When is bar-cobar duality an equivalence?

Question

Let $A$ be an augmented differential graded algebra over a field $k$. I will write $BA$ for its bar construction (whose homology is $Tor^A(k, k)$). This is a co-augmented differential graded coalgebra over $k$; write $\Omega BA$ for its cobar construction. There is a natural dga map $\Omega BA \to A$ which uses the fact that $\Omega C$ is the tensor algebra on a shift of $C$, and the projection map $BA \to A$.

I think I'm only going to highlight my ignorance here, but my question is: what assumptions on $A$ imply that the map $\Omega BA \to A$ is an equivalence?

If I am reading any of the following references correctly:

• Husemoller-Moore-Stasheff's "Differential homological algebra and homogeneous spaces" (section II.4)
• Felix-Halperin-Thomas "Rational homotopy theory" (exercise 2, section 19)
• Keller "A-infinity algebras, modules and functor categories" (Theorem 4.3, attributed to Lefevre),

then I think that the answer is supposed to be: "no assumptions are required at all."

I seem to have two possible counterexamples to this claim. But their being counterexamples seems to rely on the positive answer to another question: when is $\Omega$ a homotopy-invariant functor? That is: if $f: C \to D$ is an equivalence of dg coalgebras, is $\Omega f$ an equivalence of dga's?

Let me explain the (counter?)examples. Let $G$ be a finite group, $k[G]$ its group ring, and $k^G$ the dual Hopf algebra (the ring of functions on $G$ with values in $k$). Then both $k[G]$ and $k^G$ are augmented algebras, where the augmentation is the counit.

1. $B(k[G])$ is isomorphic to the simplicial chain complex of the bar construction $BG$ (with coefficients in $k$). If $k$ has characteristic zero, then this is is equivalent to $k$, via a transfer argument. If $\Omega$ is a homotopy invariant, then $\Omega B(k[G]) \simeq \Omega k = k$, which is certainly not $k[G]$ unless $G$ is trivial.

2. $k^G$ is isomorphic as a $k$-algebra to a product of fields $$k^G \cong \prod_{g \in G} k,$$ and the augmentation can be identified with the map which projects onto the factor corresponding to the identity $e \in G$. The kernel of the augmentation is a summand, and hence a projective $k^G$-module. From this, you can write down a very short resolution of $k$ over $k^G$ and compute $Tor^{k^G}(k, k) = k$, concentrated in $Tor_0$. So $B(k^G) \simeq k$. Again, if $\Omega$ is homotopy-invariant, then $\Omega B(k^G) \simeq \Omega k = k$, which is not equivalent to $k^G$.

So either I have misunderstood the references above, or $\Omega$ is not a homotopy invariant functor. I'm guessing it's the latter; can anyone point me to a reference which gives criteria for this to hold?

Answer 1

What the references are saying is correct, and you are right. Yes, $\Omega BA \to A$ is always a quasi-isomorphism. No, $\Omega$ does not in general take quasi-isomorphisms to quasi-isomorphisms.

A sufficient condition for $\Omega$ transforming a DG-coalgebra morphism to a quasi-isomorphism of DG-algebras is a filtered quasi-isomorphism of conilpotent DG-coalgebras. This also generalizes to CDG-coalgebras (which correspond to nonaugmented DG-algebras, and for which the conventional notion of quasi-isomorphism does not even exist, but filtered quasi-isomorphisms make perfect sense).

For a reference, see my 2011 AMS Memoir "Two kinds of derived categories, Koszul duality, and comodule-contramodule correspondence", http://arxiv.org/abs/0905.2621 , Section 6.10.

Answer 2

See Hasegawa's thesis, available here. Lemma 1.3.2.3 is what you want, on page 35 of the pdf. To see that $\Omega B A \to A$ is a quasi-isomorphism, one filters $BA$ by its primitive elements (i.e. by the length of bar elements) which then induces on $\Omega BA$ a filtration. One gives $A$ the trivial filtration (i.e. $F^0A=0$ and all higher $F^i$ are $A$) and sees the counit is a filtered morphism. The associated graded map is the identity of $A$ on degree one, and then one shows that $\text{Gr}_i(\Omega BA)$ is contractible for $i>1$.

Another way to remember why the counit is a quasi-isomorphism is that it corresponds to the trivial twisting cochain $\beta:BA\longrightarrow A$ whose total space $BA\otimes_\beta A$ is acyclic. Then one fits the counit into the following sequence

$$1\otimes \varepsilon_A : BA\otimes_\gamma \Omega B A\to BA\otimes_\beta A$$

Both twisted complexes above are acyclic so $1\otimes \varepsilon_A$ is a quasi-isomorphism. Now $BA$ is simply connected so a spectral sequence argument works to show that, because $1$ is of course a quasi-isomorphism, so is $\varepsilon_A$.

This argument fails for $\Omega$ because it decreases the connectivity of your space. Lemma 1.3.2.2 in the thesis above gives a sufficient criterion for $\Omega$ to preserve quasi-isomorphisms. The bar construction, on the other hand, preserves quasi-isomorphisms by a direct argument using the filtration by length and the Kunneth formulas. Again, no connectivity issues arise here, which renders the desired spectral sequence convergent.