Suppose given a homotopy cartesian square of (commutative) ring spectra (or (c)dgas)

$\begin{matrix}A & \to & A_1 \\ \downarrow & & \downarrow \\ A_2 & \to &A'\end{matrix}.$

Here the morphisms are morphisms of ring spectra (dgas) and by homotopy cartesian I mean that the underlying diagram of spectra (chain complexes) is homotopy cartesian (equivalently, cocartesian).

Let me write $Mod(A)$ for the stable $\infty$-category (dg category) of $A$-module spectra, and similarly for the other rings. There is an induced diagram of infinity categories

$\begin{matrix}Mod(A) & \to & Mod(A_1) \\ \downarrow & & \downarrow \\ Mod(A_2) & \to &Mod(A')\end{matrix},$

where the morphisms are given by base change. Recall the homotopy fibre product $Mod(A_1) \times^h_{Mod(A')} Mod(A_2)$; we obtain a functor $F: Mod(A) \to Mod(A_1) \times^h_{Mod(A')} Mod(A_2)$.

I want to know if there are (tractable) criteria for $F$ to be a quasi-equivalence?

Here are some remarks:

  • In http://arxiv.org/abs/1201.6118 the authors provide a nice explicit model for the homotopy fibre product of dg categories, see section 4. They also state that $F$ is a quasi-equivalence if the diagram is actually cartesian, the dgas have no positive homotopy groups (in topologists' notation), and the degree zero chains satisfy milnor descent (c/f below). I'd be interested in a more "homotopical" criterion.

  • Suppose all the rings are commutative. Let me write $P(A) \subset Ho(Mod(A))$ for the additive karoubi-closed subcategory generated by $A$. If the diagram

$\begin{matrix}\pi_0(A) & \to & \pi_0(A_1) \\ \downarrow & & \downarrow \\ \pi_0(A_2) & \to &\pi_0(A')\end{matrix}$

is cartesian and one of the lower/right maps is surjective, then we have that $P(A) \to P(A_1) \times_{P(A')} P(A_2)$ is an equivalence. This is milnor descent.

  • Using the explicit model for $Mod(A_1) \times^h_{Mod(A')} Mod(A_2)$, namely objects being triples $(M_1, M_2, \phi)$ where $M_i \in Mod(A_i)$ and $\phi: M_1 \otimes_{A_1} A' \to M_2 \otimes A_2 A'$ is a closed equivalence of degree zero, one defines a functor $R: Mod(A_1) \times^h_{Mod(A')} Mod(A_2) \to Mod(A)$ mapping $(M_1, M_2, \phi)$ to the homotopy fibre of $M_1 \oplus M_2 \to M_2 \otimes_{A_2} A'$ (viewed as a diagram of $A$-modules). I think functors $Ho(F), Ho(R)$ are adjoints and $RF$ is equivalent to the identity functor. I also think that $FR$ is equivalent to the identity functor if (and only if) $F$ has dense image, i.e. iff $Mod(A_1) \times^h_{Mod(A')} Mod(A_2)$ is generated by $F(A)$. I do not know how to make use of this criterion.

  • As a side note, in the commutative case, all categories are symmetric monoidal. The functor $F$ is symmetric monoidal, but I do not see why $R$ (or $Ho(R)$ would be (unless $F$ is a quasi-equivalence)$.

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    $\begingroup$ I believe that your description of the right adjoint and the identification of $RF$ are correct. As a result I believe that you can identify $Mod(A)$ with its essential image in the fiber product. In general, checking essential surjectivity seems to lead into delicate questions analogous to checking affineness and is often not satisfied. Did you have a specific type of "homotopical" property in mind? $\endgroup$ Jul 23, 2015 at 6:01
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    $\begingroup$ I have no particular property in mind. Well maybe asking all the rings to have vanishing negative homotopy, plus surjectivity of $\pi_0(A_1) \to \pi_0(A')$ plus something further. Perhaps "$A_1 \to A'$ effective epi" but since I know very little about what effective epi means this is just giving the problem another name... Can you elaborate on your analogy to affineness? $\endgroup$ Jul 23, 2015 at 7:59
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    $\begingroup$ If you are looking for the case (which I think is relevant for "Milnor descent") of a derived version of a pushout along closed immersions, there is a positive result with connectivity hypotheses in Theorem 7.2 of DAG IX, as well as a counterexample if these are omitted (Warning 7.3). As I think is implicit in your post, the functor $F$ is automatically fully faithful, so the question is equivalent to essential surjectivity of $F$ (or conservativity of $R$). $\endgroup$ Jul 23, 2015 at 8:18
  • $\begingroup$ Great question, but I need to say that in my paper with Block we are actually not working with Mod(A) but with a different category defined by Block. The motivating example is the Dolbeaut algebra where as we try to explain in the article, Mod(A) is the not the category one wants to look at! $\endgroup$ May 5, 2016 at 17:11

1 Answer 1


I shall call an object with vanishing negative homotopy groups (objects) -1-connected, and an object with only finitely many non-vanishing negative homotopy groups (objects) connective. (This is at odds with Lurie!)

Akhil's comment points in exactly the right direction. Indeed proposition 7.6 in DAG-IX is basically an answer to my question: If $A, A_1, A_2, A'$ are -1-connected then $Mod(A)$ etc have good t-structures (the forgetful functor to spectra is exact). Write $Mod(A)_+$ for the triangulated subcategory of connective objects, etc. Then the functor $F: Mod(A)_+ \to Mod(A_1)_+ \times^h_{Mod(A')_+} Mod(A_2)_+$ is an equivalence, provided $\pi_0(A_1) \to \pi_0(A')$ is surjective.

This statement is slightly different from Lurie's. Firstly he uses $Mod(A)_{\ge 0}$ etc, but the connective case is obtained by just shifting. Secondly he assumes that $\pi_0(A_2) \to \pi_0(A')$ is surjective as well, but this is not necessary. He only uses this assumption in the last sentence, to deduce that $\pi_n(M) = 0$. But he already knows that $N \in Mod(A_1)_{> n}$ and hence $P \simeq N \otimes_{A_1} A' \in Mod(A')_{> n}$. Thus $0 = \pi_n(P) = \pi_n(M)$. At least unless I'm missing something...

Warning 7.3 shows that the connectivity assumption cannot be removed, even if $A_1, A_2, A'$ are discrete.


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