Past exam of the mathematics course of the University of Cambridge / 2022 / iii / Paper 156 / 3

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,

Choose a generic isotopy from ${\alpha }_0$ to ${\alpha }_1$ while keeping ${\beta }_0$ fixed. Except at finitely many times, all intersections are transverse. At each exceptional time a tangency creates or removes a pair of crossings. In oriented local coordinates the two crossings have opposite signs, so their contributions cancel. The signed sum is constant throughout the isotopy, and thereforeThis is the isotopy invariance of algebraic intersection number.

At a crossing $x$, the sign is the orientation of the ordered pair of tangent vectors $(\dot{\alpha },\dot{\beta })$. Exchanging the two vectors reverses orientation, soSumming over the same finite set of crossings gives

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.