DGA formality via $A_\infty$

Question

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 $A\rightarrow H(A)$.

Answer 1

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...)