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Given an $A_\infty$-quasi-isomorphism $A\rightarrow B$ of dgas A and B how does one get a zig-zag of dga-quasi-isomorphisms $A\rightarrow \cdot \leftarrow \ldots \leftarrow \cdot \rightarrow B$? This is one implication in an equivalence on page 7 in B. Vallette, ALGEBRA + HOMOTOPY = OPERAD (https://arxiv.org/pdf/1202.3245.pdf) which allows him to conclude that formality of a dga A is equivalent to the existence of an $A_\infty$-quasi-isomorphism $A\rightarrow H(A)$.

Pavel
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1 Answers1

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This essentially follows from the "rectification" procedure that is mentioned in the paper of Vallette (see e.g. Chapter 11 of the book Algebraic Operads of Loday and Vallette for more detail).

Given an augmented associative algebra $A$, you have the bar construction $BA$. It is the cofree conilpotent coalgebra on the suspension of the augmentation ideal $\bar{A}$, with some differential. In other words, $$BA = (T^c(\Sigma \bar{A}), \partial)$$ and the differential $\partial$ is given by: $$\partial(a_1 \otimes \dots \otimes a_n) = \sum_{i=1}^{n-1} \pm a_1 \otimes \dots \otimes a_i a_{i+1} \otimes \dots \otimes a_n.$$

An $\infty$-morphism $f : A \leadsto A'$ is nothing but a morphism of dg-coalgebras $\alpha_f : BA \to BA'$. The composition of $\alpha_f$ with the projection onto $\bar{A}'$ in the bar construction is given by the maps $f_n : A^{\otimes n} \to A'$. Given the cofree nature of the underlying coalgebra of $BA'$, this is enough to specific $\alpha_f$ completely. I encourage you to check this by hand.

Then you have the cobar construction. Given a dg-coalgebra $C$, the cobar construction $\Omega C$ is a dg-algebra. It has a definition that is formally dual of the bar construction. You can find the definition in the book of Loday—Vallette for example. It's also a functor, i.e. a morphism of dg-coalgebras $C \to C'$ induces a morphism of dg-algebras $\Omega C \to \Omega C'$. Moreover, given an algebra $A$, there is a canonical quasi-isomorphism $\Omega B A \xrightarrow{\sim} A$.

So you can apply this functor to $\alpha_f : BA \to BA'$ to obtain a morphism $\Omega(\alpha_f) : \Omega BA \to \Omega BA'$. Together with the canonical quasi-isomorphisms above, you get a zigzag as expected: $$A \xleftarrow{\sim} \Omega BA \xrightarrow{\Omega(\alpha_f)} \Omega BA' \xrightarrow{\sim} A'.$$

Moreover if $f$ is an $\infty$-quasi-isomorphism, then so is $\Omega(\alpha_f)$.

(The nice thing is, of course, that this works for any Koszul operad, so you can apply this to other kinds of algebra, e.g. commutative algebras, Lie algebras...)

Najib Idrissi
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  • Thank you very much for your nice answer! The bar-cobar resolution seems like a natural solution. I was actually thinking of something like lifting the $A_\infty$-quasi-isomorphism $A\rightarrow B$ to minimal models $M_A \rightarrow M_B$ and hoping one gets a dga-quasi-isomorphism somewhere. – Pavel Apr 03 '18 at 12:55
  • @user32817 That's also possible in theory. The minimal model $M_A$ is cofibrant, so if $A$ and $B$ are quasi-isomorphic, then there must exist a direct quasi-isomorphism $M_A \to M_B$. The problem is that I can't really say how to build it from the $\infty$-morphism $A \leadsto B$. – Najib Idrissi Apr 03 '18 at 13:08
  • If we have a Zig-Zag of $A_{\infty}-$quasi-isomorphisms from a dg algebra $A$ to a dg algebra $B$, then can we deduce that there is a Zig-Zag of dg quasi-isomorphisms from $A $ to $B$? – user12580 Jan 07 '19 at 09:44
  • @user12580 Yes, just consider the level 1 part of the $A_\infty$-morphisms. – Najib Idrissi Jan 07 '19 at 09:47
  • You mean, since $A_{\infty}-$quasi isomorphisms has a homotopy inverse, then we have a direct $A_{\infty}-$quasi isomorphism from $A$ to $B$？Then I have the question: if I know the homology algebra of $A$ and $B$ are isomorphic as graded algebras, can I deduce that $A$ and $B$ are linked by quasi-isomorphisms? – user12580 Jan 07 '19 at 09:57
  • @user12580 No, I mean that if $f = (f_n){n \ge 1} : A \leadsto B$ is an $A\infty$-quasi-isomorphism, then $f_1 : A \to B$ is a quasi-isomorphism of chain complexes, by definition. And no, if two dgas have isomorphic homologies, it does not follow that they are quasi-isomorphic as dgas. It's usually a difficult question to know whether two dgas are quasi-isomorphic, even if they have isomorphic homologies (if their homologies are not isomorphic, then clearly they cannot be quasi-isomorphic). – Najib Idrissi Jan 07 '19 at 10:02
  • But Since we have $H^(A)\rightarrow A$ and $H^(B)\rightarrow B$. So the question is whether $H^(A)\rightarrow H^(B)$ is an $A_{\infty}$ morphism provided they are isomorphic as graded algebras. – user12580 Jan 07 '19 at 10:22
  • @user12580 The problem is that in $H^(A) \to A$, you do not consider $H^(A)$ with its dga structure. You consider it with an $A_\infty$-structure induced by Massey products in $A$. Same in $B$. Even if $H^(A) = H^(B)$ as algebras, it does not follow that the Massey products will be the same. – Najib Idrissi Jan 07 '19 at 10:32
  • If the two dg algebras $A$ and $B$ are derived equivalent with A sent to B, then is it true that $A$ and $B$ are quasi isomorphic? I don’t know if we can recover Massey product from the derived category. – user12580 Jan 07 '19 at 11:41