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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 5.2.2.17) 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.

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    $\begingroup$ 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
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    $\begingroup$ 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
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    $\begingroup$ 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

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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.

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  • $\begingroup$ 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

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