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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 $\mathbb{R}^2$. –  wckronholm Jul 25 '11 at 15:55
    
@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... –  Marek Jul 25 '11 at 19:28
    
@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? –  wckronholm Jul 25 '11 at 20:57
    
@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... –  Marek Jul 25 '11 at 21:24
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2 Answers 2

Considering $S^2$ as the points $(x, y, z)$ in $\mathbb{R}^3$ with $x^2 + y^2 + z^2 = 1$, then stereographic projection $S^2 - \{(0,0,1)\} \to \mathbb{R}^2$ is a homeomorphism. (You can check directly that the standard stereographic projection formula defines a continuous bijection with a continuous inverse.) This shows that all points except $\{(0,0,1)\}$ have neighborhoods homeomorphic to an open subset of $\mathbb{R}^2$. Interchanging the roles of $x$ and $z$ gives a stereographic projection $S^2 - \{(1,0,0)\} \to \mathbb{R}^2$ which is also a homeomorphism.

(This is one way of getting the result with just two "patches" in your atlas.)

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this works too. –  ncmathsadist Jul 25 '11 at 23:06
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Now rotate the sphere 90 degrees and repeat with two new poles. You have 4 patches that show $\mathbb{R}^2$ is locally homeomorphic to $S^2$.

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Is it possible to show that that 4 patches are a maximal atlas or do I need to use more than those? –  Helium Jul 25 '11 at 13:10
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I think it has you covered. –  ncmathsadist Jul 25 '11 at 13:22
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