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I am studying the proof of the $\infty$-categorical version of the Barr-Beck theorem in Lurie's Higher Algebra, but there is a step of the proof that is puzzling me. In Lemma 4.7.3.13, a simplicial object $M_{\bullet}$ in the $\infty$-category $\mathrm{LMod}_T(\mathcal C)$ is constucted as $\operatorname{Bar}_T(T,M)_{\bullet}$.

My question is: what are the correct choices of $\mathcal C^{\otimes}$, $F_0$ and $f$ in Construction 4.4.2.7 in order to obtain the $\operatorname{Bar}_T(T,M)_{\bullet}$ used in the proof of Lemma 4.7.3.13?

As far as I understand, the idea is to apply the bar construction (Construction 4.4.2.7), to the coCartesian fibration of $\infty$-operads $\mathcal A^{\otimes} \to \mathcal{LM}^{\otimes}$ (here I am using $\mathcal A$ instead of $\mathcal C$ to avoid the conflict of notation with the $\mathcal C$ of Lemma 4.7.3.13) expressing $\mathcal C$ as weakly enriched over $\mathrm{End}(\mathcal C)^{\otimes}$ (that is, with $\mathcal A^{\otimes}_{\mathfrak m} \simeq \mathcal C$ and $\mathcal A^{\otimes}_{\mathfrak a} \simeq \mathrm{End}(\mathcal C)^{\otimes}$). My problem is now how to define $F_0$ and $f$ (with the notations of Construction 4.4.2.7) in order to get $\operatorname{Bar}_T(T,M)_{\bullet}$, and even how to make sense of the "bimodule instance" of $T$ in that expression.
To be more precise, $F_0$ should determine two bimodules in $\mathcal C$, but since $\mathcal C$ is only left tensored over $\mathrm{End}(\mathcal C)^{\otimes}$, I don't know how to make sense of the construction (in fact, I don't know how to make sense of the construction for any base $\infty$-operad other than $\mathcal O^{\otimes} = \mathcal{BM}^{\otimes}$). Moreover, even interpreting $M$ as some sort of bimodule (maybe with some kind of trivial right action, but I am just guessing), I can't find a way to interpret the second instance of $T$ in the expression $\operatorname{Bar}_T(T,M)_{\bullet}$. Or maybe I am just considering the wrong coCartesian fibration of $\infty$-operads to begin with, but then I have no clue about any other possible option.

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    $\begingroup$ I would guess that the relevant input is $\mathcal A^\otimes \to \mathcal{BM}^\otimes$ encoding $\mathcal C$ as left tensored over $End(\mathcal C)$ and right tensored over $*$, the map $f$ being the retraction discussed in Example 4.4.2.12. $\endgroup$ – Marc Hoyois Jan 6 '18 at 2:30
  • $\begingroup$ @MarcHoyois Yes, this definitely works, thank you. If you add this as an answer, I will accept it. $\endgroup$ – Stefano Ariotta Jan 6 '18 at 12:25

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