Homeomorphism of the Disk

Solution 1:

This answer extends on Chris Eagles comment:

Let $D^n \subset \mathbb R^n$ denote the $n$-dimensional closed unit disk, that is $D^n = \{ x \in \mathbb R^n \;|\; |x|\leq 1 \}$, with boundary $\partial D^n = S^{n-1} = \{ x \in \mathbb R^n \;|\; |x| = 1 \}$ the $(n-1)$-dimensional sphere.

Let $f: D^n \to D^n$ be a homeomorphism that maps $x \in \partial D^n$ to $f(x) \in D^n \setminus \partial D^n$. Obviously $f$ induces a homeomorphism $\tilde{f}: D^n \setminus \{ x\} \to D^n \setminus \{ f(x) \}$.

Since $x \in \partial D^n$, we have that $D^n \setminus \{ x\}$ is convex and therefore homotopy equivalent to a point. On the other hand we can construct a homotopy equivalence $D^n \setminus \{ f(x) \} \simeq \partial D^n = S^{n-1}$ since $D^n$ is compact and radially convex wrt. a neighborhood of $f(x)$. Thus we get $\{pt\} \simeq D^n \setminus \{ x\} \cong D^n \setminus \{ f(x)\} \simeq S^{n-1}$, which is a contradiction by your technique of choice. For example $\pi_{n-1}(\{pt\}) \not \cong \pi_{n-1}(S^{n-1})$.