8
$\begingroup$

Let $\mathcal{K}$ be a category. I denote by $\mathcal{D}\mathcal{K}$ the category of all small diagrams over $\mathcal{K}$: an object is a functor $F:I\to \mathcal{K}$ from a small category $I$ to $\mathcal{K}$ and a morphism from $F:I\to \mathcal{K}$ to $G:J\to \mathcal{K}$ is a functor $\phi:I\to J$ together with a natural transformation $\mu:F\Rightarrow G\circ \phi$. If $\mathcal{K}$ is complete and cocomplete, then $\mathcal{D}\mathcal{K}$ is complete and cocomplete as well. I have two basic questions about $\mathcal{D}\mathcal{K}$:

  1. When $\mathcal{K}$ is cartesian closed, is $\mathcal{D}\mathcal{K}$ cartesian closed ?
  2. Does it exist any paper/book gathering what is known about this category which is certainly not new ? In particular, does this construction have a name ?
$\endgroup$
  • $\begingroup$ 1. seems to be a request living outside $\cal K$, a property of $\bf Cat$ instead. What am I missing? (you are building something like a "lax slice category" over $\cal K$ though) $\endgroup$ – Fosco Apr 7 '17 at 14:37
  • $\begingroup$ @FoscoLoregian Probably I am missing something too because I do not understand your comment. If $\mu:F\Rightarrow G \circ \phi$ and $\nu: G\Rightarrow H \circ \psi$ are two maps of $\mathcal{D}\mathcal{K}$, the composite is defined by the natural map $\nu_{\phi(A)}\mu_A:F(A) \to G(\phi(A)) \to H(\psi(\phi(A))$ and this composition law is associative. $\mathcal{D}\mathcal{K}$ is a category, not a "lax slice category" (what is it ?). $\endgroup$ – Philippe Gaucher Apr 7 '17 at 18:05
  • $\begingroup$ "lax slice category" is a construction, not a type. ncatlab.org/nlab/show/slice+2-category $\endgroup$ – Mike Shulman Apr 8 '17 at 9:18
  • 3
    $\begingroup$ The fibred category $DK \to Cat$ is that associated to the 2-functor $CAT(-,K):Cat^{op} \to CAT$. Some of the properties of $DK$ -- like completness and cocompleteness -- can be understood as special cases of well known results about fibred categories, so it might be worth looking at a book about those. But I don't know about cartesian closedness. $\endgroup$ – john Apr 9 '17 at 0:21
  • $\begingroup$ Some properties (which are probably too trivial to be of use to you) of a related construction are in section 3 of my pre-print: arxiv.org/pdf/1701.02250.pdf. If one considers the category of only discrete diagrams in C, I think that Joyal calls it Fam(C). $\endgroup$ – Karthik Yegnesh Apr 10 '17 at 0:21
5
$\begingroup$

If $\mathcal{K}$ is assumed both complete and cartesian closed, then $\mathcal{DK}$ is also complete and cartesian closed. Since completeness is not in question for the OP, I'll skip over that part and focus on cartesian closure, although we will need to recall the structure of finite cartesian products.

So: if $g: J \to \mathcal{K}$ and $g': J' \to \mathcal{K}$ are two objects of $\mathcal{DK}$, then their product is the composite $J \times J' \stackrel{g \times g'}{\to} \mathcal{K} \times \mathcal{K} \stackrel{\text{prod}}{\to} \mathcal{K}$. I'll denote this as $g \cdot g': J \times J' \to \mathcal{K}$. If $p: L \to \mathcal{K}$ is another object, I claim the exponential $(p: L \to \mathcal{K})^{(g': J' \to \mathcal{K})}$ is the functor $p^{g'}: L^{J'} \to \mathcal{K}$ defined by a formula given by an end of exponentials in $\mathcal{K}$,

$$p^{g'}(F) := \int_{j': J'} p(F(j'))^{g'(j')}.$$

This is a straightforward calculation: morphisms from $(g: J \to \mathcal{K}) \times (g': J' \to \mathcal{K})$ to $p: L \to \mathcal{K}$, say $(l: J \times J' \to L, \beta: g \cdot g' \Rightarrow pl)$, are clearly in natural bijection with pairs $(\hat{l}: J \to L^{J'}, \hat{\beta}: g \Rightarrow p^{g'} \circ \hat{l})$ where $\hat{l}(j) := l(j, -)$ and $\hat{\beta}$ is the natural transformation whose component at $j \in Ob(J)$ is naturally associated with the family $\beta_{j, j'}$ (indexed over $j'$) according to the evident natural bijection between morphisms indicated below:

$$\frac{g(j) \times g'(j') \stackrel{\beta_{j, j'}}{\to} pl(j, j')}{g(j) \underset{\hat{\beta}_j}{\to} \int_{j'} pl(j, j')^{g'(j')}}$$

(holding $j$ fixed and currying the family $\beta_{j, j'}$ which is natural in $j'$ yields a family dinatural in $j'$, so that we get an induced map to the end which is a universal wedge for that dinatural family).

(Perhaps this somewhat pedestrian calculation is a special case of some more abstract consideration, possibly involving a form of Artin gluing. For example, if we follow Karthik's comment and specialize to where the categories $I, J$ are discrete and $\mathcal{K}$ is say $\text{Set}$ or even a topos $E$, then $Fam(E)$ is the Artin gluing along $\Delta: \text{Set} \to E$. And indeed, if I recall correctly, Carboni and Johnstone do discuss Artin gluings in the doctrine of cartesian closed categories, so maybe something like this can be made to work. However, I am doubtful that one could dispense with completeness of $\mathcal{K}$, as the end calculation seems to mandate this.)

$\endgroup$
  • $\begingroup$ I will take your post as an answer since the second question is not really a rigorous mathematical one. $\endgroup$ – Philippe Gaucher Apr 15 '17 at 16:00

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.