How would I go about constructing a homeomorphism between $\Bbb R^2\setminus \{(0,0)\}$ and the annulus $\{\,(r,\theta)\mid 1<r<3\,\}$. I'm thinking we can represent all points in $\Bbb R^2$ as $\frac{e^{i\theta}}{r}$ we can still get to all points in the xy-plane but we can't have a radius of zero which is what we want. As far as $\theta$ goes we really only need $0\leq\theta\leq 2\pi$ and we can have a direct correspondence between the angles of points from each set, that is if I want to map a point from the plane with angle $\theta$ just map it to some point in the annulus with the same angle. The issue I am having is how to construct a function that will shrink $r\in\Bbb R^2\setminus \{(0,0)\} $ so that it lands between 1 and 3 and vice-versa.
[Math] Construct a homeomorphism between the punctured plane and the annulus
general-topology
Best Answer
One example for a homeomorphism $(0,\infty)\to (1,3)$ would be $r\mapsto \frac2{r+1}+1=\frac{r+3}{r+1}$.
A direct homeomorphism for the punctured plane and the annulus mihgt be $$(x,y)\mapsto\left(\frac{\sqrt{x^2+y^2}+3}{\sqrt{x^2+y^2}+1}\cdot \frac x{\sqrt{x^2+y^2}},\frac{x^2+y^2+3}{x^2+y^2+1}\cdot \frac y{\sqrt{x^2+y^2}}\right) $$