Consider the nerve functor $N : \mathbf{Cat} \to \mathbf{sSet}$; it is fully faithful, preserves finite limits, products and exponentials, etc. I am wondering if it additionally preserves whatever dependent products exist in $\mathbf{Cat}$.

For instance, even though $\mathbf{Cat}$ is not locally Cartesian closed, we do have right adjoints to pullback along product projections. In this case, I'm curious whether the nerve functor takes these to dependent products in the category of simplicial sets.

If these dependent products are indeed preserved, I am interested in understanding the general conditions under which other "nerve-like" situations exhibit the same behavior, meaning functors of the form $X \mapsto \mathbb{D}[i -,X]$ for $i : \mathbb{C}\to \mathbb{D}$, primarily considering the case where $i$ is a subcategory inclusion.

EDIT: I was hypothesizing that this property would follow from the following conditions:

  1. the subcategory that generates the nerve is dense, and thence the nerve is fully faithful
  2. the nerve functor has a left adjoint
  • 2
    $\begingroup$ I'm only half-remembering this and also mixing ordinary categories with $\infty$-categories, but there's a characterization of exponentiable functors: they're the Conduche functors, and there's an $\infty$-categorical analog of this. The thing is, if I recall correctly, the way one computes the dependent product for a Conduche functor is to compute it in $sSet$ and then verify that the result is in fact a (quasi)category. So I think the answer to your question is yes. $\endgroup$
    – Tim Campion
    Nov 3, 2019 at 20:19
  • $\begingroup$ I believe I have a proof in the general case, so I'll wait a day or so and post it. $\endgroup$ Nov 5, 2019 at 17:18
  • 1
    $\begingroup$ It seems to me it should be true as soon as the subcategory is dense. Does your proof need a left adjoint? $\endgroup$ Nov 6, 2019 at 4:43
  • $\begingroup$ Hi Mike, my proof uses a left adjoint but it may be that another proof can be found which doesn't use it. I just used the left adjoint a bunch to moves things between the left and the right easily. $\endgroup$ Nov 6, 2019 at 17:22

1 Answer 1


I have (I think) a proof under the assumptions I mentioned in my question, that the nerve is built from a dense subcategory and that there is a corresponding realization functor as a left adjoint. I would very much appreciate any readers of this answer to comment or edit if they find errors in my reasoning; and if you have an idea for a better proof, to post your own answer which I will accept. Thank you!

Let's fix some notation. We will consider a fully faithful, dense inclusion $K : C_K\hookrightarrow C$. This generates a nerve functor $N(c) = C[K\bullet,d] : C \to \widehat{C_K}$; because $K$ is dense, the nerve functor is fully faithful. We assume a left adjoint to $N$, the realization $|-|: \widehat{C_K} \to C$. I am only going to use this left adjoint as a crutch at some points to transpose along; it may be that there is a better proof which doesn't use it.

We will now check that a dependent product in $C$ is taken to a dependent product in $\widehat{C_K}$. We fix $f : d\to c,g : e\to d$ and wish to check that $\Pi_fg : C/c$ (if it exists) is taken to $\Pi_{Nf}Ng : \widehat{C_K}/Nc$.

We note that the slice of the presheaf category is in fact another presheaf category, the category of presheaves $\widehat{C_K/Nc}$ on the category of elements $C_K/Nc$. Therefore, we may check the isomorphism by probing against "representables", fixing $x : C_K/Nc$; in this case, $x$ is concretely a map $x : K(\partial_0x)\to c$ in $C$. Let us write $\mathcal{E}/X$ for each slice $\widehat{C_K/X}$.

First, we transpose: \begin{align*} \mathcal{E}/Nc[Y(x),N(\Pi_fg)] &\cong C/c[|Y(x)|, \Pi_fg] \end{align*}

We note that $|Y(x)| : C/c$ is actually just the map $x$ described above; this is a consequence of the density of $K$.

\begin{align*} &\cong C/c[x,\Pi_fg] \\ &\cong C/d[f^*x, g] \end{align*}

Now, because $N$ is fully faithful, we may just put it everywhere:

\begin{align*} &\cong \mathcal{E}/Nd[N(f^*x),Ng] \end{align*}

But $N$ is continuous; so we may commute it into the pullback:

\begin{align*} &\cong \mathcal{E}/Nd[Nf^*Nx,Ng] \\ &\cong \mathcal{E}/Nc[Nx,\Pi_{Nf}Ng] \end{align*}

But $Nx$ is isomorphic to $Yx$ (this follows again from the assumption that $K$ is fully faithful and dense), so we are done.


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