$\mathbb{R}^2$ is diffeomorphic to $D^2$

Question

I am trying to prove to myself that $D^2 = \{ (x,y) \mid x^2 + y^2 < 1 \}$ (the open unit ball), is diffeomorphic to $\mathbb{R}^2$. I know that I would need a differential and invertible map $f$ from $\mathbb{R}^2$ to $D^2$ in order to conclude the the two spaces are diffeomorphic. However, that is where my troubles are. I am thinking that if I use polar coordinates that this might be easier. Help would be appreciated.

Look at $x\mapsto \frac{x}{\sqrt{1+|x|^2}}$, which works in any dimension — Didier, Oct 09 '22 at 12:04

ajotatxe · Accepted Answer · 2022-10-10T18:44:44.083

2

Polar coordinates are indeed very handy here. The idea is to conserve the argument and find a differentiable function from $[0,1)$ to $[0,\infty)$ for the modulus.

$$f(\phi,r)=\left(\phi,\frac{r^2}{1-r}\right)$$

should work. (You have to make the change of coordinates first).

edited Oct 10 '22 at 18:44

answered Oct 09 '22 at 11:50

ajotatxe

65,084

This does not seem to be differentiable in $0$ (?). – Klaus Oct 09 '22 at 11:52
Right, I hope that it is corrected now. – ajotatxe Oct 09 '22 at 11:55

score 1 · Answer 2 · answered Oct 09 '22 at 12:03

Let $\phi : D^2\rightarrow H$ be $$\phi(x,y)=\left(\Re\left({i\frac{z+1}{1-z}}\right),\Im\left({i\frac{z+1}{1-z}}\right)\right)$$ where $H=\left\{(x,y)|x>0\right\}\subset\Bbb{R}^2$ is the open upper half-plane.

And let $\psi: H\rightarrow \Bbb{R}^2$ be $$\psi(x,y)=(x,\ln y) $$ then $f:D^2\rightarrow \Bbb{R}^2$, $f(x,y)=\psi\circ\phi(x,y)$ is the desired diffeomorphism, I think. I asked this question in exam as homeomorphism.

$\mathbb{R}^2$ is diffeomorphic to $D^2$

2 Answers2