# Homeomorphism from S^2 to a subset of S^2

## 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 $$g : S^2 \to g(S^2) \subset S^2$$, 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.

matt grime
Homework Helper
Closed, connected and _simply_ connected. Homotopy type is invariant under homeomorphism so its also not a contractible subset of S^2.

Closed, connected and _simply_ connected. Homotopy type is invariant under homeomorphism so its also not a contractible subset of S^2.
So we didn't actually study simply connected sets much (just in passing when talking about Poincare's Lemma), so I didn't know the definition too well when trying the problem and it wasn't obvious at the time that it would be a topological property, but now that I look at the definition, it's extremely obvious.

I don't know at all what a contractible set is, so that doesn't help.

However, since I know that g(S^2) is compact and simply connected, this tells me that it is essentially S^2 with at most 1 closed hole punched out of it.

If I assume that there is such a hole, then I can take its boundary (which I will call $$\psi$$). Then $$g^{-1}(\psi)$$ is a closed path in S^2. This divides S^2 into 2 (non-empty, open) regions. I can take 2 points, one in each region (say, p and q). Any path between them obvious intersects $$g^{-1}(\psi)$$.

I can then take g(p) and g(q). Since p and q were not in the image of $$g^{-1}(\psi)$$, g(p) and g(q) are not on the boundary of the hole ($$\psi$$). Therefore, there is a path (called $$\varphi$$) that is connecting them that does not intersect $$\psi$$.

Then $$g^{-1}(\varphi)$$ is a path that connects p and q but does not intersect $$g^{-1}(\psi)$$, which is a contradiction.

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This seems good... the only leap that I see is my conclusion that simply connected implies that there is at most 1 hole. Intuitively, this seems true, but I don't know how to prove it.