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Let $Y \to X \leftarrow Z$ be a cospan of topological spaces and let $W = Y \times_X^h Z$ be their homotopy fiber product. I am interested in sufficient conditions on the map $Y \to X$ that ensure that the cohomology of $W$ is the tensor product of the cohomologies of $Y$ and $Z$, $$H^*(W, R) = H^*(Y, R) \otimes_{H^*(X, R)} H^*(Z, R),$$ for any ring $R$ (feel free to add conditions on $R$ here, too).

Is this true, for example, when $H^*(Y, R)$ is free of finite rank over $H^*(X, R)$?

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  • $\begingroup$ You might want to assume that the spaces are simply connected. If $X$ is acyclic (same homology as a point) but not contractible, and if $Y$ and $Z$ are contractible, then $W$ will not be connected. $\endgroup$
    – Tom Goodwillie
    Commented Oct 21, 2023 at 19:22
  • $\begingroup$ Feel free to make assumptions on the spaces. $\endgroup$
    – Matthias Ludewig
    Commented Oct 21, 2023 at 20:35
  • $\begingroup$ Singular cohomology? $\endgroup$
    – Martin Brandenburg
    Commented Oct 22, 2023 at 12:45
  • $\begingroup$ Say, yes! Although I am happy to restrict to a class of spaces where this does not matter. $\endgroup$
    – Matthias Ludewig
    Commented Oct 22, 2023 at 18:00
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    $\begingroup$ Are you familiar with the Eilenberg-Moore spectral sequence? I think something like this appears in the set up, for instance in McCleary's book. $\endgroup$
    – Mark Grant
    Commented Oct 25, 2023 at 7:10

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$R$ can be arbitrary. Assume that $H^\ast(Y)$, as a module for $H^\ast(X)$, is free with a finite basis (or even just a basis whose degrees tend to infinity). Suppose also that $X$ is simply connected. I believe that then the answer is "yes".

We may as well assume that $Y\to X$ is a fibration, so that $W$ can be taken to be the strict fiber product $Y\times_X Z$.

The desired statement is that the canonical map $$ H^\ast(Y)\otimes_{H^\ast(X)} H^\ast(Z)\to H^\ast(Y\times_X Z) $$ is an isomorphism, for every $Z$ and every map $Z\to X$. I will consider the more general statement that the canonical map $$ H^\ast(Y)\otimes_{H^\ast(X)} H^\ast(Z,A)\to H^\ast(Y\times_X Z, Y\times_X A) $$ is an isomorphism, for every pair of spaces $(Z,A)$ and every map $Z\to X$.

We may assume that $(Z,A)$ is a CW pair, and by an inverse limit argument we may assume that it is a finite CW pair, and by a five-lemma argument we may reduce to the case when the pair is $(D^n,S^{n-1})$ for some $n\ge 0$. Another five-lemma argument and induction over $n$ reduces us to the case when $Z$ is a point and $A$ is empty. To put it another way, as functors of $(Z,A)$ both sides are cohomology theories, so the map is an isomorphism for all $Z$ if it is so for a point. (We are using that $H^\ast(Y)$ is flat over $H^\ast(X)$.)

So the task is to show that, under our assumptions, the cohomology of the fiber of $Y\to X$ is the tensor product $H^\ast(Y)\otimes_{H^\ast(X)} R$. I have not thought this through, but it seems that it should follow without much trouble from the Serre spectral sequence of $Y\to X$.

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