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Is there a fibrant replacement functor in the Joyal model structure which can be described non-recursively, like $Ex^\infty$ for the Quillen model structure? I believe another way to put this is to ask: is there a fibrant replacement functor in the Joyal model structure which is a right adjoint, or else a colimit of functors which are right adjoints?

What I mean by "non-recursive"

Garner's small object argument certainly provides a functorial fibrant replacement which is simple enough to describe, but the description is still recursive. I think one could hope for something better.

Let me illustrate this in the setting of the Quillen model structure. In this setting Garner's construction also provides a fibrant replacement functor $G$, but I think Kan's $Ex^\infty$ functor is clearly "simpler" than $G$. There is a ``closed-form" description of $Ex^\infty X$: an $n$-cell consists of a map $\mathrm{sd}^k\Delta^n \to X$ for some $k$, where $\mathrm{sd}$ is the subdivision functor. Whereas $GX$ can seemingly only be described recursively: an $n$-cell of $GX$ is the result of some sequence of horn-fillers being pasted in and possibly identified.

You might object that describing the $Ex^\infty$ functor does require some recursion, in that we need to consider iterates of the subdivision functor $\mathrm{sd}$. I think the crucial distinction is that this recursion is independent of $X$ -- we only need to consider iterated subdivisions in a small set of universal cases -- the simplices $\Delta^n$. Whereas to compute $GX$ we need to perform a recursion separately for each $X$ we consider.

Well -- that's a bit of a lie. I believe $G$ commutes with filtered colimits, so we really only need to compute $GX$ on the small set of all finite simplicial sets $X$, and then extend it formally. But it's already an undecidable problem to compute $GX$ for all finite $X$ because this includes the word problem for groups. So maybe the key distinction is that the recursive procedure involved in computing $Ex^\infty X$ is actually (easily!) decideable whereas the one involved in computing $G X$ is not.

Why adjointness would help

Suppose we have a fibrant replacement functor $R: \mathsf{sSet} \to \mathsf{sSet}$ which has a left adjoint $L: \mathsf{sSet} \to \mathsf{sSet}$. Then we necessarily have $(RX)_n \cong \mathrm{Hom}(L\Delta^n,X)$. So in order to compute $R$, we need only compute $L\Delta^n$ for each $n$. If $R$ is a colimit of functors $R_k$ with left adjoints $L_k$, then we have $(RX)_n = \varinjlim Hom(L_k\Delta^n, X)$, and again, we need only compute $L_k\Delta^n$ for each $k,n$. I think this would be the kind of description I'm looking for (and totally analogous to the case of $Ex^\infty$, which is the colimit of right adjoints to iterated subdivision $\mathrm{sd}^k$).

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  • $\begingroup$ Oh -- now I see this is essentially a duplicate of this question. $\endgroup$ – Tim Campion Dec 15 '16 at 13:02
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    $\begingroup$ Note that if there is a fibrant replacement functor which preserves finite limits and fibrations, then the model category is right proper. Since the Joyal model structure is not right proper, any functorial fibrant replacement which is a filtered colimit of right adjoints must fail to preserve fibrations. In particular, the Joyal model structure does not admit a fibrant replacement functor which is a filtered colimit of right Quillen functors. $\endgroup$ – Tim Campion Mar 24 '19 at 23:35
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    $\begingroup$ It could still be possible that such a fibrant replacement functor exists for the (co-)cartesian model structure on marked simplicial set right? That is, one that preserves finite limits and fibrations. $\endgroup$ – Saal Hardali Dec 11 '19 at 23:07
  • $\begingroup$ Also in what sense does computing $GX$ for finite $X$ solves the word problem for groups? Presumably you could calculate $GX$ but still find it very hard to compare to $GY$ for any other $Y$. Similarly to how difficult it is to compare arbitrary Kan complexes to each other. Most likely you meant something more subtle which I missed. $\endgroup$ – Saal Hardali Dec 11 '19 at 23:13
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    $\begingroup$ Just to record the disproof of right properness (which works either in Joyal or in $sSet^+$ and which I'll bet I learned from Alexander Campbell): $\Delta[1] \xrightarrow{d_1} \Delta[2]$ is a fibration, and $\Lambda^1[2] \to \Delta[2]$ is a weak equivalence, but the pullback $\partial \Delta[1] \to \Delta[1]$ is not a weak equivalence. $\endgroup$ – Tim Campion Feb 11 at 14:59

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