= Solution
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 $\langle x^2,[P]_2\rangle=1$. Thus
$$
\boxed{H^*(P;\mathbb F_2)=\mathbb F_2[x]/(x^3),\qquad |x|=1.}
$$
This 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 gives
$$
\boxed{H^*(P\times P;\mathbb F_2)=\mathbb F_2[x,y]/(x^3,y^3).}
$$
There 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 by
$$
\langle D\smile z,[P\times P]_2\rangle=\langle i^*z,[P]_2\rangle,
$$
where $i:P\to P\times P$ is the <diagonal embedding>. Test $z=x^2,xy,y^2$. Each diagonal pullback is $x^2$, with evaluation one. Since $x^2y^2$ is the product's top generator, comparison of coefficients gives
$$
D=x^2+xy+y^2.
$$
If the submanifold were the graph of a smooth map $f:P\to P$, its parametrization would be $j=(\mathrm{id},f)$. For some $\varepsilon\in\mathbb F_2$, $f^*x=\varepsilon x$, and therefore
$$
j^*D=(1+\varepsilon+\varepsilon^2)x^2=x^2\ne0.
$$
On 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 $j^*D=0$, a contradiction. \b[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.
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