Show that the projection of a path along a meridian from south to north pole is non-trivial in real projective space for $n\geq2$ Let $\mathbb{R}\text{P}^n$ be the real projective space where antipodal points are identified and $n\geq2$. Let $\tilde{\gamma}:[0,1]\to \mathbb{S}^n$ be the path from the south pole $-P$ to the north pole $P$ along some meridian and let $\gamma = p \circ \tilde{\gamma}$ be a loop in $\mathbb{R}\text{P}^n$, with $p:\mathbb{S}^n \to \mathbb{R}\text{P}^n$ being the quotient map which is also assumed to be a covering map. Show that $\gamma$ is non-trivial in $\pi_1(\mathbb{R}\text{P}^n)$.
My idea was to assume for sake of contradiction $\gamma$ is trivial, then $\gamma$ is homotopic to the constant loop at $[P]$ via a homotopy $H$. Since $\tilde{\gamma}$ is a lift of $\gamma$, by the homotopy lifting lemma there exists a continuous map $\tilde{H}:[0,1]\times [0,1] \to \mathbb{S}^n$ such that $H=p\circ \tilde{H}$ and $\tilde{H}(s,0)=\tilde{\gamma}(s)$ for $s\in [0,1]$. Futhermore the constant path is lifted to a constant path obtaining $\tilde{H}(s,1)=P$ for $s\in [0,1]$, where we chose $P$ without loss of generality. Now we have $\tilde{H}(0,0)= -P$ and $\tilde{H}(0,1)= P$, meaning $\tilde{H}(0,t)$ is a path between $-P$ and $P$, but we also have $p \circ \tilde{H}(0,t) = H(0,t) = [P]$ for all $t\in [0,1]$ which is a contradiction.
I couldn't find an answer to this question but since this seems to be a standard exercise I would appreciate any feedback on my proof.
 A: Your proof is correct. Actually it works for any path from the south pole to the north pole.
A more general result is this.
Let $p: \tilde X \to X$ be a covering map, $u_0, u_1 : [0,1] \to X$ be two loops based at $x_0 \in X$ and $\tilde x_0 \in p^{-1}(x_0)$. The paths $u_i$ have unique lifts to paths $\tilde u_i : [0,1] \to \tilde X$ such that  $\tilde u_i(0) = \tilde x_0$. If $u_0$ and $u_1$ are homotopic loops, then $\tilde u_0(1) =  \tilde u_1(1)$.
The proof is essentially the same as in your question.
For a homotopy $F : Y \times [0,1] \to Z$ and $t \in [0,1]$ let us define $F_t : Y \to Z, F_t(y) = F(y,t)$. Now let $H : [0,1] \times [0,1] \to X$ be a path homotopy from $u_0$ to $u_1$. It lifts to a homotopy $\tilde H : [0,1] \times [0,1] \to \tilde X$ such that$\tilde H_0 = \tilde u_0$. Clearly $\tilde H_1$ is a lift of $u_1$. The paths $\phi_i : [0,1] \to X, \phi_i(t) = H(i,t) = H_t(i)$, are constant for $i = 0,1$ (with value $x_0$) and the paths $\tilde \phi_i : [0,1] \to \tilde X, \tilde \phi_i(t) = \tilde H(i,t) = \tilde H_t(i)$, are lifts of $\phi_i$ starting at $\tilde \phi_i(0) = \tilde H_0(i) = \tilde u_0(i)$. Since also the constant paths starting at $\tilde u_0(i)$ are lifts of $\phi_i$, unique path lifting shows that the paths $\tilde \phi_i$ are constant.
This implies $\tilde H_0(i) = \tilde \phi_i(0) = \tilde \phi_i(1)  = \tilde H_1(i)$ for $i = 0,1$. Thus $\tilde H_1$ starts at $\tilde x_0$. By unique path lifting we get $\tilde H_1 = \tilde u_1$. Therefore
$$\tilde u_0(1) = \tilde H_0(1) = \tilde H_1(1) = \tilde u_1(1) .$$
Note that the above theorem says that $\pi_1(x,x_0)$ acts on the fiber $p^{-1}(x_0)$.
