Cohomology class of the diagonal (${\epsilon }_{\Delta }$)

Let $M$ be a closed oriented $n$-manifold, let $e_{p,i}$ be a homogeneous basis of $H^p(M;\mathbb{Q})$, and choose its Poincare-dual basis $e_{p,i}^{\ast }\in H^{n-p}(M;\mathbb{Q})$ by $⟨e_{p,i}{e}_{p,j}^{\ast },[M]⟩={\delta }_{ij}$. With the product-orientation convention, the Poincare dual of the diagonal is

The cohomology ring of the product of two real projective planes is ${\mathbb{F}}_2[x,y]/(x^3,y^3)$, with $|x|=|y|=1$. Pairing against $x^2$, $xy$ and $y^2$ identifies the Poincare dual of the diagonal as $x^2+xy+y^2$. For a self-map $f$ with $f^{\ast }x=\epsilon x$, the graph pullback evaluates on the mod-two fundamental class as $1+\epsilon +{\epsilon }^2=1$. Thus no graph of a map on the whole projective plane can avoid the diagonal. The argument uses a relative Thom class and does not assume transverse intersections.

For the graph map $(1,f):M\to M\times M$, the cohomology class of the diagonal satisfiesIf $f$ is fixed-point-free, its graph misses the diagonal, so the pullback vanishes. This proves the Lefschetz fixed-point theorem for closed oriented manifolds.