve that cuts the surface into two once-punctured tori. Choose a simple closed curve $\beta$ that passes from one side to the other and back, with the two crossings arranged in minimal position. The crossings have opposite signs, sobut the bigon criterion shows that they cannot be removed and henceEquivalently, $\alpha$ is separating and therefore represents zero in first homology, forcing its algebraic intersection with every curve to vanish even though its geometric intersection need not vanish.