Projective plane, fixed point

Question

How to show, that for every continuous $f: X\rightarrow X$ there exists $x \in X$, such that $f(x) = x$, where X is a real projective plane $\mathbb{R}P^2$.

In other words: every continuous map of RPP to itself has a fixed point.

EDIT

Probably it's easier to proof, that existence of fixed point in map from $S^2$ to itself implies needed fact.

Answer 1

As mentioned in the comments, this link reduces your question to the following implication, which you said you don't know how to do:

(1) for any continuous map $f : S^{2n} \to S^{2n}$ there exists $x \in S^{2n}$, such that $f(x) =x$ or $f(x) = -x$;

implies

(2) any continous map $g: \mathbb R P^{2n} \rightarrow \mathbb RP^{2n}$ has a fixed point.

To prove this implication, given a continuous $g: \mathbb R P^k \rightarrow \mathbb RP^k$, define $f : S^k \to S^k$ as follows. First, let $p : S^k \to \mathbb R P^k$ be the universal covering map. By composition we obtain $g \circ p : S^k \to \mathbb R P^k$. Applying the general lifting lemma, we obtain a continuous map $f = \widetilde{g \circ p} : S^k \to S^k$ such that $p \circ f = g \circ p$. Applying (1), we obtain $x$ such that $f(x)=x$ or $f(x)=-x$, and it follows that $$g(p(x)) = p(f(x)) = p(\pm x) = p(x) $$ and so $p(x)$ is the required fixed point of $g$.