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Let $f:S^{2n} \rightarrow S^{2n}$ be a continuous map. Show that

  1. there exists $x \in S^{2n}$, such that $f(x) =x$ or $f(x) = -x$;
  2. any continous map $g: \mathbb R P^{2n} \rightarrow \mathbb RP^{2n}$ has a fixed point;

I think 1 implies 2. So, how can I find an $x$ in 1?

Thanks for the help.

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  • $\begingroup$ What exactly is your question and what have you tried? $\endgroup$
    – Kris Williams
    Commented May 30, 2013 at 16:21
  • $\begingroup$ @KrisWilliams: Thanks for reminding me. I am trying to find the $x$ in 1. $\endgroup$
    – sunkist
    Commented May 30, 2013 at 16:27
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    $\begingroup$ Are you aware of the hairy ball theorem? $\endgroup$
    – Dan Rust
    Commented May 30, 2013 at 16:40
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    $\begingroup$ @sunkist, how did you prove that 1 implies 2? $\endgroup$
    – Sigur
    Commented Oct 30, 2014 at 16:29
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    $\begingroup$ @Sigur Given $f:RP^{2n}\to RP^{2n}$ and the canonical surjection $p:S^{2n}\to RP^{2n}$, you can lift $fp$ to a map $g:S^{2n}\to S^{2n}$ (so that $pg=fp$) because $p$ is a universal cover of $RP^2$. Then $g(x)=\pm x$ for some $x$, so that $f(p(x))=p(g(x))=p(\pm x)=p(x)$. $\endgroup$
    – Anonymous
    Commented Dec 19, 2019 at 9:47

1 Answer 1

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Hint: If $f(x)\neq -x$ for all $x$, show $f$ is homotopic to the identity, $I:S^{2n}\to S^{2n}$. If $f(x)\neq x$ for all $x$, show $f$ is homotopic to $-I$. Therefore, if $f(x)\neq \pm x$ for all $x$, then $I$ and $-I$ are homotopic.

Presumably, you know that in $S^{2n}$ that $I$ and $-I$ cannot be homotopic.

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  • $\begingroup$ Thank you very much for the answer. $\endgroup$
    – sunkist
    Commented Jun 17, 2013 at 12:51

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