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Proof of Borsuk-Ulam Theorem . In reading about the answer of Andre Mejia, he uses “compose this with a half rotation $\rho$ about $ S^1$. And if $\phi(x) = -\phi(-x) $, then $\rho \circ \phi(x) = \rho \circ \phi(-x)$.” But I couldn’t see why this rotation $\rho$ makes the terms with the opposite sign equivalent in his argument. I think there was another question about this problem, but without proper answer of this specific question.

Jiahao Fan
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Original Answer: There's only one thing a half rotation can mean in the context of $S^1$, $$f(e^{i\theta})=f^{i(\theta+\pi)} $$ which one can easily simplify to $$f(z)=-z $$

Added: But in the context of the linked answer, it turns out that "half rotation" was the incorrect wording, which has now been fixed by editing the answer. Instead $\rho$ is supposed to be the double covering map $\rho : S^1 \to S^1$ given by the formula $\rho(z)=z^2$, which has the effect $\rho(z)=\rho(-z)$.

Lee Mosher
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  • In the original answer if $f$ just flip the sign of the function $\phi(x)$, then how could this function make the $\phi(x)$ and $\phi(-x)$ equal? For which they originally have the opposite sign. – Jiahao Fan Dec 31 '19 at 15:24
  • I answered to title question of your post, but if you want to ask a question about the details of the proof of the Borsak Ulam theorem, then you should edit your question to reproduce any details from the proof that are needed for someone to understand what you are asking without having to click on an external link. Take a look here for how to write your question in that manner, with emphasis on making your question visible. – Lee Mosher Dec 31 '19 at 15:36
  • Thanks Lee, I am new here, I have already rephrased in my questions. Will be mindful in the future. – Jiahao Fan Jan 01 '20 at 06:34
  • I took a closer look at that linked answer, and I think this should clear up the issues that you raised. Thanks for catching this. – Lee Mosher Jan 01 '20 at 14:58