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Let $f: RP^2 \to RP^2$ be a continuous map such that $f$ induces a non-trivial homomorphism on the fundamental group. Show that $f$ can be lifted to a map $\overline{f}: S^2 \to S^2$ such that $\overline{f}(-x))=-\overline{f}(x)$.

I'm stuck on this problem, I try to construct first a lift from $RP^2$ to $S^2$ by using the covering proyection map $p: S^2 \to RP^2$, using this proposition from Hatcher's book:

Proposition: Suppose given a covering space $p:\overline{X} \to X$ and a map $f:Y \to X$ with $Y$ path-connected and locally path-connected. Then a lift $\overline{f}:Y \to \overline{X}$ of $f$ exists iff $f_{\ast}(\Pi_1 (Y))\subseteq p_{\ast} (\Pi_1 (\overline{X})$ . But $S^2$ has trivial fundamental group, so I don't know what to do.

I will appreciate any help.

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Here's a hint: You're using the right proposition, but you're applying it to the wrong map. Instead, apply it to the map $$f \circ p : S^2 \mapsto \mathbb R P^2 $$

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  • $\begingroup$ I got it now, thank you!. $\endgroup$
    – Giotaker
    Oct 21, 2019 at 2:08

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