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## Main Question or Discussion Point

So for my analysis final, one of the questions was to prove the smooth version of the Hairy Ball theorem (that there is no smooth, non-vanishing function f from S^2 to itself such that for all x in S^2, f(x) is non-zero and x is tangent to f(x)) (The exam was over 2 weeks ago, so I think it's safe to talk about now)

Assuming that such an f exists, you can use it to construct another function [tex]g : S^2 \to g(S^2) \subset S^2[/tex], which is a homeomorphism that is Lipschitz and whose inverse is Lipschitz.

One key part in the proof was to show that g(S^2) = S^2. However, I could not figure this part out.

I was trying to use topological properties to show this: So obviously, g(S^2) has to be compact and connected (since it is homeomorphic to S^2, which is compact and connected), but why can it not be a subset of S^2 with some closed holes punched out of it? Or can it? Is there a homeomorphism from S^2 to a proper subset of S^2? Did I need some other property of g that I was not considering (for instance, that g and its inverse are Lipschitz). S^2 is simply connected. Does this imply that g(S^2) is simply connected? This would immediately imply that g(S^2) can have at most 1 hole, but I need to show that it has none.

Assuming that such an f exists, you can use it to construct another function [tex]g : S^2 \to g(S^2) \subset S^2[/tex], which is a homeomorphism that is Lipschitz and whose inverse is Lipschitz.

One key part in the proof was to show that g(S^2) = S^2. However, I could not figure this part out.

I was trying to use topological properties to show this: So obviously, g(S^2) has to be compact and connected (since it is homeomorphic to S^2, which is compact and connected), but why can it not be a subset of S^2 with some closed holes punched out of it? Or can it? Is there a homeomorphism from S^2 to a proper subset of S^2? Did I need some other property of g that I was not considering (for instance, that g and its inverse are Lipschitz). S^2 is simply connected. Does this imply that g(S^2) is simply connected? This would immediately imply that g(S^2) can have at most 1 hole, but I need to show that it has none.