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

*

*Objects: Finite strings $[A_1, A_2, \dots, A_n]$ with $A_i \in \mathcal{A}$

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