# Is there an analog of Kan's $Ex^\infty$ functor for quasicategories?

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.

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 $$\mathrm{sd}^k$$).
• 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. – Saal Hardali Dec 11 '19 at 23:13
• 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): $\Delta \xrightarrow{d_1} \Delta$ is a fibration, and $\Lambda^1 \to \Delta$ is a weak equivalence, but the pullback $\partial \Delta \to \Delta$ is not a weak equivalence. – Tim Campion Feb 11 at 14:59