Homotopy classes of maps from projective plane to projective plane Maybe I should think a bit longer, but are there more than two homotopy classes of maps $\mathbb{RP}^2\rightarrow \mathbb{RP^2}$? I am interested in both based and unbased maps. 
 A: There are at least $\Bbb Z$-many unbased homotopy classes of maps.
First, an odd map $S^2 \to S^2$ (that is, one such that $f(-x) = -f(x)$) descends to a map $\Bbb{RP}^2 \to \Bbb{RP}^2$. I claim that the degree of the map on the level of $S^2$ is a homotopy invariant of the map on the level of $\Bbb{RP}^2$. For pick a homotopy $f_t: \Bbb{RP}^2 \times I \to \Bbb{RP}^2$. By assumption $f_0$ came from an odd map $\tilde f_0: S^2 \to S^2$. I claim that there is a lift $\tilde f_t: S^2 \times I \to S^2$ such that all of the $\tilde f_t$ are odd.
This is pretty easy: just pick a lift! I claim that any lift is automatically odd. For if $\tilde f_t$ lifts $f_t$, we necessarily have $\tilde f_t(\{x,-x\}) = \{x,-x\}$; if this is sufficiently close to an odd map, then $\tilde f_t(x)$ must be close to $-x$; so $\tilde f_t(x)$ must actually be $-x$.
So any map that descends from an odd map $S^2 \to S^2$ is only homotopic to maps that descend from odd maps, and the homotopy class of the odd map is a homotopy invariant of the map $\Bbb{RP}^2 \to \Bbb{RP}^2$.
In particular, its degree is determined by the odd map up above. Now all you need to know is that there are odd maps of arbitrary odd degree. I rather believe I once proved this but I don't remember the construction right now. I'll edit it in if I remember it.
The same thing works for even maps - but for $S^2$ an even map must be degree 0. The lift of a map $\Bbb{RP}^2 \to \Bbb{RP}^2$ is either even or odd, so we've now classified all of them.
A: Too long for a comment:
@MikeMiller: you seem to suggest that the maps with even lift are homotopically trivial on $\mathbf P^2(\mathbf R)$, which is not the case, I think: let $\pi$ be the quotient map $S^2\rightarrow \mathbf P^2(\mathbf R)$ and let $g\colon\mathbf P^2(\mathbf R)\rightarrow S^2$ be the map that contracts the $1$-cell to a point. Let $\tilde f$ be the composition $g\circ \pi$. It is clearly an even map from $S^2$ to itself.  Allthough it is of degree $0$, it is not homotopically trivial between all even maps from $S^2$ to itself. Indeed, otherwise $g$ would be homotopically trivial as well, which it is not; $g$ is the unique  homotopically nontrivial map from $\mathbf P^2(\mathbf R)$ to $S^2$. It follows that the map $f$ from $\mathbf P^2(\mathbf R)$ to itself induced by $\tilde f$, i.e., $f=\pi\circ g$, is homotopically nontrivial. 
A: Just to supplement Mike's wonderful answer with the existence of the odd odd degree maps(this didn't fit into a comment). 
See $S^2=\mathbb{R}/2\pi\mathbb{Z}\times [-1,1]/(x,\pm 1)\sim(y,\pm 1)$ (i.e. the suspension of $S^1$) . The map $f:S^2\rightarrow S^2$ being odd means
$$
f(\theta+\pi,-t)=(f_1(\theta,t)+\pi,-f_2(\theta,t)).
$$
Then a degree $(2n+1)$ map is just $f(\theta,t)=((2n+1)\theta,-t)$. This is odd
$$
f(\theta+\pi,-t)=((2n+1)\theta+(2n+1)\pi,t)=((2n+1)\theta+\pi,t)=(f_1(\theta,t)+\pi,-f_2(\theta,t)).
$$
Easier: Take $S^2\subset \mathbb C\times \mathbb R$ and define 
$f(z,t)=(z^{2n+1},t)$. Then $f(-z,-t)=( (-z)^{2n+1},-t)=-(z^{2n+1},t)$.
A: Just an addition to this nice question. Let $f:\mathbb P^2(\mathbb R)\to\mathbb P^2(\mathbb R)$ have an even lifting $\tilde f:\mathbb S^2\to\mathbb S^2$. Then this lifting descends to a map $g:\mathbb P^2(\mathbb R)\to \mathbb S^2$. Now here we know that the homotopy class of $g$ is given by its degree mod $2$. If it is $0$, then $g$ is nullhomotopic; if it is $1$, then $g$ is
homotopic to Huisman's mapping that can be realized as follows
$$
g(x_0:x_1:x_2)\mapsto\tfrac{1}{\|x\|^2}(2x_0^2-\|x\|^2,2x_0x_1,2x_0x_2).
$$
The even lift $\tilde f$ has always degree $0$, hence it is nullhomotopic. In case $g$ is not, we have an example of a homotopy of even maps (from $\tilde f$ to a constant map) which necessarily fails to be an even map (because it cannot descend to $\mathbb P^2(\mathbb R)$.
