homeomorphism of the closed disc on $\mathbb R ^2$

Question

I just found a statement that for any closed disc on $\mathbb R^2$ and any two points inside it, there is a homeomorphism taking one point to another and is identity on the boundary. But I can't write down the homeomorphism myself. Can anyone help? Many thanks in advance.

Answer 1

For each $\mathbf v_0=(x_0,y_0)\in D$, the unit disk, I will define a specific homeomorphism $f:D\to D$ so that $f_{\mathbf v_0}(\mathbf 0)=\mathbf v_0$ which fixes the boundary $S^1$.

Now, for $\mathbf v\neq 0$, we have $\mathbf v=r\mathbf s$ for a unique $(r,\mathbf s)$ for $r\in(0,1]$ and $\mathbf s\in S^1$. This is essentially representing $\mathbf v$ in polar coordinates.

Then define $$f_{\mathbf v_0}(\mathbf v)=r(\mathbf s-\mathbf v_0) +\mathbf v_0.$$

This is clearly well-defined on $D\setminus \{\mathbf 0\}$, and continuous.

Show it is one-to-one and onto $D\setminus\{\mathbf v_0\}$.

Pretty obviously, $f_{\mathbf v_0}(\mathbf v)\to \mathbf v_0$ as $\mathbf v\to \mathbf0$. So we can extend $f_{\mathbf v_0}$ to $f_{\mathbf v_0}(\mathbf0)=\mathbf v_0$. It takes a little more to prove this extension is a homeomorphism $D\to D$. It obviously fixes each $s\in D$ has $r=1$

Now if you need to send $\mathbf v_1\to\mathbf v_0$, do so by composing homomorphisms: $f_{\mathbf v_1}^{-1}\circ f_{\mathbf v_0}$.

This exact argument shows the same result in $n$ dimensional closed balls, or even any $n$-dimensional compact convex set.

Given any convex and compact set $D$ in $n$-dimension, let $B$ be the boundary, and $x\in D\setminus B$ be a point in the interior. Then every element of $D\setminus \{\mathbf x\}$ has a unique representation as $r(\mathbf b-\mathbf x)+\mathbf x$ with $r\in(0,1]$ and $\mathbf b\in B$.

This also lets you prove that any convex and compact subset of $\mathbb R^n$ with non-empty interior is homeomorphic to a closed ball.

This is related to the notion of a topolical cone. If $D$ is convex and compact, and $B$ is on the boundary, then the space $C(B)$, the cone on $B$, is homeomorphic to $D$, and we can define that homeomorphism so that the "tip" of $C(B)$ is sent to any point of $D$ we choose.