Under which conditions is the bar construction a conservative functor?

Question

The bar construction is a functor $A\mapsto Bar(A)$ from the category of augmented differential graded algebras over a commutative ring $R$ to the category of chain complexes of $R$-modules. It sends an algebra $A$ to the derived tensor product $R\otimes^{L}_AR$.

Assume that I have a map of augmented algebras $A\to B$ which induces a quasi-isomorphism $Bar(A)\to Bar(B)$. This does not necessarily implies that the original map was a quasi-isomorphism but I believe it does if some connectivity or co-connectivity assumptions are made on $A$ and $B$. Philosophically speaking, the bar construction is the algebraic analogue of the loop space construction for based spaces and this construction is conservative if one restricts to connected based spaces.

I am using the homological grading convention. My guess is that the Bar construction is conservative when restricted to simply coconnected algebras (i.e. whose homology is concentrated in degrees $\leq -2$ apart from the unit in degree zero). The analogy with spaces suggests that this might actually also be true for coconnected algebras but I'm not so sure about that. In any case, I'm interested in finding a reference for this fact.

Answer 1

$\vphantom{0pt}$ Hi Geoffroy. The natural way to approach this is by finding conditions under which the cobar functor preserves quasi-isomorphisms. If $A \to A'$ is a morphism of dg algebras, then you get a zig-zag $$ A \leftarrow \Omega BA \to \Omega BA' \to A'.$$ The two counit maps here are always quasi-isomorphisms, without connectivity/coconnectivity hypotheses. The cobar functor preserves quasi-isomorphisms of coalgebras concentrated in homological degree $\geq 2$, and also of coalgebras concentrated in degree $\leq 0$. (I hope I didn't mess up any degree shifts just now.) So natural hypotheses which imply that the bar construction reflects quasi-isomorphisms is that $A$ and $A'$ are either both connected or both coconnected.

Off the top of my head I don't have a good reference for the above, but let me tell you how the proofs go. With no assumption at all, it is true that the cobar functor takes quasi-isomorphisms to filtered quasi-isomorphisms, where the cobar construction is filtered by its natural "adic" filtration. A filtered quasi-isomorphism is a quasi-isomorphism if the filtration is complete and exhaustive. The reason the cobar construction fails to preserve quasi-isomorphisms in general is that the adic filtration is typically not complete, but if your algebra is concentrated in degrees as above then the adic filtration becomes bounded below in every homological degree, in particular complete, and you win.