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Let $k$ be an algebraically closed field of characteristic $0$.

Let $A,B$ be commutative differential graded algebras (cdga) over $k$ such that $H^{i}(A)=H^{i}(B) =0 \ (i>0)$.

Here, differentials are defined cohomologically, i.e, $$ \cdots \rightarrow A^{i-1} \overset{d_A}{\rightarrow} A^i \overset{d_A}{\rightarrow} A^{i+1} \overset{}{\rightarrow} \cdots $$ where $d_A$ is the differential of $A$. In other word, $A,B$ are connective.

We also denote the dg categories of dg $A$-modules and dg $B$-modules by $D_{dg}(A),D_{dg}(B)$, respectively.

Question If $D_{dg}(A),D_{dg}(B)$ are quasi-equivalent as dg categories, then $A, B$ are quasi-isomorphic as cdgas ?

Any comments and references are welcome. Thank you !

The same question is in MO (Please take a look at the useful comments.).

Edit (a variant of the question):

Let $\tilde{D}_{dg}(A),\tilde{D}_{dg}(B)$ be dg-derived categories, i.e, $\tilde{D}_{dg}(A) = D_{dg}(A)/ Ac(A), \tilde{D}_{dg}(B) = D_{dg}(B)/ Ac(B)$, where $Ac(A) \subset D_{dg}(A), Ac(B) \subset D_{dg}(B)$ are the full sub dg categories of acyclic complexes of $D_{dg}(A), D_{dg}(B)$, respectively.

If $\exists F: \tilde{D}_{dg}(A) \rightarrow \tilde{D}_{dg}(B)$ are quasi-equivalent of dg categories and $H^0(F)$ is a monoidal functor between monoidal categories $H^0(\tilde{D}_{dg}(A)), H^0(\tilde{D}_{dg}(B))$, then $A,B$ are quasi-isomorphic ?

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  • $\begingroup$ I think so. Here’s how youd show it. If you assume that the functor commutes with colimits(or is right exact) then the initial object of the first cat, namely A, is sent to the initial object of the second cat B. The equivalence of cats you have above restricts to the one object cat given between the one object categories on A and B. This gives maps on the morphism chain complexes which are quasi isomorphisms. But that’s just a map from A to B that’s a quasiiso. $\endgroup$
    – Hari Rau-Murthy
    Sep 19 at 4:41
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    $\begingroup$ Thank you for the comment. I think the initial object in the categories is $0$. $\endgroup$
    – Walterfield
    Sep 19 at 4:56
  • $\begingroup$ oh dang you’re right $\endgroup$
    – Hari Rau-Murthy
    Sep 19 at 5:08
  • $\begingroup$ I think what i said still works if you can show, or you can use an assumption to guarantee that the object A is sent to the object B $\endgroup$
    – Hari Rau-Murthy
    Sep 19 at 5:09
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    $\begingroup$ if you ask that the functor is monoidal (for tensor product) then A is sent to B $\endgroup$
    – Hari Rau-Murthy
    Sep 19 at 5:13

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