differential geometry – $f: B^2 rightarrow B^2$ be a continuous map such that $f(x) = x$ for every $x in S^1$, show $f$ is surjective

Let $B^2 = {x in mathbb{R}^2; ||x|| leq 1}$ be the unit disk inthe plane. Let $f: B^2 rightarrow B^2$ be a continuous map such that $f(x) = x$ for every $x in S^1 = {x in mathbb{R}^2; ||x|| = 1}$. Show that $f$ is surjective.

This my following attempt at a solution:

If $f$ isn’t surjective, then there exists a point $y in B^2$ such that $f(x) – y$ is never $0$ for all $x in B^2$. Thus $frac{f(x) – y}{|f(x) – y|}$ is well defined and exists on $S^1$.

however, I’m sure this is a dead end. Some hints would be greatly appreciated!