7
$\begingroup$

Given a locally finitely presentable category $\mathcal{C}$ it is well-known that every functor category $[\mathcal{A},\mathcal{C}]$ (where $\mathcal{A}$ is a small category) is also locally finitely presentable. Is there a concrete description of the finitely presentable objects in such categories?

If this is not possible in general, what about special cases like $\mathcal{A}=\mathcal{Set}_{fin}^{op}$?

$\endgroup$
4
  • 1
    $\begingroup$ Coequalizers of finite coproducts of representables. $\endgroup$ Jan 20, 2014 at 21:53
  • 1
    $\begingroup$ @FernandoMuro What representables? $\mathcal{C}$ is not $\mathbf{Set}$. $\endgroup$
    – Todd Trimble
    Jan 21, 2014 at 2:52
  • $\begingroup$ Well, your conditions ensure that C is equivalent to [B, Set] for B the category of finitely presentables in C, so you end up with [AxB, Set] $\endgroup$ Jan 21, 2014 at 8:01
  • $\begingroup$ Sorry, preserving finite limits in B. $\endgroup$ Jan 21, 2014 at 8:33

1 Answer 1

6
$\begingroup$

When $\mathcal{A}$ is a finite category, the finitely presentable objects in $[\mathcal{A}, \mathcal{C}]$ are precisely the diagrams that are componentwise finitely presentable: see e.g. Proposition 2.23 here. The general case is harder to describe. There are two steps:

  1. First, determine the finitely presentable objects in $[\operatorname{ob} \mathcal{A}, \mathcal{C}]$; these will contain e.g. "finitely supported" families of finitely presentable objects.
  2. Then, determine the left adjoint of the restriction functor $[\mathcal{A}, \mathcal{C}] \to [\operatorname{ob} \mathcal{A}, \mathcal{C}]$; by general nonsense, the restriction functor is finitely accessible and monadic, and the finitely presentable objects of $[\mathcal{A}, \mathcal{C}]$ will be the objects in the smallest full subcategory that is closed under finite colimits and that contains the free "algebras" generated by the finitely presentable objects in $[\operatorname{ob} \mathcal{A}, \mathcal{C}]$.
$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

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