Let $p:S^2\to P^2$ be the double covering of the real projective plane. Let $g:P^2 \to P^2$ be a map such that its induced homomorphism on fundamental group is not trivial.
I'd like to show that there exists $T:S^2\to S^2$ covering $g$ with $T(-x)=-T(x)$.
Given $x\in S^2$ take $\bar x=p(x)\in P^2$ and $\bar y=g(\bar x)\in P^2$. It is clear that the fiber over $\bar y$ has two points $y_1,y_2\in S^2$.
I guess that $T(x)$ should be one of them. Also I guess that the action of $\pi_1(P^2)$ should be relevant to choose between $y_1,y_2$. But I'm not sure how.
I tried to take $\alpha\in\pi_1(P^2,\bar x)$ a generator and considered its image $\beta=g_*(\alpha)\in\pi_1(P^2,\bar y)$. Then I tried to take a lifting. But what should be the criteria to choose the necessary points?