Fundamental class of diagonal of $\mathbb{R}P^n \times \mathbb{R}P^n$ with coefficients $\mathbb{Z} / 2$ I am having trouble to understand the fundamental class of the diagonal $\Delta \subset \mathbb{R}P^n \times \mathbb{R}P^n$ with coefficient $\mathbb{Z} / 2$ as in the class I only gets a rough idea how to do this using the cohomology mod torsion. Could someone give me a direction how to proceed with such problem?
 A: The question is not so clear, I assumed that the OP is asking for a description of the class of the diagonal in the cohomology ring.
First of all lets look at the cohomology ring of $\mathbb{R}P^n \times \mathbb{R}P^n$ by kunneth theorem we have (all cohomology groups are taken with $\mathbb{F}_2$ coefficients):
$$H^*(\mathbb{R}P^n \times \mathbb{R}P^n) \cong H^*(\mathbb{R}P^n) \otimes H^*(\mathbb{R}P^n) \cong \mathbb{F}_2 [\alpha,\beta]/(\alpha^{n+1},\beta^{n+1})$$
The class of the diagonal whatever it is can be expressed as a linear combination:
$$[\Delta]= \Sigma^{n}_{j=0} c_j\alpha^j\beta^{n-j}$$
To determine the $c_j$ we take the cup product $[\Delta] \cap\alpha^{n-k}\beta^{k}$. Plugging the expression above we see that for dimensional reasons everything else dies except $c_k \alpha^n\beta^n=c_k [\mathbb{R}P^n \times \mathbb{R}P^n]$ (a multiple of the fundamental class).
We can interpret $\alpha$ and $\beta$ as duals to the hyperplanes $H \times \mathbb{R}P^n$ and $\mathbb{R}P^n \times H$ respectively. But then $\alpha^{n-k}$ is dual to a linear subspace of codimension $n-k$ denoted $L \subset \mathbb{R}P^n$ and $\beta^k$ is dual to a linear subspace of codimension $k$ denoted $K \subset \mathbb{R}P^n$.
Calculating the intersection numbers mod 2 we get:
$$c_k= I(\Delta , L\times K)= I(L,K) = 1 $$ 
Where the last equality is because intersecting linear subspaces of complementary dimensions just gives a point. All in all we got the following formula:
$$[\Delta] = \Sigma_j \alpha^j \beta^{n-j}$$
