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 $\mathrm{sd}^k$).

no, the there can be no fibrant replacement functor on $sSet^+$ which preserves finite limits and fibrations. $\endgroup$1more comment