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.