Compute the cohomology of the real projective plane $\Bbb R P^2$ This is the problem from Tu's introduction to maniflods,I'm tring to learn some basic de Rham cohomology from the book.(Actually I haven't formally learned some algebraic topology,so some advanced knowledge in your answer may be not familiar to me)
My attempt:choose the punctured plane $\Bbb R P^2-\{p\}$ and a small disc containing $p$ to cover the manifold,we use $U$ and $V$ to represent them respectively.
So we may apply Mayer-Vietoris sequence and we get the chain(we omit the $dR$ to simplify the notation):$$0\to H^0(\Bbb R P^2)\to H^0(U)\oplus H^0(V)\to H^0(U\cap V)$$$$\to H^1(\Bbb R P^2)\to H^1(U)\oplus H^1(V)\to H^1(U\cap V)\to....$$
Since $\Bbb R P^2-\{p\}$ and $\Bbb R P^1$ have the same homotopy type and $\Bbb R P^1$ is homeomorphic to $S^1$,so we get $H^*(U)\simeq H^*(S^1)$.
We denote that $U\cap V$ is connected and has the homotopy of $S^1$,we may get$$H^k(U\cap V)=H^k(S^1)=\begin{cases}
\Bbb R,  & \text{if $k$=0,1,} \\
0, & \text{if $k$ $\gt$1. }
\end{cases}$$
Also,$V$ is a disc,so by using poincare lemma we get $$H^k(V)=\begin{cases}
\Bbb R,  & \text{if $k$= 0,} \\
0, & \text{if $k$ $\gt$0. }
\end{cases}$$
As a consequence,we get the equivalent sequence:$$0\to H^0(\Bbb R P^2)\to \Bbb R\oplus \Bbb R\to \Bbb R\to H^1(\Bbb R P^2)\to \Bbb R\to \Bbb R$$
So I don't know whether my inference is true or not,and if it's true,can you tell me what's the next step?
 A: This looks (at a quick glance) fine to me. You have
$$0\to H^0(\Bbb R P^2)\to H^0(U)\oplus H^0(V)\to H^0(U\cap V)\to H^1(\Bbb R P^2)\to H^1(U)\oplus H^1(V)\to H^1(U\cap V)\to\ldots$$
$$0\to H^0(\Bbb R P^2)\to^a \Bbb R\oplus \Bbb R\to^b \Bbb R\to^c H^1(\Bbb R P^2)\to^f \Bbb R\to^g \Bbb R$$
but now it's time to look at what the maps actually are. I've added (misplaced) names for them so that I can talk about them. The map $b$ is roughly defined by $(x, y) \mapsto x - y$, so its kernel (which is $H^0$) is just $\Bbb R$. Because $b$ is surjective (just hold $y = 0$ in my previous formula!) the map $c$ sends all of $\Bbb R$ to $0$, so we can just split the sequence there, to get
$$0 \to^c H^1(\Bbb R P^2)\to^f \Bbb R\to^g \Bbb R$$
That means that $H^1$ is either $0$ or $\Bbb R$, depending on whether $g$ has rank 1 or rank 0. 
Now $U$ is just a Mobius band, which retracts to $S^1$ and $U \cap V$ is its boundary circle (roughly). On the first, the 1-form $d\theta$ is a generator. On the second, $d \phi$ (where $\phi$ is the angular coordinate on the boundary circle) is a generator. Integrating $d\theta$ over the centerline of $U$ gives $2\pi$; integrating it over the boundary circle gives $4\pi$. So if $i$ is the inclusion map from $U \cap V$ to $U$, we see that $i^{*} d\phi = \frac12 d\theta$ (I think I have that right...the $\frac12$ might be a $2$, but either way, it's a nonzero constant!) Hence the map between the last two items you've written, which is just $(i^{*}, 0)$, is multiplication by some nonzero constant, i.e., $g$ has rank $1$, so its kernel is $0$, and we conclude that $H^1(\Bbb R P^2) = 0$. 
