Suppose that $A$ is an affine algebraic variety and $P$, and $Q$ are subvarieties. It is easy to see that the coordinate rings $C[P]$ and $C[Q]$ of $P$ and $Q$ are modules over $C[A]$ and $C[P]\otimes_{C[A]} C[Q]$ is frequently the coordinate ring of $P\cap Q$. For instance, if the sum of the ideals of $P$ and $Q$ is its own radical.
If $X$ is a topological space, and $P$ and $Q$ are subspaces then their singular cohomology groups $ H^{\*}(P) $ and $ H^{\*}(Q) $ are modules over $H^{|*}(X)$.
To make it simple, assume that $X$ is a smooth manifold and $P$ and $Q$ are smooth submanifolds. Are there simple conditions that imply that $H^{\*}(P\cap Q)=H^{\*}(P)\otimes_{H^{\*}(X)} H^{\*}(Q)?$
Here is an example. Let $X=(S^2)^4$. Let $P\subset X$ of all pairs of the form $(x,x,y,y)$. Let $Q\subset X$ of all pairs of the form $(x,y,y,x)$. Notice that $P\cap Q$ is all pairs of the form $(x,x,x,x)$ so it is homeomorphic to $S^2$.
With a little work you can see that
$H^{\*}(X)=\mathbb{Z}[a,b,c,d]/(a^2,b^2,c^2,d^2)$,
that is, integer polynomials in
$4$ variables where the square of any variable is zero.
We get $H^{\*}(P)$ is the quotient of $H^{\*}(X)$ by the ideal generated by $a-b,c-d$ and
$H^{\*}(Q)$ is the quotient of $H^*(X)$ by the ideal generated by $a-d,b-c$.
Its easy to check that
$$H^{\*}(P)\otimes_{H^{\*}(X) H^{\*}(Q)=\mathbb{Z}[a,b,c,d]/(a^2,b^2,c^2,d^2,a-b,b-c,c-d)$$$$H^{\*}(P)\otimes_{H^{\*}(X)} H^{\*}(Q)=\mathbb{Z}[a,b,c,d]/(a^2,b^2,c^2,d^2,a-b,b-c,c-d)$$
which is just $\mathbb{Z}[x]/(x^2)$ which is the cohomology group of the sphere.
