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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.

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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 $$(x,y)\mapsto\left(\frac{\sqrt{x^2+y^2}+3}{\sqrt{x^2+y^2}+1}\cdot \frac x{\sqrt{x^2+y^2}},\frac{x^2+y^2+3}{x^2+y^2+1}\cdot \frac y{\sqrt{x^2+y^2}}\right) $$

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  • $\begingroup$ Is the way you went about constructing the first example was to find a function whose limits as it approaches both bounds falls between range we want? $\endgroup$ Sep 9 '16 at 3:19
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    $\begingroup$ @Diehardwalnut In principle, yes. Reciprocals would bring $\infty$ to finiteness; however, one first has to translate away from $0$ in order to avoid the other end go to infinity instead. The rest is scaling and translation to adjust $\endgroup$ Sep 9 '16 at 4:54
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Note that $x\mapsto \tan x$ is homeomorphism of the bounded interval $(0,\pi/2)$ with the unbounded interval $(0,\infty)$. With some linear changes the domain can be changed to any open interval (of finite length). That is, the maps of the form $x\mapsto \tan (ax+b)$ for various real numbers $a,b, a\ne0$.

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