# Lift a continuous map $f:RP^2 \to RP^2$

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.

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 $$f \circ p : S^2 \mapsto \mathbb R P^2$$