Sorry if I put zero effort in this but I'm really struggling with algebraic topology and when I read an excercise I never know how to begin from.
Let $p$ be the standard projection from the unit circle to the real projective line:
$$p: S_1 \rightarrow \mathbb{PR}^1$$
Show that there cannot be a continuous function $q$ such that $p\circ q$ is the identity map of the real projective line. I know that the two spaces are homeomorphic, that $p$ is a 2-fold covering map that induces an epimorphism between their first foundamental groups (for every base point) $\mathbb{Z}$ however from now on I'm totally lost.
Here the idea is to see that although $\mathbb{R}P^1$ is homeomorphic to $S^1$, the map $p$ wraps $S^1$ twice around itself and so the image of a generator of the fundamental group is sent to twice itself: it can't be surjective.
Now to make this more precise: let $c: S^1\to S^1$ be the squaring function $z\mapsto z^2$. Then $c(z)=c(-z)$ for all $z$, hence $c$ factors through the quotient $S^1\to S^1/(\forall z, z=-z)\to S^1$.
But this quotient is $\mathbb{R}P^1$ up to a canonical homeomorphism so we get that $c$ factors through $p$: $S^1 \to \mathbb{R}P^1\to S^1$. Call the second map $r$: $r\circ p =c$. But now $r$ is injective, continuous, from a comapct set to a separated set: it's thus a homeomorphism onto its image. But $c$ is surjective, hence so is $r$: $r$ is a homeomorphism $\mathbb{R}P^1\to S^1$, that's what I meant when I said that the projection map was essentially wrapping $S^1$ twice around itself.
Now assume $q: \mathbb{R}P^1\to S^1$ satisfies $p\circ q = id_{\mathbb{R}P^1}$. Then $r= r\circ p\circ q = c\circ q$. Let's look at the induced maps in homotopy: $r$ is a homeomorphism so $r_\star$ is surjective. This implies $c_\star$ is surjective (remember that $r_\star = c_\star\circ q_\star$). However, if $[\gamma : t\mapsto e^{2i\pi t}]$ is the generator of $\pi_1(S^1,1)$, then $c_\star([\gamma]) = [\gamma]^2$ and so $c_\star$ can't be surjective !
We have reached a contradiction: such a $q$ can't exist.