Let $f:S^{2n} \rightarrow S^{2n}$ be a continuous map. Show that
- there exists $x \in S^{2n}$, such that $f(x) =x$ or $f(x) = -x$;
- 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.