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(This question was originally asked https://math.stackexchange.com/questions/725421/an-isomorphism-of-categories, with no affirmative answer there.)

Let $C$ be an (finite) extensive category with terminal object $1$. Let $I$ be an index category. Let $j: \mathrm{FinSet}\to C$ be the coproduct-preserving functor induced by $*\mapsto 1$. Let $f$ be a finite sets-valued presheaf on $I$, that is a functor $I^{op}\to \mathrm{FinSet}$. Let $F$ be the associated category of elements.

There is a canonical functor $[I^{op},C]→[F^{op},C]$, induced by the projection $F\to I$. It has a left adjoint that sends the terminal object in $[F^{op},C]$ to $j(f)$. So there is a canonical functor

$$ [F^{op}, C]\to[I^{op}, C]/j(f). $$

My question: is this an isomorphism of categories? I have a not-so-vigorous argument why this should hold, I would also like to known if this is proven some where (at least for $C=Sets$).

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    $\begingroup$ In the case $\mathcal{C} = \mathbf{Set}$ this is well known and probably due to Grothendieck. $\endgroup$
    – Zhen Lin
    Mar 31, 2014 at 14:38
  • $\begingroup$ @ZhenLin, MaMing, I have no idea what you are talking about. First, there are typing errors in the question: a) an adjoint to $[F^{op},C] \rightarrow [I^{op},C]$ has type $[I^{op},C] \rightarrow [F^{op},C]$, b) you cannot apply $j$ to $f$, so perhaps you mean $j\circ f$, right?, c) it is not clear which category of elements you have in mind (i.e. you cannot compose $I^{op} \rightarrow C$ with $F \rightarrow I$; do you compose with the dual of the projection?). (cont) $\endgroup$ Mar 31, 2014 at 16:40
  • $\begingroup$ Second, how is $j$ defined? Does it send everything from $\mathrm{FinSet}$ to the terminal $1$ (what does $*$ mean?)? If so, then $j$ is terminal in $[\mathrm{FinSet}, C]$ and $j\circ f$ is terminal in $[I^{op}, C]$. Thus $[I^{op}, C]/{(j\circ f)} = [I^{op}, C]$ --- I do not understand how, in the general case, could $[I^{op}, C]$ be isomorphic to $[F^{op}, C]$. For example, if $I = 1$ then $F$ is just an arbitrary finite set. $\endgroup$ Mar 31, 2014 at 16:41
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    $\begingroup$ @MichalR.Przybylek There is lately a bad habit of writing $*$ for terminal objects. There is a unique (up to unique isomorphism) coproduct-preserving functor $\mathbf{FinSet} \to \mathcal{C}$ that also preserves the terminal object. $\endgroup$
    – Zhen Lin
    Mar 31, 2014 at 16:49
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    $\begingroup$ @ZhenLin Thanks for pointing out, I found it was asked mathoverflow.net/questions/23887/slices-of-presheaf-categories and mathoverflow.net/questions/89635/… $\endgroup$
    – Ma Ming
    Mar 31, 2014 at 17:03

2 Answers 2

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I think I have finally understood your question. $\require{AMScd} \newcommand{\mor}[3]{#1 \colon #2 \rightarrow #3}% \newcommand{\catl}[1]{\mathbb{#1}}% \newcommand{\catw}[1] {\mathbf{#1}}%$

Since $\catw{FinSet}$ is a free finite coproduct cocompletion of the terminal category $1$, every functor $\mor{A}{1}{\catl{C}}$ to a category $\catl{C}$ with finite coproducts, uniquely extends to a coproduct-preserving functor $\mor{(-) \otimes A}{\catw{FinSet}}{\catl{C}}$, which assigns to a finite set $X$ the tensor $X \otimes A = \coprod_X A$. In particular if $\catl{C}$ has terminal object $1$, there is a functor: $$\mor{(-) \otimes 1}{\catw{FinSet}}{\catl{C}}$$ which was denoted by $j$ in the question.

I shall rephrase your statement in terms of copresheaves, because I think it is a more natural setting (to move back to your original question just put $\catl{J} = \catl{I}^{op}$). So, let: $$\mor{f}{\catl{J}}{\catw{FinSet}}$$ be a copresheaf on finite sets. There is a corresponding discrete opfibration given by the (op)Grothendieck construction: $$\mor{\pi_f}{\int f}{\catl{J}}$$ The total category $\int f$ may be expressed via the coend: $$\int f = \int^{j\in \catl{J}} f(j) \times j/\catl{J}$$ We have to show that: $$[\int f, \catl{C}] \approx [\catl{J}, \catl{C}]/{(-) \otimes 1}$$ By universal properties, there are isomorphisms: $$[\int^{j\in \catl{J}} f(j) \times j/\catl{J}, \catl{C}] \approx \int_{j\in \catl{J}}[f(j) \times j/\catl{J}, \catl{C}] \approx \int_{j\in \catl{J}}[j/\catl{J}, \catl{C}^{f(j)}]$$ If we assume that $\catl{C}$ is finitely extensive, then $\catl{C}^{f(j)} \approx \catl{C}/{\coprod_{f(j)} 1} = \catl{C}/{f(j) \otimes 1}$, and so: $$\int_{j\in \catl{J}}[j/\catl{J}, \catl{C}^{f(j)}] \approx \int_{j\in \catl{J}}[j/\catl{J}, \catl{C}/{f(j) \otimes 1}] \approx [(-)/\catl{J}, \catl{C}/{f(-) \otimes 1}]$$

There should be a high-level argument showing that: $$[(-)/\catl{J}, \catl{C}/{f(-) \otimes 1}] \approx [\catl{J}, \catl{C}]/{f(-) \otimes 1}$$ but I am not seeing it at the moment. In fact, there is a subtle problem here --- the functor: $$j \mapsto \catl{C}/{f(j) \otimes 1}$$ from the definition of extensivity is not the usual slice functor, which has type $\catl{J} \rightarrow \catw{Cat}$. Our functor has type $\catl{J}^{op} \rightarrow \catw{Cat}$ and is defined as a component-wise right adjoint to the usual slice functor.

Nonetheless, it is relatively easy to verify the claim.

In one direction let us assume, that we are given a functor $\mor{G}{\catl{J}}{\catl{C}}$ and a natural transformation $\mor{\alpha}{G}{f(-)\otimes 1}$. We shall define a natural family of functors $\mor{H_X}{X/\catl{J}}{\catl{C}/{f(X)\otimes 1}}$ as follows:

  • on objects: $H_X(X \overset{h}\rightarrow A) = (\catl{C}/{f(h)\otimes 1})(\lambda_A)$
  • on morphisms $A \overset{s}\rightarrow B$ such that $s \circ h = k$ under object $X$: as the unique factorisation of morphism $G(s)$ through pullback $H_X(k)$

In the other direction, let us assume that a natural family of functors $\mor{H_X}{X/\catl{J}}{\catl{C}/{f(X)\otimes 1}}$ is given. One may construct a functor $\mor{G}{\catl{J}}{\catl{C}}$:

  • $G(A) = \mathit{dom}(H_X(\mathit{id}_X))$
  • $G(A \overset{s}\rightarrow B) = \pi_{G(B)} \circ H_A(s)$, where $\pi_{G(B)}$ is the pullback projection on $G(B)$

and a natural transformation $\mor{\alpha}{G}{f(-)\otimes 1}$ by putting $\lambda_X = H_X(\mathit{id}_X)$.

Another way to see the above equivalence, is to apply the Grothendieck construction to functor $(-)/\catl{J}$ and to functor $\catl{C}/{f(-) \otimes 1}$. The first functor yields the domain fibration $\mor{\delta_0}{\catl{J}^2}{\catl{J}}$, and the second functor yields a "subfibration" $\mor{\delta_1}{\int \catl{C}/{f(-) \otimes 1}}{\catl{J}}$ of the codomain fibration $\catl{C}^2 \rightarrow \catl{C}$. Natural transformations $[(-)/\catl{J}, \catl{C}/{f(-) \otimes 1}]$ are tantamount to fibred functors $H$: $$\begin{CD} \catl{J}^2 @>{H}>> \int\catl{C}/{f(-) \otimes 1}\\ @V\delta_0VV @VV\delta_1V \\ \catl{J} @= \catl{J} \end{CD}$$ Functor $H$ maps a diagram: $$\begin{CD} X @>{h}>> A\\ @V{\mathit{id}_X}VV @VV{\mathit{id}_A}V \\ X @>{h}>> A \end{CD}$$ to the diagram: $$\begin{CD} G(X) = H(X) @>{G(h) = H(h)}>> G(A) = H(A)\\ @V{\lambda_X = H(\mathit{id}_X)}VV @VV{\lambda_A = H(\mathit{id}_A)}V \\ f(X) \otimes 1 @>{f(h) \otimes 1}>> f(A) \otimes 1 \end{CD}$$ which induces a functor $\mor{G}{\catl{J}}{\catl{C}}$ and a natural transformation $\mor{\lambda}{G}{f(-)\otimes 1}$.

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In the $Set$ case the assertion says

$$[I^{op}, Set]/f \cong [F^{op}, Set]$$

where $F$ is the category of elements of $f$ per your notation. This isomorphism is easy to see if you switch to discrete fibrations via grotherndieck construction. For, than it becomes

$$\textbf{Dfib}(I)/(F \to I) \cong \textbf{Dfib}(F)$$

For any discrete fibration $F \to I$, this follows from properties of discrete fibrations: Given $f\circ g = k$, if $f$ and $g$ are d.f. then so is $k$, and if $f$ and $k$ are d.f. then so is $g$.

Anyway this is probably what the answer in one of your links explains.

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