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Give Complex projective space its complex orientation. In the homology cross product basis of $H_{2n}({\mathbb{CP}}^n\times {\mathbb{CP}}^n;\mathbb{Z})$, the diagonal has the displayed class. If $a,b$ are the two normalized degree-two hyperplane classes, its Poincare dual is ${\sum }_{j=0}^n{a}^j{b}^{n-j}$. Indeed every degree-$2n$ monomial $a^r{b}^{n-r}$ restricts on the diagonal to the top power of the hyperplane class, whose evaluation on the fundamental class is one. These evaluations determine all coefficients.

Work over an algebraically closed field of characteristic different from two. A smooth projective plane curve $B\subset {\mathbb{P}}^2$ of degree $2d$ is the zero set of a global section $s$ of $L^{\otimes 2}$, where $L={\mathcal{O}}_{{\mathbb{P}}^2}(d)$. The finite morphism $\pi :X\to {\mathbb{P}}^2$ defined locally by $w^2=s$ is a double cover, branched along $B$. The Jacobian criterion shows that $X$ is smooth: away from $B$ the derivative in $w$ is nonzero, while on $B$ some derivative of $s$ is nonzero.

The ramification divisor $R$ satisfies ${\mathcal{O}}_X(R)={\pi }^{\ast }L$. The determinant of the differential giveswhere $H$ is the hyperplane class. Thus $K_X$ is ample for $d\ge 4$, so $X$ is a surface of general type; its self-intersection number is $K_X^2=2(d-3{)}^2$. A smooth octic branch curve gives $K_X^2=2$. For $d=3$, the same formula instead gives a trivial canonical line bundle of a smooth variety, illustrating why the branch degree changes the classification. The assumption on characteristic is needed for this local double-cover construction and its smoothness argument.

The cellular homology of complex projective space gives one integral generator $[{\mathbb{CP}}^j]$ in each degree $2j$, for $0\le j\le 2$, and zero in odd degrees. Since these groups are free, the Künneth theorem gives the homology cross product basisThus 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 $⟨h^2,[{\mathbb{CP}}^2]⟩=1$, and write $a=p_1^{\ast }h$, $b=p_2^{\ast }h$. The cohomology ring of complex projective space and the Künneth theorem give $H^{\ast }({\mathbb{CP}}^2\times {\mathbb{CP}}^2;\mathbb{Z})=\mathbb{Z}[a,b]/(a^3,b^3)$. If $d$ is the diagonal embedding, then $d^{\ast }a=d^{\ast }b=h$. 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 givesEquivalently its Poincare dual is $a^2+ab+b^2$. The signs are positive with the complex and product orientations.

A cohomology operation is stable when it commutes with the suspension isomorphism on reduced cohomology:Products of positive-degree reduced classes on a suspension vanish, but stable operations can still connect its nonzero cohomology classes. The Steenrod squares and, at odd primes, Steenrod reduced powers are examples.

Let $x\in H^2({\mathbb{CP}}^2;{\mathbb{F}}_2)$ be the hyperplane class modulo two. It satisfies ${\mathrm{Sq}}^{2}x=x^2\ne 0$. Stability gives ${\mathrm{Sq}}^2(\sigma x)=\sigma (x^2)\ne 0$ on $\Sigma {\mathbb{CP}}^2$. The suspension of $S^2\vee S^4$ has the same additive groups and vanishing positive-degree products, but its corresponding degree-three class has zero ${\mathrm{Sq}}^2$, by pullback from $S^3$. Thus these suspensions have different cohomology operations even though their additive groups and cup products agree.^[ref]