I am trying to compute the cohomology for the complex projective space $P_{\mathbb{C}}^2$ which we know is a manifold of dimension $4$. Recall that $P_{\mathbb{C}}^2 = {\mathbb{C}}^3 - \{0\}/\sim$ where the equivalence relation is given by $\omega \sim \eta$ iff there exists $\lambda \in {\mathbb{C}}^*$ such that $\omega = \lambda \eta$. We denote an element $[z_0,z_1,z_2] \in P_{\mathbb{C}}^2$. Now, since $P_{\mathbb{C}}^2$ is connected, we already know $H^0(P_{\mathbb{C}}^2)={\mathbb{R}}$. For the remaining cohomologies we use the Mayer-Vietoris sequence. Indeed, we choose $$U = \{[z_0,z_1,z_2] \in P_{\mathbb{C}}^2 : z_2 \neq 0\},$$and $$V = \{[z_0,z_1,z_2] \in P_{\mathbb{C}}^2 : z_2 = 0, \ \ z_i \neq 0 \text{ for some } i\}.$$
Now, we can identify $U = {\mathbb{C}}^2$ with the map $[z_0,z_1,z_2] \mapsto (z_0/z_2,z_1/z_2)$? Another way of seeing it is that since $z_0 \neq 0$ then any element in $U$ can be identified by $(z_0,z_1,1)$ since any other element can be obtained by multiplication of $\lambda \in {\mathbb{C}}^*$. I am not sure about this identification but I used in the exericse the fact that $H^K(U) = H^k({\mathbb{C}}^2)$.
Also, another identification that I am not really understanding is the one given by $V$ with $P^1_{\mathbb{C}}$ (if anyone could elaborate on these identifications I would appreciate). So that $H^k(V) = H^k(P^1_{\mathbb{C}})$.
Furthermore, this is easy to see that $U \cup V = P^2_{\mathbb{C}}$ but again why does $U \cap V = {\mathbb{C}}^2 - \{0\}$? From here we see that $H^k(U \cup V) = H^k({\mathbb{R}}^{2n}-\{0\})$.
Recall that:
$H^0(P^1_{\mathbb{C}}) = H^2(P^1_{\mathbb{C}}) = {\mathbb{R}}$ with $H^1(P^1_{\mathbb{C}}) = 0$ (this is obtained since $P^1_{\mathbb{C}} = S^2$);
$H^0({\mathbb{C}}^2) = {\mathbb{R}}$ and $H^k({\mathbb{C}}^2) = 0$ for all other $k$;
$H^0(S^3) = H^3(S^3) = {\mathbb{R}}$ with $H^1(S^3) = H^2(S^3) = 0$;
Going for the Mayer-Vietoris sequence, the only surviving terms are $$0 \rightarrow {\mathbb{R}} \rightarrow {\mathbb{R}} \oplus {\mathbb{R}} \rightarrow {\mathbb{R}} \rightarrow H^1(P_{\mathbb{C}}^2) \rightarrow 0,$$ $$0 \rightarrow H^2(P^2_{\mathbb{C}}) \rightarrow 0 \oplus \mathbb{R} \rightarrow 0,$$ $$0 \rightarrow {\mathbb{R}} \rightarrow H^4(P^2_{\mathbb{C}}) \rightarrow 0.$$Using the fact that the sum of the alternating sum of dimensions of an exact sequence of finite dimensional spaces must be zero we obtain $$H^1(P^2_{\mathbb{C}}) = H^3(P^1_{\mathbb{C}}) =,0$$and $$H^0(P^2_{\mathbb{C}}) = H^2(P^2_{\mathbb{C}}) = H^4(P_{\mathbb{C}})^2 = {\mathbb{R}}.$$