This question is turning out to be a little long so let me start off with the headline. Given a differential graded algebra $A$, we can recover $A$ from its homology $HA$ if we know "the" $A_\infty$-structure of $HA$. Vaguely, I would like to know:

Question 0: Can this be done "functorially" in some sense?

Ok, now for the long version.

For a dg algebra $A$, the association $A\mapsto HA$ gives a functor from the category of dg algebras into the category of graded algebras. Kadeishvili's theorem states that there is a unique(ish) $A_\infty$-structure on the homology $HA$ with certain nice properties. In this way we can think of the association $A\mapsto HA$ as having values in the category of $A_\infty$-algebras. Unfortunately, there seems to be two problems when trying to make this into a functor:

  1. Given a dg morphism $f:A\to B$ the induced graded morphism $Hf:HA\to HB$ may not be an $A_\infty$-morphism
  2. One can get an $A_\infty$-morphism $f_*:=q\circ f\circ j:HA\to HB$ where $j:HA\to A$ and $q:B\to HB$ are $A_\infty$-quasi-isomorphisms, but then the association $f\mapsto f_*$ is no longer functorial.

This brings us to:

Question 1: Is there any way to fix this? Eg., can we somehow view homology as an $A_\infty$-functor or some other sort of "functor up to homotopy"?

Similarly, for a dg $A$-module $M$ there is an $A_\infty$-$HA$-module structure on $HM$ having nice properties.

Question 2: Can we recover the category of dg $A$-modules from the category of $A_\infty$-$HA$-modules, i.e., is there a functor $A$-mod$\to$ $HA$-$A_\infty$-mod (or better yet in the other direction) giving some sort of equivalence?

References also would be much appreciated.

  • $\begingroup$ Have you looked at Shipley's paper homepages.math.uic.edu/~bshipley/zdga17.pdf ? $\endgroup$ Commented Mar 26, 2013 at 21:48
  • 1
    $\begingroup$ The answer to the three questions is YES in some sense, somehow... $\endgroup$ Commented Mar 26, 2013 at 22:59
  • $\begingroup$ @Fernando: yes, I believe so but I would like to know what the precise statements are. $\endgroup$
    – Steve
    Commented Mar 28, 2013 at 15:26
  • $\begingroup$ @Sean: What's the relevance? The "HA" there is different from the "HA" in the question, if that's what you were thinking... $\endgroup$ Commented Apr 2, 2013 at 7:54
  • $\begingroup$ I am also learning in this area recently, please correct me if I am wrong. I think the setup of the question itself needs to be clarified/modified: Kadeishvili's uniqueness theorem is stated up to quasi-iso, which is always the problem in A∞-theory (IMO). There can be different choices of A∞-model on $HA$, which are quasi-iso to each other but with different higher multiplications! Now view back to your question, you need to make sense of "taking value in the category of A∞-algebras"; this seems to be the reason that leads to the problems you stated. $\endgroup$
    – Aaron Chan
    Commented Apr 2, 2013 at 12:59

1 Answer 1


The following answer does not address question 1 in full, but it shows, I think, what one would need to think through.

Take a dg algebra $(A,d)$ over a commutative ring $k$. Specify a splitting of the cocycles as cohomology plus coboundaries: $$ \mathrm{ker} (d) = HA \oplus \mathrm{im}(d) $$ (such a splitting exists provided that $HA$ is projective), and let $i\colon HA\to A$ be the resulting inclusion. We can then construct canonically

(i) An $A_\infty$ structure on $HA$ with differential $\mu^1=0$;

(ii) an $A_\infty$ morphism $\mathcal{I} \colon HA \to A$ whose first term is the given inclusion $i$.

So $HI \colon HA \to HA$ is the identity map. This is Kadeishvili's construction.

These structures are defined by explicit recursive formulae. As such, they already have some desirable functoriality properties. For instance, if a group $G$ acts by automorphisms on $A$, and if the summand $i(HA)$ is $G$-invariant, then the $A_\infty$ data will be $G$-equivariant.

Now suppose we specify in addition a splitting of $A$ as $\mathrm{ker} (d) \oplus A'$. We then have a projection $p\colon A \to HA$, and this extends canonically to

(iii) an $A_\infty$ morphism $\mathcal{P}\colon A\to HA$, with

(iv) a nullhomotopy of $\mathcal{P}\circ \mathcal{I}- \mathrm{id}_{HA}$.

Moreover, there exists a nullhomotopy of $\mathcal{I}\circ \mathcal{P}-\mathrm{id}_A$, but I'm not sure how canonical this nullhomotopy is. A reference for these assertions is Paul Seidel's book Fukaya categories and Picard-Lefschetz theory, chapter 1. In general, $A_\infty$ quasi-isomorphisms induce quasi-equivalences of their module-categories, and this gives an affirmative answer to question 2.

Now take a dg morphism $f\colon A \to B$, and suppose we're given splittings of $A$ and of $B$ as coboundaries plus cohomology plus complement and that $f$ respects these summands. Then we can construct an $A_\infty$ morphism $$ \mathcal{H}f = \mathcal{P}_B \circ (Hf) \circ \mathcal{I}_A \colon A\to B, $$ as indicated in the question. Under composition $g\circ f$ of splitting-respecting dg maps, there is a homotopy $\mathcal{H}g\circ \mathcal{H}f \simeq \mathcal{H}(g\circ f)$. The homotopy comes from the existence of a homotopy $I_B \circ P_B \simeq \mathrm{id}_B$.`

So we get a functor from the category of dga with splittings to the category of $A_\infty$-algebras and homotopy classes of morphisms. Presumably, if one can establish just how canonical the homotopy $I_B \circ P_B \simeq \mathrm{id}_B$ is, one can sharpen the functoriality statement.


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