# When does cohomology take pullbacks to pushouts?

I've encountered a simple situation where one has a pullback diagram of topological spaces and taking cohomology takes it into what I believe is a pushout diagram in the category of rings. I'm not sure if I can make diagrams here, but I have maps $X \overset{f}\to Z \overset{g}\leftarrow Y$ and the fiber product $$X \times_Z Y = \big\{(x,y) \in X \times Y : f(x) = g(y)\big\}$$ given the subspace topology. When I take singular cohomology with rational coefficients, in this case, I get an isomorphism $$H^*(X \times_Z Y) \cong H^*(X) \otimes_{H^*(Z)} H^*(Y),$$ where the $H^*(Z)$-algebra structures on $H^*(X)$ and $H^*(Y)$ are given by pullback along $f$ and $g$.

How general is this situation? What needs to go right for it to work? What needs to go wrong for it to fail?

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I found that article (again, incidentally) about ten minutes after posting my question, and felt very unresourceful. –  jdc Feb 20 '14 at 0:38
I have very little idea how to compute the double-subscripted Tor, which seems kind of daunting. Is that worth posting another question about? –  jdc Feb 20 '14 at 0:46
The double-graded Tor is not as bad as it sounds. You still compute Tor the same way (using a projective resolution), but since your ring and modules are graded the resolution can inherit its own grading too. It's worth trying a couple of examples (e.g. $S^1 = S^3 \times_{S^2} *$ for the Hopf fibration) to get a feel forhow the grading goes. –  Tyler Lawson Feb 20 '14 at 2:15

## 1 Answer

Well, your question is exactly what motivates people to study homotopy theory. I would advise you to look at Model Category Theory.

To answer your question, I think it always works when $f$ or $g$ is a fibration (and if your ring of coefficents is a field).

Let me also give you a counterexample. Let $X$ and $Y$ be points, $Z$ be the closed interval, and $f$ and $g$ be the inclusions on $0$ and $1$. Then the fiber product is empty… (The right thing to do here would have been to take the homotopy fiber product). But $f$ and $g$ are homotopy equivalences, so $H^\ast(X) \simeq H^\ast(Z) \simeq H^\ast(Y)$.

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So what is the homotopy fiber product? In my trivial examples, $f$ is a fibration (actually, $g$ is too), the fiber is totally nonhomologous to zero,'' if I'm using that term right, and the Serre spectral sequence for $X \times Y \to Y$ collapses at $E_2$. Books seem to be telling me that the Eilenberg&ndash;Moore spectral sequence breaks down in that case. I'm kind of wondering how much room I have. –  jdc Feb 20 '14 at 0:45
I've seen it claimed elsewhere, too, that if $f$ or $g$ is a fibration, the EMSS breaks down at $E_2$: do you know of any citations for this? –  jdc Mar 26 '14 at 4:55