Is the map $q: \mathbb{RP}^n \to \mathbb{RP}^n / \mathbb{RP}^{n-1}$ nullhomotopic? I've been exploring maps from $\mathbb{RP}^n$ to $S^n$ lately, and I came across the quotient map $q:\mathbb{RP}^n \to \mathbb{RP}^n/\mathbb{RP}^{n-1} \approx S^n$. I haven't been able to find any information about this map; is it nullhomotopic? Is there an easy way to find its degree (as in, the homomorphism $q_{\ast}: \pi_{n}(\mathbb{RP}^n) \to \pi_{n}(S^{n})$)?
I'm sorry for not being able to post any progress, I just have no idea how to approach this. 
Maybe this could be useful: I know that, when we look at it as a map of topological pairs, $q: (\mathbb{RP}^n, \mathbb{RP}^{n-1}) \to (\mathbb{RP}^n/\mathbb{RP}^{n-1}, \ast)$, I know that $q_{\ast}: \pi_{k}(\mathbb{RP}^n, \mathbb{RP}^{n-1}) \to \pi_{k}(\mathbb{RP}^n/\mathbb{RP}^{n-1}, \ast)$ is an isomorphism for $k \leq n-1$, and an epimorphism for $k=n$.
I guess my question reduces to:
If a map $f:X \to Y$ is nullhomotopic, $A \subseteq X$, $B \subseteq Y$, does that necessarily mean that $f_{\ast}:\pi_{k}(X, A, a_{0}) \to \pi_{k}(Y, B, b_{0})$ is zero?
If the answer to my sub-question is negative, how do I find the solution in this specific case?
**EDIT: ** I think that the answer to my sub-question is "no", because $f$ being nullhomotopic doesn't imply that $f$ is nullhomotopic as a mapping of topological pairs. But I'm not sure if I could find a counterexample.
 A: To prove that $q$ is not nullhomotopic, notice that you have a cofiber sequence $\mathbb{R}P^{n-1}\to \mathbb{R}P^n\to S^n$ so that for any pointed CW-complex $Z$ the following is an exact sequence of pointed sets $\langle S^n, Z\rangle \to^{q^*} \langle \mathbb{R}P^n, Z\rangle \to\langle\mathbb{R}P^{n-1},Z\rangle$; where $\langle, \rangle$ denotes the set of pointed homotopy classes of pointed maps. For $Z$ simply-connected, this is the same as the set of homotopy classes $[-,Z]$.
So if $q$ were nullhomotopic, then $q^*$ would be the trivial map, sending everything to the basepoint, thus for simply-connected $Z$ (I have to make this assumption, otherwise I would have to assume that $q$ is pointed-nullhomotopic), any map $\mathbb{R}P^n\to Z$ which restricts to a nullhomotopic map $\mathbb{R}P^{n-1}\to Z$ is nullhomotopic. 
Right, but we can prove that this isn't true [mhm there should be some simpler counterexample, I seem to be forced to use some machinery but I think there should be simpler]. Say $n\geq 2 $ for starters, in which case $S^n$ is simply connected. 
Then $[\mathbb{R}P^n,S^n]\simeq H^n(\mathbb{R}P^n)\neq 0$ because $\mathbb{R}P^n$ is $n$-dimensional, and $S^n$ is the $n$th Postnikov stage for a $K(\mathbb{Z},n)$, and similarly $[\mathbb{R}P^{n-1},S^n] = H^n(\mathbb{R}P^{n-1}) = 0$, so there is a non nullhomotopic map $\mathbb{R}P^n\to S^n$, but none $\mathbb{R}P^{n-1}\to S^n$, which contradicts what we said before. 
As for the question of the degree, we know that the projection $S^n\to \mathbb{R}P^n$ induces an isomorphism on $\pi_n$, so we're just looking at the degree of a map $S^n\to S^n$ and we have to figure out what it is. But if you look closely, you see that killing $\mathbb{R}P^{n-1}$ is something you could have done above, in $S^n$ already : it corresponds to the pinch map; so you get a wedge of two spheres, and you can just follow up by identifying them in the right way. Now we have to see what "the right way" is: is it to identify the sphere to itself without twist, or do we have to twist ? In the first case, by definition of addition on $\pi_n$, this will just be adding things, so the degree would be $2$; and in the other case the degree would be $0$. 
Now to find out which one it is you either have to be cleverer or go through all the identifications we made to see if it's actually $2$ or $0$. I suspect it should be $2$, but right now I don't have a clever way of seeing that (perhaps using the fact that it's not nullhomotopic) and I don't have the courage to go through all the identifications myself to see which one it is (but for sure if you are motivated enough there is no inspiration there, you just have to go through the details and after a finite -maybe long- time you'll get there)
ADDED : Here's a way to get the degree, by using the Hurewicz map; let $h_1$ denote said Hurewicz map for $\mathbb{R}P^n, h_2$ for $S^n$. Now we can prove with a bit of work that $H_n(\mathbb{R}P^n)/h_1(\pi_n(\mathbb{R}P^n)) \simeq H_n(K(\mathbb{Z/2},1))$. Now note that $\mathbb{R}P^\infty $ is a $K(\mathbb{Z/2},1)$ and $H_n(\mathbb{R}P^\infty)$ is $\mathbb{Z/2}$ if $n$ is odd, $0$ else. So if $n$ is odd, the Hurewicz map has image precisely $2\mathbb{Z}$ (and the Hurewicz morphism is $+-(n\mapsto 2n)$). 
From this we get that the degree is exactly twice the "homological degree" of $f$, so we just have to get said homological degree. But for $n$ odd still, $H_{n-1}(\mathbb{R}P^n) = 0$ so $H^n(\mathbb{R}P^n)\simeq \mathbb{Z}$ and the homological degree is the same as the cohomological degree. But now we see that $[S^n,S^n]\to[\mathbb{R}P^n,S^n]$ is not a constant map : it sends the identity to $q$, which we know from above is not nullhomotopic. But this map is, again for dimension reasons $H^n(S^n)\to H^n(\mathbb{R}P^n)$, and the fact that it's nonconstant as a map of sets means it's nonzero as a map of groups : the degree is nonzero, therefore the overall degree must be $2$ (because we saw it was $2$ or $0$)
If $n$ is even, $H_n(\mathbb{R}P^n)=0$, and $h_2$ is an isomorphism, from which it follows by naturality of the Hurewicz map, that the degree is $0$.
In conclusion: $\deg(q) = 0$ if $n$ is even, $2$ otherwise.  
For $n=1$, you have to say what you mean more specifically by $\mathbb{R}P^0$, but in any case it should be easy to deal with since $\mathbb{R}P^1$ is just $S^1$ so you just have to compute an element of the fundamental group. 
Then for your claim about relative homotopy groups, as you see the problem is that the map may not be nullhomotopic from $A$ to $B$. Take for instance the identity $\mathbb{R^3\to R^3}$, and $A=B=S^2$. Then you if the map were zero on $\pi_3$, then $\pi_3(\mathbb{R}^3,S^2)$ would be $0$. However, we have an exact sequence $pi_3(\mathbb{R}^3)\to \pi_3(\mathbb{R^3},S^2)\to \pi_2(S^2)\to \pi_2(\mathbb{R}^3)$ which of course leads to an isomorphism $\pi_3(\mathbb{R}^3,S^2)\to \pi_2(S^2)$
A: I think I can elaborate on Max's calculation of the degree. Like he said we can just look at the composition of the projection with the quotient map and calculate the degree of that.
The degree of this map can be calculated through the homotopy groups, homology groups, or cohomology groups. Here we use the latter:
First we inspect what $q^*$ does to the cohomology. Thinking of the map as one from $[S^n , S^n] \rightarrow [\mathbb{R}P^n,S^n]$ we see it is nontrivial since the image of the identity of $S^n$ is $q$ which is not nullhomotopic.
The next map we look at is $p^* : [\mathbb{R}P^n, S^n] \rightarrow [S^n, S^n]$. The image of this map is $0$ for $n$ even and the even degree maps for $n$ odd (because that is how self maps sending antipodal points to the same point work).
Looking at the composition of these two we can see the degree is $0$ for $n$ even and $2$ for $n$ odd. The latter is by Max's observation that the only two options for the degree are $0$ and $2$ and it can't be $0$. Alternatively, if you could prove that $q$ generates the cohomology of $\mathbb{R}P^n$ then you could immediately say the degree is $2$ in odd dimensions.
