Understanding certain deformation retractions I'm trying to prove the following:
$1)$ Let $p:S^1\to S^1$ be the covering $p(z)=z^n$, $z\in S^1\subset \mathbb{C}$. Show that there does not exist a continuous map $q:S^1\to S^1$ such that $p\circ q$ = id$_{S^1}$.
$2)$ Let $\pi:S^n\to \mathbb{RP}^n$ be a projection. Show that there does not exist a continuous map $\sigma:\mathbb{RP}^n\to S^n$ such that $\pi \circ \sigma =$ id$_{\mathbb{RP}^n}$
$3)$ Show there is no retraction from $r:\mathbb{RP}^2\to \mathbb{RP}^1$
$\textbf{My attempt:}$ 
For $1)$, I'm not exactly sure how to proceed. 
For part $3)$ my argument is as follows:
The inclusion map $i$ induces a homomorphism $\pi(i):\pi(Y,x_0)\to \pi(X,x_0)$ such that:
$$\pi(r\circ i) = \pi(r)\circ \pi(i) = id_{\pi_1(Y)}.$$
In particular, the map is injective. However,
$$\pi(\mathbb{RP}^2, x_0) :\to\pi(\mathbb{RP}^2, x_0)$$
is not injective due to the fact that $\pi(\mathbb{RP}^2, x_0)=\mathbb{Z}/2$ and $\pi(\mathbb{RP}, x_0)$ = $\mathbb{S}^1$
For part $2)$ there is similar argument but I'm having trouble putting the pieces together.
 A: For $1, n>1$, $p$ is a covering,  this implies that if $q$ exists, it is open, thus its image is open (since $p$ is a local homeomorphism, $q$ is a local homeomorphism) and closed since $S^1$ is compact, so $q$ must be surjective, $p$ is not injective. $p(x)=p(y), x\neq y$ there exist $u,v$ with $q(u)=x, q(v)=y$ implies that $p(q(u))=p(q(v))$ contradiction.
For $2, n>1$ remark that $\pi$ is a local homeomorphism since it is a covering, this implies that if $q$ exists, it is open, thus its image is open and closed since $\mathbb{R}P^n$ is compact thus it is $S^n$, apply the idea used in $1$.
A: 1) By taking degrees, we have
$$\mathrm{deg}(p)\cdot \mathrm{deg}(q)=1. $$
Thus, $\mathrm{deg}(p)=\pm 1$, and then $n$ must be $1$ or $-1$. These cases are indeed possible, so the question is wrong as written.
2) If $n=1$, $\pi$ has degree $2$, and a similar reasoning as in item 1) yields that such a map is impossible. (This can be generalized to deal with any $n$ if you know the degree mod $2$ of a map between compact manifolds: $\pi$ has $0$ degree mod $2$ regardless of $n$, and thus  the same reasoning holds again.)
If $n>1$, the induced map $\sigma_{\#}$ on the fundamental group is zero, since $S^n$ is simply connected. By functoriality, $(\pi \circ \sigma)_{\#}$ is also zero, thus it is impossible for it to be the identity since $\pi_1(\mathbb{R}P^n)$ is non-zero.
3) You solved it.
