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I try solving the following excercise:

Show by stereoscopic projection that the $S^2$ sphere is locally homeomorphic to $R^2$.

I tried to solve this by using the cotangens function on the two angles defining the surface of the ball by the projection:

$\phi : S^2 \rightarrow \mathbb{R}^2, \ (\alpha, \beta) \mapsto \left(\cot \alpha, \cot \frac{\beta}{2}\right)$ for $\alpha \in (0, \pi), \beta \in (0, 2\pi)$

However the poles and one of the connecting lines ($\beta = 0$) is not defined by this map. Is there a neat trick to get this out of the way?

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    It seems like you're already labeling points on the sphere with two real numbers, and so at least implicitly already are viewing the sphere as locally homeomorphic to $\mathbb{R}^2$.2011-07-25
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    @wckronholm: you can label all points of any subset of $\mathbb R^2$ with two real numbers but this surely doesn't assume anything about local homeomorphisms...2011-07-25
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    @Marek: Of course, but points on $S^2$ aren't points in a subset of $\mathbb{R}^2$. Presumably, Helium is thinking of the sphere as the units in $\mathbb{R}^3$ and so the $\alpha$ and $\beta$ need to be explained somehow. If these come from spherical coordinates, then what work is there really to do?2011-07-25
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    @wckronholm: well, if we are to prove anything we need to have some definition of $S^2$. As this is obviously an elementary question it's clear from the context that $S^2$ is understood as a subset of $\mathbb R^3$ (not to mention that OP talks about the stereographic projection) and not as a CW-complex with 1 0-cell and 1 2-cell (say); which requires some definition of spheres anyway...2011-07-25

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