Cohomology of complex projective plane

How can I compute Cohomology of complex projective plane $CP^2$? Any magic like the one here?

There are a lot of different approaches depending upon what you know, what precise incarnation of cohomology you're using, and so forth.

Off the top of my head, the following seems easiest: for any $n \in \mathbb{Z}^+$, complex projective $n$-space $\mathbb{C} \mathbb{P}^n$ has the natural structure of a CW-complex with exactly one cell in each degree $2k$ for $0 \leq k \leq n$ and no cells in any other degrees. (You can see this by writing projective $n$-space as affine $n$-space union projective $n-1$ space.) This makes the co/homology work out about as simply as it gets: all the attaching maps are zero, so we have $H_i(\mathbb{C} \mathbb{P}^n, \mathbb{Z})$ is $\mathbb{Z}$ if $0 \leq i \leq 2n$ and $i$ is even and $0$ otherwise. The vanishing of all the odd homology groups also makes the Universal Coefficient Theorem easy to apply, and one gets exactly the same answer for the cohomology groups.

If you wanted to use some other method -- or, if you need to know the ring structure of the cohomology (short answer: it is a truncated polynomial ring) -- then please let me know.

Let $f: \mathbb CP^n \to \mathbb R$ be given by: $$[z_0, \dots, z_n] \mapsto \frac{|z_1|^2 + 2|z_2|^2 + \dots + n|z_n|^2}{|z_0|^2 + |z_1|^2 + \dots + |z_n|^2}$$ One can show that $f$ is a morse function with morse polynomal $P_f(t) = \sum_{k=0}^n t^{2k}$, which satisfies the gap condition and therefore $f$ is a perfect morse function. We conclude $H^k(\mathbb CP^n,\mathbb R) = \mathbb R$ for $k = 0,2,\dots,2n$ and $H^k(\mathbb CP^n,\mathbb R) = 0$ otherwise.