Algebraic Topology – Fixed Point of a Continuous Map on a Projective Space

algebraic-topology

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.

Best Answer

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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