Use the universal cover to prove $\gamma * \gamma$ is nullhomotopic if $\gamma$ is a loop in the projective plane
Well, If I was not asked to prove it this way, I could have argued like : $\pi_1(\mathbb{R}P^2 ) = \{e,\gamma\}$ and obviously $\gamma * \gamma = e$ and $e*e=e$, hence we're done.
I know that the universal cover of the projective plane is the $2$-sphere $S^2$, Should I look at the lift of this loop upstairs ?
My attempt (I would be very thankful if somebody could correct or verify my chain of thought) Let's choose a base point downstairs, say $x_0$, and let $p:S^2 \to \mathbb{R}P^2$ be the covering map. then $p^{-1}(x_0)=\{x_0,-x_0\}\subset S^2$. A loop $[\gamma]$ based at $x_0$ downstairs , lifts to either :
$1)$ a path connecting $x_0$ and $-x_0$ in $S^2$, Which means $p_*^{-1}([\gamma* \gamma])$ is a loop based at $x_0$ (or $-x_0$) in $S^2$, and since $S^2$ is simply connected, $p_*^{-1}([\gamma* \gamma])$ is nullhomotopic.
$2)$ a loop based at $x_0$ in $S^2$, which means that $p_*^{-1}([\gamma* \gamma])$ is again a loop based at $x_0$ in $S^2$ and is nullhomotopic.
$3)$ a loop based at $-x_0$ in $S^2$ which means that $p_*^{-1}([\gamma* \gamma])$ is again a loop based at $-x_0$ in $S^2$ and is nullhomotopic.
Are these observations sufficient to prove that $p_*(p_*^{-1}([\gamma* \gamma]))=[\gamma* \gamma]$ is nullhomotopic ? What is a theorem that addresses the relation between fundamental groups of a cover (not necessarily universal) space and a base space under the map $p_*$ ?
Thank you in advance,
This is enough. All you need now is that the map on fundamental groups induced by the covering map is injective. I mean, if $f\colon \tilde{X}\to X$ is a covering, then $f_*\colon \pi_1(\tilde{X})\to \pi_1(X)$ is injective. This is a standard fact, see for example Proposition 1.31 in Hatcher's book.