I've been trying to read Jacob Lurie's approach to the bar-cobar constructions (Higher algebra §5.2 in the 2014-09 version) but I don't recognize what I know about these constructions. I wonder if anybody has already digested this.

Before I can ask my question, I need to recall some elements of Lurie's work.

In 5.2.1 he considers the twisted arrow category $TwAr({\cal C})$ for ${\cal C}$ an $\infty$-category. Essentially the objects are maps $f:X\to Y\in {\cal C}$ and maps from $f$ to $f':X'\to Y'$ are pairs of maps $(X\to X',Y'\to Y)$ such that the obvious square commutes.

$(f:X\to Y) \mapsto (X,Y)$ defines a functor $TwAr({\cal C})\to {\cal C}\times {\cal C}^{op}$. This functor is the right fibration classified by the mapping space functor : $Map:{\cal C}^{op}\times {\cal C}\to {\cal S}$ (where ${\cal S}$ is the $\infty$-category of Kan complexes).

Then, Lurie develops the formalism of bimodules and adjunctions representing them. A pairing is a functor $M:{\cal C}\times {\cal D}\to {\cal S}$, we say that it is representable in the first variable if there exists a functor $f:{\cal C}\to {\cal D}$ such that $M(X,Y) = Map(f(X),Y)$. Same thing for the second variable with a functor $g:{\cal D}\to {\cal C}$. If $M$ is representable in both variables, the functor $f$ and $g$ are adjoint.

In 5.2.2, ${\cal C}$ is assumed symmetric monoidal and Lurie considers monoids in $TwAr({\cal C})$. Such a monoid is a map $f:A\to C$ from a monoid $A$ to a comonoid $C$ satisfying some unusual condition (*): essentially, the map $\Delta_Cfm_A:A\otimes A\to A\to C\to C\otimes C$ must be equal to $f\otimes f$.

Then, essentially because the functor $Map$ is always lax monoidal, we get a pairing $Mon(TwAr({\cal C}))\to Mon({\cal C})\times Mon({\cal C}^{op})$. The main result (Thm is that this pairing is representable in both variables when ${\cal C}$ satisfies some mild assumptions. The corresponding adjunction is the bar-cobar adjunction for monoids and comonoids in ${\cal C}$. It is written $$ Map_{Mon({\cal C})}(A,Cobar(C)) = Map_{coMon({\cal C})}(Bar(A),C). $$

Remark : this result is perfectly symmetric in ${\cal C}$ and ${\cal C}^{op}$ replacing one by the another gives the exact same adjunction.

Now, here is what I find strange: classically the bar-cobar adjunction is going the other way (Bar is right adjoint and Cobar left adjoint) so why the change here ?

Variation on the same question : the classical the bar-cobar adjunction is related to "twisting cochains" which are Maurer-Cartan elements in the convolution dg-algebra $[C,A]$, hence some kind of maps from the comonoid $C$ to the monoid $A$. But in Higher Algebra the adjunction is related to maps from $A$ to $C$. Is there a way to relate twisting cochains $\alpha:C\to A$ to maps $f:A\to C$ satisfying the condition (*) ?

Last remark : a version of the convolution algebra does appear in Lurie's approach, provided we consider comonoids in $TwAr({\cal C})$ instead of monoids. They correspond to maps $f:C\to A$ from a comonoid $C$ to a monoid $A$ which are idempotent for the convolution product in the (external) convolution algebra $Map(C,A)$ (this is the condition dual to (*)).

I'm puzzled... I'll be glad if anybody has some insight into this.

  • 1
    $\begingroup$ I think in "where $\mathcal C$ is the $\infty$-category of Kan complexes" you mean $\mathcal C$ to be $\mathcal S$. $\endgroup$ Dec 18, 2014 at 1:30
  • 3
    $\begingroup$ I don't know how correct this is, but suppose you only want to think about maps from a cofibrant object to a fibrant object. We have cofibrant replacement functors ($Cobar\ Bar$ and $id$) and fibrant replacement functors ($id$ and $Bar\ Cobar$) and so using these we have $$Hom(Bar(A),Bar\ Cobar(C)) = Hom(Cobar\ Bar(A), Cobar(C))$$ which uses the regular adjunction where $Cobar$ is left and $Bar$ is right to give a homotopy adjunction where the roles are reversed. $\endgroup$ Dec 18, 2014 at 4:13
  • $\begingroup$ To Gabriel. Thanks for your answer. You are using that the BarCobar adjunction is a Quillen equivalence (hence both functors are left and right adjoint). I know such result when dg-coalgebras are assumed conilpotent (Lefevre-Hasegawa PhD) and this require fancy weak equivalence on coalgebras (in particular this is different then comonoids in the category of complexes up to qis). I don't know any result involving non-conilpotent coalgebras. Lurie does not use any conilpotency hypothesis and does not prove that the BarCobar adjunction is an equivalence. So I don't think your argument works here. $\endgroup$ Dec 18, 2014 at 11:24
  • $\begingroup$ @MathieuANEL I agree about conilpotency. But in the conilpotent case you can view $bar$ between $(alg, qi)$ and $(coalg, qi)$ as the composition of $(alg, qi)\xrightarrow{bar}(coalg, fancy)\xrightarrow{id}(coalg,id)$ where the first arrow is a Quillen equivalence and the second arrow is a left Quillen functor (obviously likewise for cobar). So I think in the conilpotent case what I said might still be okay. $\endgroup$ Dec 18, 2014 at 12:32
  • 1
    $\begingroup$ I am interested in figuring out which algebras still satisfy CobarBar = id. I believe certain kinds of complete algebras will still satisfy this condition. $\endgroup$
    – Joey Hirsh
    Dec 28, 2014 at 23:09

1 Answer 1


I really like this question, I've been trying to sort out some of these ideas for a little while. I don't know the answer to your questions about conilpotence and twisting morphisms vs twisted arrows. I do have reason to believe that twisted arrows between A and C are the same as the twisted arrows from A to conil(C) but I don't know how to prove that.

I think Gabriel's answer is worth expanding on. Lurie is describing an adjunction between infinity categories: Alg and Coalg. The bar and cobar construction you mention are between categories---let me denote them by ALG and COALG [and I mean conilpotent coalg]---and so must be equipped with weak equivalences in order to induce functors on the infinity categories.

To model Alg, we equip ALG with quasi-isomorphisms. To model Coalg [or rather, conilCoalg], we must equip COALG with quasi-isomorphisms too. However, the classical bar and cobar construction are not homotopical between these relative (or model) categories.

However, we have a second class of weak equivalences on COALG---you called them fancy---that makes this adjunction into a Quillen pair, and as you point out, this Quillen pair is an equivalence. Gabriel's point, though, is that (COALG, fancy) left localizes to (COALG, quasi-iso). Conjugating this localization by the ``bar-cobar as equivalence between (ALG, quasi-iso) and (COALG, fancy)" will show you that this left adjoint from (COALG, fancy) to (COALG, quasi-iso) models Lurie's infinity left adjoint from Alg to Coalg, and it is defined by something that looks like the classical bar construction.

  • $\begingroup$ Thank to you & Gabriel for your answers. It sounds like a good explanation. But is it obvious that the localisation from (COALG,fancy) to (COALG,qis) is a left localisation ? $\endgroup$ Dec 21, 2014 at 1:20
  • $\begingroup$ I think these references explain better than I could in this box: To see that fancy weak equivs are contained in quasi-isos, check out arxiv.org/abs/1411.5533 Proposition 2.5. To see that both structures have the same cofibrations, check out arxiv.org/abs/1411.5526 Lemma 3.10. $\endgroup$
    – Joey Hirsh
    Dec 28, 2014 at 23:06

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.