homeomorphism of the interior of an annulus on the plane and the whole plane without one point

Question

I would like to establish the homeomorphism of the interior of an annulus on the plane and the whole plane $R^2$ without one point.

Let the interior of annulus be $A=\{(x,y) \in \mathbb{R}^2 : a < \sqrt{x^2+y^2} < b\}$

Simply the first thing that occurred to me is to use a composition of homeomorphisms, and using the ordinary coordinates to set homeomorphisms: $\varphi :=\varphi_2 \circ \varphi_1$

where $\varphi_1(x,y)=(a\frac{\sqrt{x^2+y^2}-a}{b-a}\frac{x}{\sqrt{x^2+y^2}},a\frac{\sqrt{x^2+y^2}-a}{b-a}\frac{y}{\sqrt{x^2+y^2}})$

and $\varphi_2(x,y)=\frac{(x,y)}{a-\sqrt{x^2+y^2}}$

Then $\varphi :=\varphi_2 \circ \varphi_1$ is needed homemorphism.

Could someone check it, please?

Answer 1

You have the right idea, but a lot of unnecessary things: $$\varphi_1(x,y)=(\frac{\sqrt{x^2+y^2}-a}{b-a}x,\frac{\sqrt{x^2+y^2}-a}{b-a}y)$$ followed by $$\varphi_2(x,y)=\frac{(x,y)}{b-\sqrt{x^2+y^2}}$$ is enough. You can also combine them into a single nice formula without any "$ b-a $"...

Answer 2

This is easier in polar coordinates. Map
(r,$\theta$) to (r-a,$\theta$) for a < r < b and then map
(r,$\theta$) to (tan pi.r/2(b-a),$\theta$) for 0 < r < b-a.