Let $f:S^{2k+1}\rightarrow S^{2k+1}$ be a map that factors through $\mathbb{RP}^{2k+1}$, then what can be said about the about the degree of $f$?
In the even dimensional case, we have that there exists a unique $\phi:\mathbb{RP}^{2k}\rightarrow S^{2k}$ such that $f=\phi\circ \pi$. It follows that $f_*=\phi_*\circ \pi_*$, but since the top homology of $\mathbb{RP}^{2k}$ is trivial we must have that $\pi_*=0$, so $f_*=0$ and the degree of the map must be zero.
In the odd dimensional case the top homology of $\mathbb{RP}^{2k+1}$ is $\mathbb Z$, so not much can be immediately said about the degree of $\pi$, and I am not exactly sure how to actually go about calculating the degree of $\pi$ in this case. Even if I could say something about the degree of $\pi$, I still have that $\phi_*$ is a map $\mathbb{Z}\rightarrow \mathbb{Z}$, so all I could say is that the degree of $f$ is some multiple of the degree of the $\pi$ right?
I also tried looking at the cohomology rings of $\mathbb{RP}^{2k+1}$ and $S^{2k+1}$, but in both caes the ring structures essentially separates the top cohomology from the rest ring. In other words, there are no elements of lower degree that can multiply together to yield a generator of the top cohomology ring, so I don't think I can deduce anything from that...
I think my best bet is then to just calculate the degree of the $\pi_*$, but how exactly do I do that?
From homology you can see quickly that the degree has to be even. Conversely, an even-degree map factors through a degree-2 map, which is turn factors through the projection $S^{m} \to \mathbb{R}P^{m}$ via the map $\mathbb{R}P^{m} \to S^m$ collapsing the $m-1$-skeleton.