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I'm looking to prove that continuous functions $f: S^1 \rightarrow S^1$ satisfying $f(x) = -f(-x)$ for all $x \in S^1$ represent injections on homology.

I'm trying to prove this fact on the way to proving the Borsuk-Ulam theorem, and I really don't know where to start (although I do know alternative proofs of Borsuk-Ulam).

I'm guessing the map $g(x) = \frac{f(x) - f(-x)}{\|f(x) - f(-x)\|}$ will appear at some point. Also, we know the homology groups of $S^1$, namely $\mathbb{Z}$ at $0$ and $1$, and $0$ elsewhere. But as for the actual statement we're trying to prove, I don't see how to begin.

Ted Shifrin
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user95612
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  • I've edited the question into the body of my post. Also, I mean $f(x) = -f(-x)$ for all $x \in S^1$. Sorry about that. – user95612 Sep 18 '13 at 22:55
  • No problem, thanks :-) – Stefan Hamcke Sep 18 '13 at 22:55
  • A proof of a generalization of this fact is presented in Hatcher's Algebraic Topology book. The idea is to prove that a map which such property has odd degree, and so cannot be null. Anyway the proof require the use of homology with coefficient in $\mathbb Z/2\mathbb Z$. Are you looking for some more direct proof? – Giorgio Mossa Sep 18 '13 at 23:04
  • Wait, is it sufficient to show that $f$ is not nullhomotopic? – user95612 Sep 19 '13 at 17:42

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