$B^2 = \{x \in \mathbb{R}^2; ||x|| \leq 1\}$ is the unit disk inthe plane, and 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!
Note that since $y \notin S^1$, $ty \notin S^1$ for all $t \in [0,1]$. This implies that $H(x,t)= \frac{x-ty}{|x-ty|}$ is a homotopy between $1_{S^1}$ and the map $g$ that you defined composed with $\iota : S^1 \rightarrow B^2$. Furthermore, this homotopy fixes the point $y/|y| \in S^1$. Thus, we have that $$S^1 \xrightarrow{\iota} B^2 \xrightarrow{g}S^1$$ is (pointed) homotopic to $1_{S^1}$. Now apllying we the fundamental group we have that the identity
$$1_\mathbb{Z}: \mathbb{Z} \rightarrow 0 \rightarrow\mathbb{Z}$$ factors through the zero group (since homotopic maps induce the same group morphism), a contradiction.