# On the coalgebraic homotopy transfer theorem

Let $$A$$ be a dg algebra, say over a field. The Homotopy Transfer Theorem says that $$H(A)$$ can noncanonically be given the structure of $$A_\infty$$-algebra, extending the induced multiplication on $$H(A)$$, in such a way that $$A$$ and $$H(A)$$ are quasi-isomorphic as $$A_\infty$$-algebras.

If $$C$$ is instead a dg coalgebra then we can also transfer the coalgebra structure to a quasi-isomorphic $$A_\infty$$-coalgebra structure on $$H(A)$$. Unfortunately the relation of quasi-isomorphism is less well behaved for coalgebras than for algebras and often one wants to consider instead the notion of weak equivalence: a coalgebra morphism $$C \to C'$$ is called a weak equivalence if the induced map on cobar constructions $$\Omega C\to\Omega C'$$ is a quasi-isomorphism. Is any dg coalgebra weakly equivalent to its cohomology as an $$A_\infty$$ coalgebra?

• "If $C$ is instead a dg coalgebra then we can also transfer the coalgebra structure to a quasi-isomorphic $A_\infty$-coalgebra structure on $H(A)$." -- is this true? – Leonid Positselski Nov 26 '18 at 0:45
• "Is any dg coalgebra weakly equivalent to its cohomology as an $A_\infty$ coalgebra?" -- is seems that this can't be true. Say, suppose that we are working in the categories of noncounital (or equivalently, coaugmented strictly counital) dg and $A_\infty$-coalgebras. Then there exists a dg-coalgebra wihich is not weakly equivalent to zero, but whose cohomology coalgebra is zero. But there is only one (namely, zero) $A_\infty$ coalgebra structure on a zero vector space. – Leonid Positselski Nov 26 '18 at 0:54
• Thanks, Leonid. That settles it pretty decisively. Regarding your question, I've seen this stated without proof in at least two places, and I imagined it would be easy: the way the HTT is proven in Loday-Vallette is that a "homotopy retract" from a chain complex V to W induces a map of co-operads $B\mathrm{End}(V)\to B\mathrm{End}(W)$ given by a sum over trees, from which the result follows. But "turning the picture upside down" should give instead a map $B\mathrm{coEnd}(V)\to B\mathrm{coEnd}(W)$ between bar constructions on coendomorphism operads. But I haven't checked anything carefully. – Dan Petersen Nov 26 '18 at 8:09

We worked out some answers to this question in our paper: arXiv:1904.03585 (Edit: the following answer refers to v1 of the paper on the arXiv!)

Here's the short version. There are two possible natural definitions of an $$A_\infty$$-coalgebra:

(A) It is a graded vector space $$V$$ with maps $$V \to V^{\otimes n}$$, $$n \geq 1$$, satisfying identities exactly dual to those defining an $$A_\infty$$-algebra.

(B) It is a graded vector space $$V$$ and a square zero derivation of the tensor algebra on $$V$$.

They are not equivalent; (B) is stronger than (A). In the paper we call them naive and genuine $$A_\infty$$-coalgebras. There is a homotopy transfer theorem for naive $$A_\infty$$-coalgebras: for any naive $$A_\infty$$-coalgebra $$C$$ there is a transfered naive $$A_\infty$$-structure on $$H(C)$$ which is quasi-isomorphic to the one on $$C$$. This is false for genuine $$A_\infty$$-algebra structures in general.

We say that an $$A_\infty$$-coalgebra $$C$$ is conilpotent if it admits an exhaustive filtration of the form $$0 = F_0 C \subseteq F_1C \subseteq F_2C \subseteq \ldots$$ which is compatible with the coalgebra structure. We call such a filtration a positive filtration. In the conilpotent case, the notions of genuine and naive $$A_\infty$$-coalgebra coincide. Any filtered quasi-isomorphism of positively filtered $$A_\infty$$-coalgebras is a weak equivalence. If $$C$$ is a conilpotent $$A_\infty$$-coalgebra equipped with a positive filtration, and $$C \to V$$ is a filtered quasi-isomorphism of chain complexes, then it's possible to transfer the $$A_\infty$$-coalgebra structure on $$C$$ to an $$A_\infty$$-structure on $$V$$, so that the $$A_\infty$$-morphism $$C \rightsquigarrow V$$ is a filtered quasi-isomorphism and hence a weak equivalence. It follows that although $$C$$ is in general not weakly equivalent to $$H(C)$$ for any transferred structure, there is always a weak equivalence from $$C$$ to $$H(\operatorname{Gr} C)$$ with a transferred $$A_\infty$$-coalgebra structure on $$H(\operatorname{Gr} C)$$. If we fix a positive filtration on $$C$$ then the $$A_\infty$$-coalgebra $$H(\operatorname{Gr} C)$$ with its transferred structure is uniquely determined up to a noncanonical filtered $$A_\infty$$-isomorphism, and it deserves to be called the filtered minimal model of $$C$$.

• Usually (A) also comes with the condition that the operators are locally of finite support, meaning that for each $v\in V$, only finitely many co-operations are non-zero. This is indeed the definition (B). I agree that (A) is naive, but do people really use that definition more than (B)? – Pedro Tamaroff Jan 27 at 17:09
• All three notions - naive, genuine, and conilpotent infinity-coalgebras - can be found in the literature being referred to as infinity-coalgebras without any modifiers. This was a bit surprising to me. In the updated version we changed the term "genuine" to "locally finite", and removed the modifier "naive" completely. It seemed to imply a moral judgement which wasn't really necessary, and it turned out that in the updated version we get the most traction out of the "naive" definition... – Dan Petersen Jan 28 at 20:10
• Agreed. The "naive" one is obtained by taking the completed tensor algebra. Perhaps that helps with a name. I also agree removing the modifier is a good idea. At any rate, kudos for a really interesting paper. :) – Pedro Tamaroff Jan 28 at 20:37