Calculating $\mathbb{R}P^1$ fundamental group. Well, I am trying to use the fact that $S^1$'s fundamental groups is free and generated by on element ($\mathbb{Z}$), denoting $\pi_1(S^1) = \langle [\gamma] \rangle$. When $\gamma$ is a loop starts at $(0,1)$ and going clockwise through $S^1$ . Using the quotient map  $q:S^1 \rightarrow \mathbb{R}P^1$ ($x \mapsto \{x,-x\}$) as a covering map, one get that $[\gamma] \mapsto q_*([\gamma]) = [q\circ\gamma]= \cases{\alpha(s) : s\in[0,\pi] \\ \alpha(s-\pi)  : s\in[0,2\pi]}$ when $\alpha$ is the loop in $\mathbb{R}P^1$ which starts at $(0,1)$, and going clockwise till $(0,-1) \sim (0,1)$.
$q\circ \gamma$ is homotopic to $\cases{\alpha(2s) : s\in[0,\pi] \\ \alpha(2s-2\pi)  : s\in[0,2\pi]}$ by the homotopy $F_t(s) = \cases{\alpha(2(1+t)s) : s\in[0,\pi] \\ \alpha((1+t)(s-\pi))  : s\in[0,2\pi]}$ which in turn is homotopic to $\alpha * \alpha$ (just composing a function to change the domain from $[0,2\pi]$ to $[0,1]$.
So one may conclute that $[q\circ \gamma] = [\alpha * \alpha]$ , but I don't succeed in formally proceeding to the conclusion that $\mathbb{R}P^1$ is generated by $\alpha$ which is what I wished to achieve. Any help, and other ideas would be appreciated!  
 A: We know $q$ is a double cover. It follows that $q_*(\pi_1(S^1))$ (which is isomorphic to $\pi_1(S^1)$, since $q_*$ is injective) is an index-2 subgroup of $\pi_1(\mathbb{R}P^1)$. 
Since we know that $\pi_1(\mathbb{R}P^1) \simeq \mathbb{Z}$ $^{(1)}$ and we have that $\alpha^2$ is the image of a generator of $\pi_1(S^1)$, it follows that it generates $q_*(\pi_1(S^1))$ (which is the index-2 subgroup). Therefore, $\alpha$ is a generator of $\pi_1(\mathbb{R}P^1)$. 
$^{(1)}$It is easy to prove that $\mathbb{R}P^1 \simeq S^1$ directly, or we can simply resort to the classification of compact $1$-manifolds (which also says that it must be $S^1$).
In a previous version of this answer, I claimed that from the fact that we had the existence of an index-2 subgroup (isomorphic to $\mathbb{Z}$) alone we could infer that $\pi_1(\mathbb{R}P^1) \simeq \mathbb{Z}$. This is, of course, false. We could have $\pi_1(\mathbb{R}P^1) \simeq \mathbb{Z} \oplus \mathbb{Z}/2\mathbb{Z}$. Indeed, those are the only two possibilities if we know that $\pi_1(\mathbb{R}P^1)$ is abelian (which we do... because $\mathbb{R}P^1$ is $S^1$! Or if you don't want to use that, then we could know that because it is a Lie group, which is almost cheating, but not quite).
One way to see that those are the only two possibilities is to use the fact that we have the exact sequence
$$0 \to \pi_1(S^1) \stackrel{q_*}{\to} \pi_1(\mathbb{R}P^1)\to \pi_1(\mathbb{R}P^1)/q_*(\pi_1(S^1)) \simeq \mathbb{Z}/2\mathbb{Z} \to 0, $$
and thus the possibilities of $\pi_1(\mathbb{R}P^1)$ are restricted by $\mathrm{Ext}(\mathbb{Z}/2\mathbb{Z},\mathbb{Z}) \simeq \mathbb{Z}/2\mathbb{Z}$, thus $\mathbb{Z}$ and $\mathbb{Z} \oplus \mathbb{Z}/2\mathbb{Z}$ are indeed the only possibilities. 
It would be nice to know a way to discard the "possibility" of $\mathbb{Z} \oplus \mathbb{Z}/2\mathbb{Z}$ from the covering map directly for example, not resorting to knowing that $\pi_1(\mathbb{R}P^1)$ is indeed $\mathbb{Z}$ beforehand.
But it is much better to evade all this and use what is in the beginning of this "footnote".
A: If you know $\pi_k(\mathbb{S}^1)$, you know $\pi_k(\mathbb{RP}^1)$!! Because $\mathbb{RP}^1 \cong \mathbb{S}^1$, homeomorphic!!
For yet another way to compute the fundamental group of a space through a covering map of it, check my answer here:
Computation of the fundamental group of the projective plane without Van Kampen theorem.
