To give some context these two exercises come from a past exam of my undergraduate course in differential geometry, course which followed loosely Milnor's "Topology from the differentiable viewpoint". I have no knowledge about the theory of Homology/Cohomology and only basic notions in algebraic topology. The exercises are the following.
Prove or give a counterexample for the following statements:
$i$) all smooth maps $h:S^2\rightarrow \mathbb{RP}^2$ have $\deg_2(h)=0$ ($\deg_2$ being the degree mod $2$).
$ii$) There exist a smooth map $r:S^3\rightarrow\mathbb{RP}^3$ such that $\deg(r)=6$
There are two facts which I thought could be useful here, the first being that $\mathbb{RP}^n$ is homeomorphic to $ {S^n}/{v\sim -v} $ , while the second one is that we can construct smooth maps of arbitrary degrees between $S^n$ and itself, I'm not aware of a standard construction of this but it seemed reasonable to me that we could construct a smooth vector field on $v:\mathbb{R}^{n+1}\rightarrow\mathbb{R}^{n+1}$ with $k$ non degenerate zeroes $\{x_1,...,x_k\}$ contained in $B_1(0)$ such that $\det{Jv}_{x_j}\gt 0$ and then consider the map $\widehat{v}=\frac{v}{||v||}$ restricted on the set ${\partial B_1(0)}\sim S^n$. This would give us a smooth map from $S^n$ to itself with degree $k$. Equivalently, we can require the opposite sign for the determinant of the Jacobian of $v$ if we wish to obtain a $-k$ degree smooth map.
With this in mind, for the second point, if we constructed a smooth map $R:S^3\rightarrow S^3$ such that $\deg{R}=3$ we could then consider $q:S^3\rightarrow \mathbb{RP}^3$ given by $q(x)=\left[x\right]$ the equivalence class of $x$, then $\deg{(q\circ R)}=\deg{R}\cdot\deg{q}=3\cdot2=6$, of course this argument requires $q$ to be smooth.
What can we say about point $\left(i\right)$? Are there flaws in my reasoning regarding point $\left(ii\right)$?