A loop $\gamma$ in $\mathbb {RP}^2$ is not nullhomotopic if and only if $p^{-1}(\gamma(0))=x_0$ and $p^{-1}(\gamma(1))=-x_0$

Question

I'm working on the problem that if $f:\mathbb{RP^2}\rightarrow\mathbb{RP^2}$ is non trivial in the fundamental group, then there is a lift $F$ of $f$ such that $F(-x)=-F(x)$, and I was able to prove it besides one step that I can't justify. In one point of the proof I took a curve $\gamma$ in $S^2$ such that $\gamma(0)=x_0$ and $\gamma(1)=-x_0$, and I want that $p\circ\gamma$ is not trivial in $\pi_1(\mathbb {RP^2})$.

So what I thinked is that I can choose this $\gamma$ to be a surjective curve, and then if $\gamma$ were trivial, I would be able to find a homotopy between $\gamma$ and a point via lifiting the homotopy in $\mathbb {RP^2}$, and then $S^2$ and then $S^2$ would contract to a point (this part I'm not sure). And then I would be able to choose a $\gamma$ in the way I wanted before.

I also feel like that if I take loop $\gamma$ in $\mathbb {RP}^2$ is not nullhomotopic if and only if $p^{-1}(\gamma(0))=x_0$ and $p^{-1}(\gamma(1))=-x_0$, of course that if $\gamma(0)=\gamma(1)$, then it would be trivial in $\mathbb {RP^2}$, but the for the other implication (the one that solve my problem) I cant find a proof.

So what I did is correct? And is true this last statement?

Thanks in advance

Answer 1

$S^2$ is the universal covering space of $RP^2$. Let's take a basepoint $y_0\in RP^2$. The covering map $p:S^2\to RP^2$ is the universal covering map and $p^{-1}(\{y_0\})=\{x_0,-x_0\}$ say. The fundamental group $\pi_1(RP^2,y_0)$ has order $2$

Each loop in $RP^2$ based at $y_0$ lifts uniquely to a path starting at $x_0$. But it can finish at $x_0$ or $-x_0$, according to the homotopy class of the look in $\pi_1(RP^2,y_0)$. So if the loop is trivial, the lifted path ends at $x_0$, and if it is nontrivial, it ends at $-x_0$.

Answer 2

You should use the fact maps from a square to $\mathbb{RP}^2$ can be lifted to $S^2$, in a unique way if you fix prescribe the lift of a point. That implies that a contractible loop in the projective plane lifts to a closed path in $S^2$ ( go around the edge of the square to see that the ends coincide).