Cohomology Group of $CP^2 \wedge CP^2$ Calculate the cohomology group of  $CP^2 \wedge CP^2$
To do this, at first I am trying to calculate the homology group and then use Universal Coefficient Theorem.
To do this, at first I have calculated homology groups of $CP^2 \times CP^2$ and $CP^2 \vee CP^2$ then consider the long exact sequence of the pair $(CP^2 \times CP^2 , CP^2 \vee CP^2).$
But I can't calculate the homology group of $CP^2 \wedge CP^2$ for dimension $< 6.$
 A: $\newcommand\Z{\mathbb{Z}}$
Let $P = \mathbb{CP}^2 \times \mathbb{CP}^2$ be the cartesian product and $S = \mathbb{CP}^2 \vee \mathbb{CP}^2$ be the wedge sum. Then the smash product is $\mathbb{CP}^2 \wedge \mathbb{CP}^2 =  P / S$. Since $S \subset P$ is a sub-CW-complex, $\tilde{H}^n(P/S) \cong H^n(P,S)$.
It is well known that $H^*(\mathbb{CP}^2) = \mathbb{Z}[x]/(x^3)$. Thus, respectively by Mayer–Vietoris and Künneth:
$$H^*(S) = \Z[x,y]/(x^3, y^3, xy) \\
H^*(P) = \Z[x,y]/(x^3, y^3)$$
where $x$ corresponds to the first factor and $y$ to the second.
The long exact sequence of the pair $(P,S)$ in cohomology gives:
$$0 \to H^0(P,S) \to H^0(P) \to H^0(S) \to H^1(P,S) \to \dots\\ \to H^n(P,S) \to H^n(P) \to H^n(S) \to H^{n+1}(P,S) \to \dots$$
Since $\dim S = 4$, $H^n(S) = 0$ for $n > 4$, thus for $n > 5$ you find $H^n(P,S) \cong H^n(P)$. And in low dimensions you have:
$$0 \to H^0(P,S) \to \Z \xrightarrow{=} \Z \to H^1(P,S) \to 0 \to 0 \to H^2(P,S) \to \Z x \oplus \Z y \xrightarrow{=} \Z x \oplus \Z y \to H^3(P,S) \to 0 \to 0 \to H^4(P,S) \to \Z xy \oplus \Z x^2 \oplus \Z y^2 \to \Z x^2 \oplus \Z y^2 \to H^5(P,S) \to 0 \to 0 \to H^6(P,S) \to \Z x^2y \oplus \Z xy^2 \to 0 \to H^7(P,S) \to 0 \to 0 \to H^8(P,S) \to \Z x^2y^2 \to 0 \to 0 \to \dots$$
So finally:
$$H^n(\mathbb{CP}^2 \wedge \mathbb{CP}^2) = \begin{cases}
\Z & n = 0 \\
0 & n = 1,2,3 \\
\Z & n = 4 \\
0 & n = 5 \\
\Z \oplus \Z & n = 6 \\
0 & n = 7 \\
\Z & n = 8
\end{cases}$$
