# Why are these two functors adjoint pairs?

In the Braided Tensor Categories paper of 1993, Joyal and Street make a nontrivial claim with no proof. It is critical to their work and I can't figure out why it's true.

Let $$\mathbb{P}$$ be the permutation category; the objects are natural numbers, and $$\text{Hom}_{\mathbb{P}}(n,n) = S_n$$ (the symmetric groups). All other homsets are empty. There are two types of categories we can construct. We can construct this category (I'm not sure what to call it).

Let $$\mathcal{A}$$ be a category and suppose suppose $$\mathcal{D} \in \textbf{Cat}/\mathbb{P}$$. Define the category $$\mathcal{D}\int\mathcal{A}$$ where

1. Objects: Finite strings $$[A_1, A_2, \dots, A_n]$$ with $$A_i \in \mathcal{A}$$
2. For two strings $$[A_1, \dots, A_n]$$ and $$[B_1, \dots, B_n]$$, denoted as $$[A_i]$$ and $$[B_i]$$, $$\text{Hom}_{\mathcal{D}\int\mathcal{A}}\Big([A_i],[B_i]\Big) = \Big\{(\alpha, f_1, \dots, f_n) \mid f_i \in \text{Hom}_{\mathcal{A}}(A_i, B_{\alpha(i)}) \Big\}$$ Here $$\alpha \in S_n$$ is a morphism obtained as the image of some morphism $$f: X \to X$$ in $$\mathcal{D}$$ via $$\Gamma$$. Finally, we allow no morphisms between two different strings of different length.

For Joyal and Street, the purpose of the above category is to set $$\mathcal{D} = \mathbb{B}$$, the braid category, which does in fact have a functor $$\Gamma: \mathbb{B} \to \mathbb{P}$$. It sends braids to their underyling permutations. We can also form the category below.

Let $$\mathcal{A}, \mathcal{B}$$ be categories. Define the "generalized functor category" $$\{\mathcal{A}, \mathcal{B}\}$$ as the category with objects $$(n, F: \mathcal{A}^n \to \mathcal{B})$$ whose morphisms are $$\text{Hom}_{\{\mathcal{A}, \mathcal{B}\}}((n, T), (m,S)) = \begin{cases} \{(\sigma, \eta: \sigma\cdot T \to S) \} & \text{if } n = m\\ \varnothing & \text{if } n \ne m. \end{cases}$$ Here $$\sigma \in S_n$$, and $$\eta: \sigma \cdot T \to S$$ is a natural transformation from the functor $$\sigma \cdot T$$, defined pointwise as $$\sigma \cdot T(A_1, A_2, \dots, A_n) = T(A_{\sigma(1)}, \dots, A_{\sigma(n)})$$ to the functor $$S$$. Note that $$\{\mathcal{A}, \mathcal{B}\}$$ is always equipped with a functor $$\Gamma: \{\mathcal{A},\mathcal{B}\} \to \mathbb{P}$$ where $$\Gamma(n, T: \mathcal{A}^n \to \mathcal{B}) = n \qquad \Gamma(\sigma, \eta: \sigma\cdot T \to S) = \sigma.$$

Apparently, these constructions may be phrased as functors: $$(-)\int A : \textbf{Cat}/\mathbb{P} \to \textbf{Cat} \qquad \{A, (-) \}: \textbf{Cat}\to \textbf{Cat}/\mathbb{P}$$ and the claim is that these functors are adjoint pairs; it's extremely critical to their work. Does anyone have a way of seeing these are adjoint pairs? I have no idea why and I can't seem to figure it out. I'm thinking there are some deep category theory tricks which JS use to justify this but I don't know, and they just point to G.M. Kelly's work. But if what they're saying is true, there should be a proof; I can't accept "Kelly did it" as a proof because that's not very evident: Kelly's work is way, way, WAY more general (on p. 74, 75, which are the pages JS cite) than what they're doing here.

Here's an answer, we can just verify that $$(-)\int \newcommand\A{\mathcal{A}}\A$$ is left adjoint to $$\newcommand\set{\left\{{#1}\right\}}\set{\A,(-)}$$ by direct computation.

Side note, the definition of $$\mathcal{D}\int A$$ as given doesn't make sense for general $$\mathcal{D}$$, we need to be keeping track of which object of $$\mathcal{D}$$ over $$n\in \mathbb{P}$$ we're at, so objects should be pairs $$(X,[A_i])$$ of an object over $$n$$ and a string of $$n$$ objects of $$\A$$.

Suppose $$F: \newcommand\D{\mathcal{D}}\D\int \A\to \newcommand\B{\mathcal{B}}\B$$ is a functor. By definition, this consists of the following data, for each pair $$(D,[A_i])$$ of an object of $$\D$$ and a string of length $$n$$, an object $$F(D,[A_i])$$ in $$\B$$, and for each morphism $$(\alpha, f_1,\ldots,f_n) : (D,[A_i])\to (D',[B_i])$$, an appropriate morphism $$F(\alpha,f_1,\ldots,f_n)$$ such that the composition law is satisfied: $$F(\beta,g_1,\ldots,g_n) \circ F(\alpha,f_1,\ldots, f_n) = F(\beta\alpha,g_{\alpha(1)}f_1,\ldots,g_{\alpha(n)}f_n).$$

Side note: This looks a lot like a categorified wreath product to me.

In particular, this is a sort of twisted product, so we should expect it to be left adjoint to some sort of twisted hom, which is what $$\set{\A,\B}$$ should be.

Let $$\pi : \D\to \mathbb{P}$$ be the structure map for $$\D$$.

On the other hand, if $$G : \D\to \set{\A,\B}$$ is a functor of categories over $$\mathbb{P}$$, then this consists of for each $$D\in \D$$, a choice of an object $$GD = (n,G_D : \A^n\to \B)$$, where we must have $$n = \pi D$$, and for each $$\alpha :D\to D'$$, a morphism $$(\sigma, \eta_\alpha : \sigma \cdot G_D\to G_{D'})$$, where we are forced to have $$\sigma = \pi(\alpha)$$.

Now the functors $$G_D$$ for each $$D$$ themselves consist of the following data, for each string $$[A_i]$$ of $$n$$ objects of $$\A$$, an object $$G_D([A_i])$$ of $$\B$$, and for each morphism $$(f_1,\ldots,f_n) : [A_i]\to [B_i]$$ a morphism $$G_D(f_1,\ldots,f_n) : G_D([A_i])\to G_D([B_i])$$, subject to the composition rule.

The $$\eta_\alpha$$s are a family of morphisms $$G_D([A_{\sigma(i)}])\to G_{D'}([A_i])$$ such that for all $$(f_1,\ldots,f_n)$$ the following diagram commutes $$\require{AMScd} \begin{CD} G_D([A_{\sigma(i)}]) @>G_D(f_{\sigma(1)},\ldots,f_{\sigma(n)})>> G_D([B_{\sigma(i)}]) \\ @V\eta_{\alpha}VV @VV\eta_{\alpha}V \\ G_{D'}([A_i]) @>G_{D'}(f_1,\ldots,f_n)>> G_{D'}([B_i]) \\ \end{CD}$$

Of course the diagram still commutes if we permute the maps by $$\sigma^{-1}$$, and we get $$\require{AMScd} \begin{CD} G_{D}([A_i]) @>G_{D}(f_1,\ldots,f_n)>> G_{D}([B_i]) \\ @V\eta_{\alpha}VV @VV\eta_{\alpha}V \\ G_{D'}([A_{\sigma^{-1}(i)}]) @>G_{D'}(f_{\sigma^{-1}(1)},\ldots,f_{\sigma^{-1}(n)})>> G_{D'}([B_{\sigma^{-1}(i)}]) \\ \end{CD}$$

However, there's actually an indexing discrepancy between the indices of the $$B_i$$ here and the $$B_i$$ in the definition of $$\D\int \A$$. Namely, the maps from $$[A_i]$$ to $$[B_i]$$ in $$\D\int\A$$ are $$f_i: A_i\to B_{\sigma(i)}$$.

Thus we see that if we reindex accordingly, for a fixed choice of $$(\alpha,f_1,\ldots,f_n)$$, with $$f_i:A_i\to B_{\sigma(i)}$$, the diagonal map is a map $$G_D([A_i])\to G_{D'}([B_i])$$.

In other words, we have recovered the data of a functor $$F:\D\int\A\to \B$$, for each object $$(D,[A_i])$$ we define $$F(D,[A_i])=G_D([A_i])$$, and for each $$(\alpha,f_1,\ldots,f_n)$$, we define $$F(\alpha,f_1,\ldots,f_n)$$ to be the diagonal map constructed above.

Conversely, if we start with such a functor $$F$$, we can go backwards and produce the data of a functor $$G$$. We define $$G_D$$ to be $$F(D,-)$$, with $$G_D$$ defined on morphisms by $$G_D(f_1,\ldots,f_n) = F(1_D,f_1,\ldots,f_n)$$. Then the natural transformations $$\eta_\alpha$$ should be $$F(\alpha,1_{A_{\sigma^{-1}(1)}},\ldots,1_{A_{\sigma^{-1}(n)}})$$. (I think, not quite sure if the $$\sigma^{-1}$$ is correct, but it's getting a bit late, so I'll just let you check that detail.)

This establishes a bijection of collections of functors. It shouldn't be too bad to show that it is a natural bijection.

End Note I feel like there ought to be a general categorical notion underlying both constructions, and if someone knows what that is, I'd love to hear about it.

• The construction is indeed sometimes called a "categorical wreath product". – Zhen Lin Jul 18 '20 at 14:19
• This looks great to me! I really appreciate you taking the time to answer this as well as figuring out what was going on despite my typos. Reading JS is hard, and this is very helpful, so thank you! – trujello Jul 19 '20 at 18:25