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.
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$B(k[G])$ 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.
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$k^G$ 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?
Best Answer
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.