# an homeomorphism from the plane to the disc

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 $x\mapsto tang(\pi x/2)$ 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?

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Let $f\colon\mathbb R^{\ge0}\to[0,1)$ be an order preserving bijection (this means that $f$ is a homeomorphism, since both sets have the order topology).

For $z\in\mathbb C$ we can write $z=\rho e^{i\theta}$. Map $z\mapsto f(\rho) e^{i\theta}$.

For example, $f(x)=\dfrac{x}{1+x}$ is continuous on $\mathbb R^{\ge 0}$, it is strictly increasing (note that we have $f'(x) = \dfrac{1}{(1+x)^2}>0$ everywhere) and $\displaystyle\lim_{x\to\infty} f(x) = 1$, as wanted.

Therefore the map $\rho e^{i\theta}\mapsto \dfrac{\rho}{1+\rho}e^{i\theta}$ should do the trick.

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I find the map that just radially shrinks the plane into the disc more natural. Explicitly, this is (after identifying $\mathbb{C}$ with $\mathbb{R}^2$), $$(x,y)\mapsto \frac{1}{\|(x,y)\|+1} (x,y)$$

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In terms of complex numbers, you can write this as $z \mapsto [|z|/(1 + |z|)] \times z.$ –  Gerben Oct 28 '11 at 10:54
let $f(x,y)=\frac{\|(x,y)\|}{\|(x,y)\|+1} (x,y)$ then why $f(x,y)$ is in the disc? indeed $\|f(x,y)\|=\frac{\|(x,y)\|^2}{\|(x,y)\|+1}$ and this need not be less then $1$. –  palio Oct 28 '11 at 11:01
Right, my mistake. It's fixed now. –  Miha Habič Oct 28 '11 at 15:28