# 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!