So this is a qual problem and I can't think of an easy obvious map that will do this. Any help will give you good karma I'm sure.
Give an example of a local homeomorphism $R^2 → S^2$ which is surjective.
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1If you show us what you have tried perhaps someone can use your work to provide a hint or answer that is based upon your work. – Jay Dec 31 '13 at 21:38
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Try to think of the problem geometrically instead of analytically. If it helps, you can replace $R^2$ by an open disk. – Jim Belk Dec 31 '13 at 23:55
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@JimBelk But it has to be surjective. – kingkongdonutguy Jan 01 '14 at 00:28
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1@kingkongdonutguy Can you imagine a disk wrapping around a sphere in this fashion? What would that look like? – Jim Belk Jan 01 '14 at 01:32
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This answer might be a little vague but I hope it helps. Now $\mathbb{R}^2$ is homeomorphic to two joined open rectangles. For e.g.
$U$ = {$(x,y) \in \mathbb{R}^2 | -1 < x\leq0, -0.01< y< 0.01$} $\cup$ {$(x,y) \in \mathbb{R}^2 | \enspace 0 < x < 2, -1 < y < 1$}
So the problem is equivalent to having a local homoemorphism from $U$ to $S^2$ which is surjective.
Let us call the first rectangle A and second one B. Now A is a thin rectangle (which we consider a strip) and B is a nice big rectangle. Now we wrap B around the sphere so that it covers everything except the north and the south pole. Now the only part left is the strip and so we use it to cover the north and the south pole.
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