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Let $\Lambda \stackrel{F}{\to} \Omega \stackrel{G}{\leftarrow} \Gamma$ be a diagram of groupoids and functors and $\Gamma \times_\Omega \Lambda$ the homotopy pullback. We will regard all these groupoids as spaces and compute the cohomology with coefficients in some field.

There should be a map $$ C^*(\Gamma) \stackrel{\mathbb{L}}{\otimes}_{C^*(\Omega)} C^*(\Lambda) \to C^* (\Gamma \times_\Omega \Lambda)$$ from the derived tensor product to the cohomology of the homotopy pullback.

Is this map an equivalence?

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    $\begingroup$ I think so, that's Eilenberg-Moore, isn't it? $\endgroup$ Commented Feb 4, 2018 at 21:39
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    $\begingroup$ According to Wikipedia, we would need that $\Omega$ is simply connected. $\endgroup$ Commented Feb 4, 2018 at 21:42
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    $\begingroup$ You're right, local coefficients come in so trivial actions would be necessary to recover this for ordinary cochains. $\endgroup$ Commented Feb 4, 2018 at 21:50
  • $\begingroup$ So are you saying the statement is false? $\endgroup$ Commented Feb 4, 2018 at 21:57
  • $\begingroup$ I don't have any counterexample. $\endgroup$ Commented Feb 4, 2018 at 21:59

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No, consider the multiplication by $2$ map on the integers. The homotopy fiber is two points. But the derived tensor product (for $\mathbb Q$ coefficients) is $\mathbb Q \otimes^{L}_{\mathbb Q[x]} \mathbb Q[x] = \mathbb Q$.

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  • $\begingroup$ Nice counterexample. Any clue under which requirements the about statement will be true for general spaces (other requirements that $\Omega$ simply connected)? $\endgroup$ Commented Feb 6, 2018 at 8:10
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    $\begingroup$ I saw somewhere that the Eilenberg Moore SS converges strongly iff the fundamental group of the base acts unipotently on the homology of the fiber-- so maybe that hypothesis will help. $\endgroup$ Commented Feb 6, 2018 at 13:36
  • $\begingroup$ Yeah I am pretty sure this is exactly controlled by the convergence of the EMSS. I think Brooke Shipley has a nice paper about this, IIRC. $\endgroup$ Commented Feb 17, 2018 at 18:55

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