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Use cohomology with coefficients in ${\mathbb{F}}_2={\mathbb{Z}}_2$. The cellular homology of real projective space has one cell in degrees zero, one and two, and its boundary coefficients become zero modulo two. Hence there is a one-dimensional cohomology group in each of those degrees. Let $x\in H^1(P;{\mathbb{F}}_2)$ be the Poincare dual of a projective line. Two distinct projective lines intersect transversely in one point, so $⟨x^2,[P{]}_2⟩=1$. ThusThis is the mod-two cohomology ring of real projective space in dimension two. The Künneth theorem over a field makes the cross product an isomorphism of graded rings. Writing $x,y$ for the factor pullbacks givesThere are no additional signs over ${\mathbb{F}}_2$.

The obstruction to a graph avoiding the diagonal is the mod-two diagonal class of the real projective plane. Its dual class $D\in H^2(P\times P;{\mathbb{F}}_2)$ is determined bywhere $i:P\to P\times P$ is the diagonal embedding. Test $z=x^2,xy,y^2$. Each diagonal pullback is $x^2$, with evaluation one. Since $x^2{y}^2$ is the product's top generator, comparison of coefficients givesIf the submanifold were the graph of a smooth map $f:P\to P$, its parametrization would be $j=(\mathrm{id},f)$. For some $\epsilon \in {\mathbb{F}}_2$, $f^{\ast }x=\epsilon x$, and thereforeOn the other hand, the Poincare dual of the diagonal comes from its relative Thom class in $H^2(P\times P,(P\times P)\setminus \Delta ;{\mathbb{F}}_2)$. It restricts to zero on the complement. A graph disjoint from the diagonal would factor through that complement and force $j^{\ast }D=0$, a contradiction. Such a submanifold cannot be the graph of a smooth map $P\to P$. The relative Thom-class argument also applies to continuous self-maps.