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Let $T$ be a monad on a concrete category $\mathcal{C}$, and $A$ an algebra over $T$. The bar construction is a simplicial object in the category $\mathcal{C}^T$ of algebras which we can think of a sort of "resolution" of $A$. Some of the arrows look like the following diagram: $$ \begin{array}{ccc} \cdots TTA \rightrightarrows TA \to A \end{array} $$ (I unfortunately cannot draw more arrows here. See the link above for a better picture.)

Now, is such a simplicial object a Kan complex, or at least a quasicategory? Is there a filling condition for horns, in general? If not, what would be a counterexample?

Any reference would also be welcome.

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  • $\begingroup$ Are you assuming that the objects of your category have underlying sets? $\endgroup$
    – Fernando Muro
    Commented Apr 5, 2018 at 15:22
  • $\begingroup$ @FernandoMuro I could. What would that imply? $\endgroup$
    – geodude
    Commented Apr 5, 2018 at 15:22
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    $\begingroup$ I don't know, maybe the answer depends on the properties of the 'underlying set' functor. It's definitely yes if this functor factors through the category of groups. Actually, I cannot make sense of your question unless you take underlying sets, since I don't know what the Kan condition is for simplicial sets in an arbitrary category. $\endgroup$
    – Fernando Muro
    Commented Apr 5, 2018 at 15:27
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    $\begingroup$ If C=Sets and T is the monoid for pointed sets, and $A=(A,a_0)$, then the Bar construction, as a simplicial set, is the union of $A \times \Delta[0]$ (i.e. const. simp. set on A) with $\Delta[1]$ along $\Delta[0]$ identified as $a_0 \in A$. It is not Kan, but it is a quasicategory. $\endgroup$
    – Chris Schommer-Pries
    Commented Apr 5, 2018 at 16:26
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    $\begingroup$ This paper gives a definition of the Kan condition for simplicial objects in any regular category, and proves that every simplicial object is Kan if and only if the category is Mal'cev, i.e. if and only every reflexive relation is an equivalence relation. This generalises Fernando Muro's observation about the functor factorising through the category of groups. $\endgroup$
    – Arnaud D.
    Commented Apr 5, 2018 at 19:19

1 Answer 1

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Here is one sufficient (not necessary) condition to get a quasi-category: if $\mathcal{C}$ has pullbacks and $T$ is cartesian. In particular, this implies that given any algebra $a:TA\to A$ in $\mathcal{C}$, $$ \begin{array}{ccc} TTTA & \rightarrow & TTA\\ \downarrow^\mu & & \downarrow^\mu \\ TTA & \rightarrow & TA \end{array} $$

is a pullback diagram, which by its universal property gives a unique composition (not even up to homotopy). Therefore the bar construction for cartesian $T$ is the nerve of a category.

More interesting discussions can be found in the comments of this post on the n-category café.

More answers are welcome!

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