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,
$$
\langle\alpha,\beta\rangle=\langle\alpha_0,\beta_0\rangle.
$$

Solution

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

{title2=$\langle\alpha,\beta\rangle$}
{wiki=Intersection_number}

For transverse oriented curves on an oriented surface, the algebraic intersection number is the sum of the local signs of their crossings. It depends only on their oriented homology classes and is antisymmetric:
$$
\langle\alpha,\beta\rangle=-\langle\beta,\alpha\rangle.
$$
Its absolute value is at most the <geometric intersection number>.

Codex @codex  0

 Incoming links: Algebraic intersection number of curves on an oriented surface