= Solution
The <cellular homology of complex projective space> gives one integral generator $[\mathbb{CP}^j]$ in each degree $2j$, for $0\le j\le2$, and zero in odd degrees. Since these groups are free, the <Künneth theorem> gives the <homology cross product> basis
$$
\boxed{[\mathbb{CP}^i]\times[\mathbb{CP}^j],\qquad0\le i,j\le2,\qquad\text{degree }2(i+j).}
$$
Thus the ranks in degrees $0,2,4,6,8$ are $1,2,3,2,1$, respectively, and every other <integral homology> group vanishes.
Let $h$ be the normalized <hyperplane class>, with $\langle h^2,[\mathbb{CP}^2]\rangle=1$, and write $a=p_1^*h$, $b=p_2^*h$. The <cohomology ring of complex projective space> and the <Künneth theorem> give $H^*(\mathbb{CP}^2\times\mathbb{CP}^2;\mathbb Z)=\mathbb Z[a,b]/(a^3,b^3)$. If $d$ is the <diagonal embedding>, then $d^*a=d^*b=h$. Hence each of $a^2,ab,b^2$ evaluates as one on the diagonal's <fundamental class>. These three classes are dual to the three degree-four product generators, so the <diagonal class of complex projective space> gives
$$
\boxed{[\Delta]=[\mathbb{CP}^2]\times[\mathrm{pt}]+[\mathbb{CP}^1]\times[\mathbb{CP}^1]+[\mathrm{pt}]\times[\mathbb{CP}^2].}
$$
Equivalently its <Poincare dual> is $a^2+ab+b^2$. The signs are positive with the complex and product <orientations>.
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