Intersection of two open sets in the projective plane

Question

I want to compute the cohomology groups of the real projective plane, $P^2$, using Mayer Vietoris exact sequence. Now $H^0(P^2)=\mathbb{R}$, $H^2(P^2)=0$ being $P$ not orientable, so my problem really is compute $H^1(P^2)$. In order to do that, I choose $U_i=\{(x_1:x_2:x_3)\in P^2|x_i \neq 0\}\quad i=1,2,3$, then I set $U=U_1=\mathbb{R^2}$ and $V=U_2\cup U_3=P^2-\{(1:0:0)\}=P^1$. And that's ok. But who is $U\cap V$?

Answer 1

$U \cap V$ consists of all triples whose first coordinate is nonzero, but whose second and third coordinate are not both zero.

A point $(x_1 : x_2 : x_3)$ corresponds to the points $\pm s(x_1, x_2, x_3)$ in $S^2$, where $s = 1/\sqrt{x_1^2 + x_2^2 + x_3^2}$.

Now look at the portion $V_1 = \{(a, b, c) \in S^2 : a > 0 \}$. This is in bijective correspondence with $U_1$ by the map $(a, b, c) \to {(a:b:c)}$, so I'm going to look at $V_1$ instead of $U_1$. I'm going to also place my coordinates so that $(1,0,0)$ is the north pole, and hence $V_1$ is the northern hemisphere.

All points of $V_1$ correspond to points of $U^2 \cup U^3$ (i.e., either 2nd coord or 3rd coord is nonzero) except the point $(1, 0, 0)$. So as seen in $V_1$, the set $U \cap V$ is the (open) northern hemisphere minus the north pole. This admits an obvious deformation retraction onto any northern line of latitude, such as the circle defined by $a = 1/2$.

So it's got the homology/cohomology groups of a circle.