Is every map from $\mathbb{R}P^2 \to S^2$ nullhomotopic? I think the answer is no, but I'm not sure. Consider the CW-structure on $\mathbb{R}P^2$ given by one $0$-cell $x$, one oriented $1$-cell $a$ attached to $x$ at both ends, and one $2$-cell whose boundary is glued along the loop $a^2$. Let $X^1 \subset \mathbb{R}P^2$ be the $1$-skeleton. My spidey senses are suggesting the map $\mathbb{R}P^2 \to \mathbb{R}P^2/X^1 \cong S^2$ given by crushing the $1$-skeleton to a point should be homotopically nontrivial. Is this true? If so, how does one prove this?
 A: There is a cofibration sequence 
$$S^1\xrightarrow{2} S^1\rightarrow\mathbb{R}P^2\xrightarrow{q}S^2\rightarrow\dots$$
where $q$ is the map pinching to the top cell. This cofibring leads to an exact Puppe sequence of homotopy sets
$$\dots\leftarrow\pi_1S^2\leftarrow [\mathbb{R}P^2,S^2]\xleftarrow{q^*}\pi_2S^2\xleftarrow{\times 2}\pi_2S^2\leftarrow\dots$$
Here $[\mathbb{R}P^2,S^2]$ is only a set with $\pi_2(S^2)$-action, but the rest of these objects are (abelian) groups. Likewise the map $q^*$ is a map of sets, but the $\times 2$ map is a homomorphism. 
Now since $\pi_1S^2=0$, we identify $[\mathbb{R}P^2,S^2]\cong\pi_2(S^2)/(2\cdot \pi_2(S^2))\cong\mathbb{Z}_2$ as sets with $\pi_2(S^2)$-action. Here we have used that $\pi_2S^2\cong\mathbb{Z}$. In particular there are two homotopically distinct maps $\mathbb{R}P^2\rightarrow S^2$. One is clearly the trivial map, and exactness of the above sequence tells us that the other is $q^*id_{S^2}=q$, since $id_{S^2}$ generates $\pi_2S^2$. Hence the pinch map $q$ is homotopically non-trivial.
At this point we would like to get our hands dirty and see all this in another way. We want to use the Hopf Degree Theorem for non-orientable manifolds, which, as I understand it, states that two maps $f,g:M\rightarrow S^n$ from a boundaryless n-dimensional non-orientable manifold $M$ are homotopic only if they have the same Brouwer-Hopf degree modulo 2.
The real projective plane $\mathbb{R}P^2$ is a 2-dimensional closed non-orientable manifold, which is a quotient of $S^2\subseteq\mathbb{R}^3$ by the equivalence relation $x\sim -x$. We write its points $[x,y,z]=[-x,-y,-z]$, where $(x,y,z)\in S^2$.
There are equivalent definitions of the degree of a map $f:\mathbb{R}P^2\rightarrow S^2$. The first takes the mod 2 fundamental class $[\mathbb{R}P^2]\in H_2(\mathbb{R}P^2;\mathbb{Z}_2)\cong\mathbb{Z}_2$ and studies its image $f_*[\mathbb{R}P^2]\in H_2(S^2;\mathbb{Z}_2)\cong\mathbb{Z}_2$. From this definition it is clear that the trivial map $\ast:\mathbb{R}P^2\rightarrow S^2$ has mod 2 degree 0.
The second definition takes a smooth representative for the homotopy class of $f:\mathbb{R}P^2\rightarrow S^2$, finds a regular value $p\in S^2$ for it, and sets $deg_2(f)=|f^{-1}(p)|\mod 2$. The preimage $f^{-1}(p)$ should be a finite set, so this makes sense.
Now we will calculate $deg_2(q)$, the degree of the pinch map. Recall that the subspace $\{[x,y,0]\}\subseteq \mathbb{R}P^2$ is exactly $\mathbb{R}P^1\cong S^1$. Hence $q$ sends all of this subspace to the basepoint of $S^2$. This is handy, because we have a chart $\phi:U\rightarrow \mathbb{R}^2$ defined over $U=\{[x,y,z]\mid z\ne0\}$ by
$$\phi[x,y,z]=(x/z,y/z).$$
Viewing $S^2$ as the 1-point compactification of $\mathbb{R}^2$ by using stereographic projection, this chart gives the pinch map
$$q[x,y,z]=\begin{cases}(x/z,y/z)\in S^2&z\neq 0\\\infty&z=0 \end{cases}$$
This is clearly smooth, and you can verify that any point away from $\infty$ is regular. In particular, the point $(0,0)\in S^2\cong (\mathbb{R}^2)^+$ is regular. hence
$$deg_2(q)=|q^{-1}(0,0)|=|\{[0,0,1]\}|=1$$
It follows that $\deg_2(q)=1\neq0=deg_2(\ast)$, and therefore that $q$ is homotopically non-trivial.
Returning to the first definition, we see that this is in line with Lord Shark's hint, for $q_*[\mathbb{R}P^2]=1\cdot[S^2]\in H_2(S^2;\mathbb{Z}_2)$.
