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

• LukeD
In summary, the student was trying to show 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). However, they were not able to figure out how to do this using topological properties.f

#### LukeD

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.

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.