Is quasi-isomorphism of $A_{\infty}$ algebras invertible

Question

Let $A_1$, and $A_2$, be two $A_{\infty}$-algebras over a field $k$. Suppose, $f=\{f_i\} : A_1 \to A_2$ is $A_{\infty}$ quasi-isomorphism of $A_{\infty}$-algebras , that is a map of $A_{\infty}$-algebras such that $f_1$ is quasi-isomorphism. Under such assumptions, is $f$ invertible? More precisely, is there an $A_{\infty}$-map $g: A_2 \to A_1$ such that compositions $f\circ g$ and $g \circ f$ induce identity map on cohomology?

Answer 1

In Kellers "Introduction to $A_\infty$-algebras" this is essentially part b) of the last theorem of chapter 3 (directly over header of chapter 4) as there it says that it is an quasi-iso if and only if it is a homotopy equivalence, i.e. one only needs to kill homotopy, and not adjoin new inverses, or add roofs. ] In particular you find an $g$ such that $f \circ g \sim id \sim g \circ f$, where sim induces homotopy equivalence. Which is precisely what you asked for.

The hands on proof might be tricky, but I think one can circumvent the calculations by using model categories. So if you read through Lefevre "Theorie de l'homotopie des $A_\infty$-algebres", that just pops out immediately (assuming your ground structure is semisimple, which is clearly is if you are working over a field).

However that might need that $k$ is a field, but I think you should be happy with that.

This is very similar to a quasi-isomorphism over field being the same as a homotopy equivalence. And also that is what is used in the proof, i.e. that every object and morphism has a representative.