Someone asked me to give an explicit homeomorphism between $\mathbb C$ and the unit disc. I gave him the following answer:
we look at $\mathbb C$ as $\mathbb R^2$. The map $x\mapsto \tan (\pi x/2)$ is an homeo from $(-1,1)$ to $\mathbb R$ which induces an homeo between $ (-1,1)\times(-1,1)$ and $\mathbb R^2$. it remains to show that $(-1,1)\times(-1,1)$ is homeo to the disc $ D=\{(x,y) \;|\; x^2+y^2<1\}$ and this is true since we have the following homeo f:
$f:D\longrightarrow (-1,1)\times(-1,1)$ such that $f(0,0)=(0,0)$ and if $(x,y)\not = (0,0)$ then $ f(x,y)=((x^2+y^2)*x/m, (x^2+y^2)*y/m)$ where $ m= max(|x|,|y|)$.
is there a more elegant/direct answer to this question?
Best Answer
I find the map that just radially shrinks the plane into the disc more natural. Explicitly, this is (after identifying $\mathbb{C}$ with $\mathbb{R}^2$), $$(x,y)\mapsto \frac{1}{\|(x,y)\|+1} (x,y)$$