Calculation of de Rham complex for real projective space

Question

I've been studying de Rham cohomology and the use of the Mayer-Vietoris sequence to compute the de Rham complex for the real projective space $\mathbb P^d$. First, I divided $\mathbb P^d$ in two open sets: a set $U=\{[x^0,...,x^d]|x^d\neq0\}$ and a set $V=\mathbb P^d\setminus\{[0,...,0,1]\}$. Then, I proved that $U$ has the homotopy type of a disk (so it is contractible to a point), $V$ has the homotopy type of $\mathbb P^{d-1}$ (so I will have to use induction) and $U\cap V$ has the homotopy type of $S^{d-1}$. I know the classes for the spheres.

I already know the result, $H^k(\mathbb P^d)\simeq\mathbb R$ if $k=0$ or $k=d$ with $d$ odd, $0$ otherwise. I want to prove that by induction on $d$. I think it should be easy enough to proceed with the inductive step.

What I'm not managing to do is the first step. For all $d$ we have $H^0(\mathbb P^d)\simeq\mathbb R$, as all projective spaces are connected. For $\mathbb P^1$, well, this is diffeomorphic to $S^1$, so it has its same homotopy type and its same classes, namely $H^0(\mathbb P^1)\simeq H^1(\mathbb P^1)\simeq\mathbb R$. Now I want to try the first non trivial case, $d=2$: in this case I have the Mayer-Vietoris sequence (here with $p$ I indicate a manifold with only a point) $$ 0\to H^0(\mathbb P^2)\to H^0(p)\oplus H^0(\mathbb P^1)\to H^0(S^1)\to\\ \to H^1(\mathbb P^2)\to H^1(p)\oplus H^1(\mathbb P^1)\to H^1(S^1)\to\\ \to H^2(\mathbb P^2)\to H^2(p)\oplus H^2(\mathbb P^1)\to H^2(S^1)\to 0. $$ The sequence actually stops earlier, as $H^2(p)\oplus H^2(\mathbb P^1)\simeq H^2(S^1)\simeq 0$. The problem is that there are two classes that I don't know, $H^1(\mathbb P^2)$ and $H^2(\mathbb P^2)$. As $\mathbb P^2$ is not orientable, I can't use Poincarè's lemma. I know all other cohomology classes: $H^0(\mathbb P^2)\simeq H^0(S^1)\simeq H^1(p)\oplus H^1(\mathbb P^1)\simeq H^1(S^1)\simeq \mathbb R$, $H^0(p)\oplus H^0(\mathbb P^1)\simeq \mathbb R^2$. As no one of those is vanishing, I'm not able to isolate the unknowns between zeroes and compute the dimensionality though alternating sum of dimensions.

But I have discovered something: the map bringing $H^0(S^1)$ in $H^1(\mathbb P^2)$ must have rank $0$. This I obtained by just noting that, for every object in the sequence (but the starting and finishing object), the dimension is equal to the sum of the ranks of the map that has range in the object and the map that has domain in the object (and the first map is injective). This also restricts the dimension of $H^1(\mathbb P^2)$ to be $0$ or $1$, and the dimensions of $H^1(\mathbb P^2)$ and $H^2(\mathbb P^2)$ are zero. Only, I don't know how to conclude from that that the dimensions are zero.

It is my first exercise with that stuff, so I may be missing something very obvious.

How can I continue from here? The inductive step seems to be easy, I'm stuck on the starting step.

EDIT: I've found answers using some other divisions of $\mathbb P^2$. I'd like to know if an answer is possible from my splitting.

Answer 1

I have found a solution, so I'm posting an answer to my own question.

There is a theorem (that I had to prove in a previous homework) stating that, if a connected manifold $M$ of dimension $d$ is not orientable, then $H^d_c(M)\simeq 0$.

With this, it becomes quite trivial. As we know, $\mathbb P^d$ is connected, and it is not orientable for $d$ even. Our case, $\mathbb P^2$, is one of such cases. Furthermore, all projective spaces are compact, so the compact cohomology and the standard cohomology are the same. We can then state $H^2(\mathbb P^2)\simeq 0$. From this, $H^1(\mathbb P^2)\simeq 0$ follows by counting alternating dimensions.

I guess there is no way to obtain that result without invoking some external theorem or chaning my division of $\mathbb P^d$.