Cohomology groups of an intersection

Question

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

$$H^{\*}(P)\otimes_{H^{\*}(X)} H^{\*}(Q)=\mathbb{Z}[a,b,c,d]/(a^2,b^2,c^2,d^2,a-b,b-c,c-d)$$

which is just $\mathbb{Z}[x]/(x^2)$ which is the cohomology group of the sphere.

Answer 1

If $P$ and $Q$ are closed subspaces of $Y$ and $Y$ is their union, we have a short exact sequence of sheaves on $Y$ $0\rightarrow\mathbb Z\rightarrow i_\ast\mathbb Z\bigoplus j_\ast\mathbb Z\rightarrow k_\ast\mathbb Z\rightarrow0$, where $i$, $j$ resp. $k$ are the inclusions of $P$, $Q$ and $P\bigcap Q$ respectively. This gives a long exact sequence of Cech cohomology $$ \cdots\rightarrow\check H^i(Y,\mathbb Z)\rightarrow \check H^i(P,\mathbb Z)\bigoplus \check H^i(Q,\mathbb Z)\rightarrow \check H^i(P\bigcap Q,\mathbb Z)\rightarrow\cdots $$ and when $Y$, $P$, $Q$ and $P\bigcap Q$ are well-behaved Cech cohomology coincides with singular cohomology and we get a similar sequence in singular cohomology. Hence $H^\ast(P\bigcap Q,\mathbb Z)$ is close (but not always equal to because of non-triviality of the boundary map) the additive pushout $H^\ast(P,\mathbb Z)\bigoplus_{H^\ast(Y,\mathbb Z)} H^\ast(Q,\mathbb Z)$. It seems in any case clear that anything we can hope to know about the cohomology of of $P\bigcap Q$ in terms of $P$ and $Q$ and some ambient space $X$ should be obtained through the Mayer-Vietoris sequence. If we compare that with the multiplicative pushout $H^\ast(P,\mathbb Z)\bigotimes_{H^\ast(Y,\mathbb Z)} H^\ast(Q,\mathbb Z)$ we see essentially that it is close to the additive pushout only in very simple cases. It is true that the multiplicative pushout could hit more of $H^\ast(P\bigcap Q,\mathbb Z)$ because the image is the subring generated by the images of $H^\ast(P,\mathbb Z)$ and not just $H^\ast(Q,\mathbb Z)$ but on the other hand all multiplicative relations would only rarely be coming from $H^\ast(Y,\mathbb Z)$.

Now, replacing $Y$ by some ambient space $X$ doesn't seem to help as we have a natural surjection $$ H^\ast(P,\mathbb Z)\bigotimes_{H^\ast(X,\mathbb Z)}H^\ast(Q,\mathbb Z) \rightarrow H^\ast(P,\mathbb Z)\bigotimes_{H^\ast(Y,\mathbb Z)}H^\ast(Q,\mathbb Z) $$ so if $H^\ast(P,\mathbb Z)\bigotimes_{H^\ast(X,\mathbb Z)}H^\ast(Q,\mathbb Z)$ works then so does $H^\ast(P,\mathbb Z)\bigotimes_{H^\ast(Y,\mathbb Z)}H^\ast(Q,\mathbb Z)$.

To be a little bit more specific, if $H^\ast(X,\mathbb Z)\rightarrow H^\ast(P,\mathbb Z)$ and $H^\ast(X,\mathbb Z)\rightarrow H^\ast(P,\mathbb Z)$ are not surjective, then it seems that $H^\ast(P,\mathbb Z)\bigotimes_{H^\ast(X,\mathbb Z)}H^\ast(Q,\mathbb Z)\rightarrow H^\ast(P\bigcap Q,\mathbb Z)$ should not be injective except in some very trivial cases. However, even when they are we easily get into trouble: Let $P$ and $Q$ be lines in the complex projective plane $X$. Then $H^\ast(X,\mathbb Z)=\mathbb Z[x]/(x^3)$ and $H^\ast(P,\mathbb Z)=H^\ast(Q,\mathbb Z)=\mathbb Z[x]/(x^2)$ so that $H^\ast(P,\mathbb Z)\bigotimes_{H^\ast(X,\mathbb Z)}H^\ast(Q,\mathbb Z)=Z[x]/(x^2)$ while $P\bigcap Q$ is a point.