# Does the right adjoint of the category of simplices functor is “homotopicaly inverse” to the category of simplices functor?

### Short Version (the question)

Let $\text{Cat}$ be the category of (small) categories and $\text{sSet}$ the category of simplicial sets. There is a functor $\Gamma:\text{sSet}\to \text{Cat}$ that takes every simplicial set to its category of simplices. This functor has a right adjoint, say, $T$.

Question 1: Given a simplicial set $X$, is it true that $T(\Gamma(X))$ is weakly equivalent to $X$?

Question 2: Is the unit $\epsilon : 1\to T\circ \Gamma$ of the adjonction a weak equivalence of simplicial sets?

### Long Version (background, motivation, partial results etc.)

To fix notation, let $\Delta$ be the simplex category whose objects are the non-empty finite ordinals $$[n]=(0<1<...<n)$$ and morphisms are the weakly monotone maps. There is the nerve functor $$N:\text{Cat}\to \text{sSet}$$ $$N(C)_n = Func([n],C)$$

Where $[n]$ is viewed as a poset category. The nerve functor has a left adjoint $\tau_1$ and the composition $\tau_1 \circ N$ is naturally isomorphic to the identity. On the other hand, the composition $N\circ \tau_1$ loses a lot of information and in particular is not weakly equivalent to the identity. One solution to this "problem" is to apply barycentric subdivision twice before taking $\tau_1$. Thomason has shown that this does not lose homotopic information and used this to transport the Quillen model structure of $\text{sSet}$ to $\text{Cat}$.

Another "solution" is to use a different way to pass from $\text{sSet}$ to $\text{Cat}$, given by the category of simplices functor $\Gamma:\text{sSet}\to \text{Cat}$. It is known that there is a natural weak equivalence $N(\Gamma(C))\to C$. Hence, $\Gamma$ is in some vague sense inverse to $N$ on the homotopical level. Now, the functor $\Gamma$ itself has a right adjoint which is given by the following explicit formula $$T(C)_n = Func(\Delta /[n], C)$$

Where we use the notation $\Delta/[n]$ for the over-category and the simplicial maps come from the functoriality of the over-category construction (namely, by pullback).

Now, one way one might hope to answer the first question affirmatively, is to compare $N(\Gamma(C))$ with $T(\Gamma(C)$. In fact, one might hope that $N(C)$ is weakly equivalent to $T(C)$ for every category $C$. A candidate for such a weak equivalence is given by the Bausfield-Kan map $\Delta/[n] \to \Delta$ which can be viewed as a morphism of cosimplicial categories. This gives a map $N(C)\to T(C)$ which one might hope to be a weak equivalence.

Some evidence that this might be true in general, come from the special case where $C$ is a poset. In this case, every functor $\Delta/[n] \to C$ factors uniquely through the "posetization" of $\Delta/[n]$, which is the poset category of non-degenerate simplices of $\Delta^n$. The nerve of this category is just the barycentric subdivision of the n-simplex $\Delta^n$. Thus, we obtain

$$T(C)_n = Func(\Delta/[n],C) = Func(Pos(\Delta/[n],C) = Hom(sd(\Delta^n),N(C))= Hom(\Delta^n, Ex(N(C)) = Ex(N(C))_n$$

and it follows that $T(C)$ is just the Kan extension of $N(C)$ which is known to be weakly equivalent to $N(C)$ (this was a bit sketchy, but hopefully clear enough)

This indeed true and is discussed in depth by Latch, Thomason and Wilson in a paper called Simplicial Sets from Categories. Your question is answered in Corollary 4.7 and relies on Theorem 4.1, the main result of the paper which says that $N \to T$ is indeed a weak equivalence. The argument can also be adapted to give a very neat combinatorial proof that $\mathrm{id} \to \mathrm{Ex}$ is a weak equivalence.
• I don't know, but my guess would be not. A necessary condition (and essentially sufficient) is that if $A \to B$ is a monomorphism of simplicial sets, then every pushout along $\Gamma A \to \Gamma B$ in $\mathrm{Cat}$ is a homotopy pushout. There are counterexamples for this in the case of $c \mathrm{Sd}$ and I suspect that some of them will also fail for $\Gamma$ which is only a fattened version of $c \mathrm{Sd}$ after all. – Karol Szumiło Aug 18 '14 at 7:45