TITLE: Does every continuous map $f : \mathbb{R} \mathbb{P}^n \to \mathbb{R} \mathbb{P}^n$ have a fixed point for $n = 2$? $n = 3$? and $n = 4$?
QUESTION [6 upvotes]: I am trying to solve this question but I don't have any idea on how to start. Thanks!
Does every continuous map $f : \mathbb{R} \mathbb{P}^n \to \mathbb{R} \mathbb{P}^n$ have a fixed point for $n = 2$? $n = 3$? and $n = 4$?

REPLY [7 votes]: (Just for fun: a solution without Lefschetz theorem)
Any map $f:\mathbb RP^n\to\mathbb RP^n$ lifts to a map of universal covers, $\tilde f:S^n\to S^n$. But $\chi(S^n)$ is non-zero for even $n$, so $\tilde f$ has either a fixed point or point mapping to its antipode (otherwise $\tilde f(x)-x$ gives a non-vanishing vector field). So $f$ has a fixed point.

And for $n=2k+1$ there is even a free $S^1$-action: $S^1\subset\mathbb C^\times$ acts on $S^{2k+1}\subset\mathbb C^{k+1}$ — so $S^1\cong S^1/\{\pm1\}$ acts on $S^{2k+1}/\{\pm1\}=\mathbb RP^{2k+1}$.