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For all $x\in S^2$, let $-x$ denote its antipode. Let $f:S^2\rightarrow S^2$ be a continuous map such that $f(x)\neq f(-x)$ for all $x\in S^2$. Show that $f$ must be surjective.

I'm working through an old practice exam, and I've been stuck on this question for hours.

I think this has something to do with the Borsuk Ulam Theorem, or maybe Brouwer degree but I am not sure how to construct a function $S^2\rightarrow \mathbb{R}$ that makes $f$ not being surjective a contradiction, or that contradicts anything about degrees.

I've found a somewhat similar question here, but not sure how to adapt it to this problem.

I know that if $f$ is not surjective, $\exists y\in S^2$ such that $y\notin f(S^2)$, so I can construct a well-defined map $$g:S^2\rightarrow S^2,\hspace{2cm} g(x)=\frac{f(x)-y}{|f(x)-y|}$$ that's homotopic to $f$. Not sure how that helps.

I also thought of the map $$h:S^2\rightarrow S^2,\hspace{2cm}h(x)=\frac{f(x)-f(-x)}{|f(x)-f(-x)|}$$ which we can well-define, and is homotopic to both $f$ and $f\circ A$ where $A:x\rightarrow -x$.

I also know that if $f$ is not surjective then its degree is $0$, and that the degree of homotopic maps is equal, so the degree of $g$, $h$ and $f\circ A$ is zero, but not sure if there's anything wrong with that either.

The most elementary answer possible would be most appreciated.

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It sounds like you have the right idea! Particularly in thinking about the Borsuk-Ulam theorem. Here's a hint:

Remember that $S^2 \cong \mathbb{R}^2 \cup \{ \infty \}$ (this is stereographic projection).

So (towards a contradiction) say we're given a map $f : S^2 \to S^2$ isn't surjective. Well without loss of generality we can say it misses $\infty$, so that composing with stereographic projection gets us a map $f' : S^2 \to \mathbb{R}^2$.

Now, since $\forall x . f(x) \neq f(-x)$, we see that the same must be true for $f'$...

Let me know if you need more than this, but I'll leave it here so it's still technically a hint :P Do you see how to get a contradiction from here?


I hope this helps ^_^

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