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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.

Paul
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  • @DietrichBurde in that question there are two statements, and author says that first implies second, and there is proof of the first one. But I'm not sure if I understand the implication. – Paul May 13 '20 at 12:46
  • This follows easily from the argument of "Anonymous". But anyway, there is no problem to find proofs here. Start searching. I found the next one. – Dietrich Burde May 13 '20 at 12:54

1 Answers1

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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 $g$.

Lee Mosher
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  • But why $p(\pm x) = p(x)$? – Paul May 13 '20 at 13:25
  • Well, $p(x)=p(-x)$ for all $x$, almost by definition of $p$. Would it be clearer if I wrote this? $$p(f(x)) \in {p(x),p(-x)} = {p(x)}$$ – Lee Mosher May 13 '20 at 13:52
  • Not really, I'm new in topology:(. I understand the geometry behind $p$, but I don't get this kind of symmetry. Could you explain, how it follows from the definition? – Paul May 13 '20 at 13:59
  • The definition of what? Do you know the definition of $S^k$? of $\mathbb R P^k$? of $p : S^k \to \mathbb R P^k$? – Lee Mosher May 13 '20 at 14:12
  • Yeah, I know them all. Still unclear, why ()=(−) for all , almost by definition of . Then probably I don't understand definition of $S^2 \rightarrow \mathbb{R}P^2$. – Paul May 13 '20 at 14:17
  • What definition of $\mathbb R P^2$ do you understand? – Lee Mosher May 13 '20 at 14:22
  • I'll just add, the first definition of $\mathbb R P^2$ that students learn can vary highly, depending on what kind of course they first see it in. So the definition of $p$ therefore also varies highly. – Lee Mosher May 13 '20 at 14:25
  • It's from here https://en.wikipedia.org/wiki/Projective_plane#Classical_examples – Paul May 13 '20 at 14:32
  • Then, using the language of that link: any pair of antipodal points $x,-x \in S^2$ lie on a unique a line $L$ in $\mathbb R^3$; that line $L$ passes through the origin of $\mathbb R^3$ and is therefore a 1-dimensional subspace of $\mathbb R^3$; that line $L$ therefore represents a unique point in $\mathbb R P^2$; and the projection map $p : S^2 \to \mathbb RP^2$, by definition, takes each of the two points $x,-x$ to the point of $\mathbb R P^2$ represented by the line $L$. – Lee Mosher May 13 '20 at 14:40
  • Finally got it, thanks a lot! – Paul May 13 '20 at 14:43
  • Glad to be of help. – Lee Mosher May 13 '20 at 14:44