2
$\begingroup$

Let $\mathbf C$ and $\mathbf D$ be small categories. $\mathrm{Ind}(\mathbf C)$ is an accessible category (by definition), and is locally finitely presentable (i.e. cocomplete, or equivalently complete) iff $\mathbf C$ has finite colimits. Let $\mathbf C$ and $\mathbf D$ have finite colimits, and consider a functor $F : \mathbf C \to \mathrm{Ind}(\mathbf D)$. By the universal property of $\mathrm{Ind}$, this extends to a finitary functor $\tilde F : \mathrm{Ind}(\mathbf C) \to \mathrm{Ind}(\mathbf D)$.

In terms of $F$, assuming such a characterisation exists:

  1. When is $\tilde F$ continuous? (Equivalently, when does $\tilde F$ have a left adjoint?)
  2. When is $\tilde F$ cocontinuous? (Equivalently, when does $\tilde F$ have a right adjoint?)

I imagine (2) should hold when $F$ preserves finite colimits, though I wasn't able to find a reference in Locally presentable and accessible categories.

Improve this question
$\endgroup$
9
  • 2
    $\begingroup$ Your "iff" is not quite correct, I think. You should assume idempotents split. $\endgroup$
    – Zhen Lin
    Commented Sep 12, 2020 at 22:47
  • $\begingroup$ @ZhenLin: just to clarify, what would be the corrected statement? $\endgroup$
    – varkor
    Commented Sep 13, 2020 at 12:19
  • 2
    $\begingroup$ Assuming idempotents in $\mathcal{C}$ split, $\textbf{Ind}_\kappa (\mathcal{C})$ has colimits if and only if $\mathcal{C}$ has $\kappa$-small colimits. $\endgroup$
    – Zhen Lin
    Commented Sep 13, 2020 at 14:03
  • 1
    $\begingroup$ No, if we don't know that idempotents in $\mathcal{C}$ split then we cannot conclude that $\mathcal{C}$ has $\kappa$-small colimits even if $\textbf{Ind}_\kappa (\mathcal{C})$ is cocomplete. $\endgroup$
    – Zhen Lin
    Commented Sep 13, 2020 at 22:06
  • 2
    $\begingroup$ That’s right. Every locally $\kappa$-presentable category is equivalent to the category of $\kappa$-ind objects of some $\kappa$-cocomplete category, but that may be a different category from what you start with. (Not too different, in reality: it is the Cauchy completion.) $\endgroup$
    – Zhen Lin
    Commented Sep 14, 2020 at 1:16

1 Answer 1

Reset to default
4
$\begingroup$

Allow me to generalise to $\kappa$-accessible categories for infinite regular cardinals $\kappa$. Your guess for (2) is correct: if $F$ preserves $\kappa$-small colimits then $\tilde{F}$ preserves colimits. The proof is a little bit indirect.

Proposition. Let $\mathcal{I}$ be a category and let $\mathcal{C}$ be a small category with $\kappa$-small colimits. If $\mathcal{I}$ is $\kappa$-small, then the comparison functor $$\textbf{Ind}_\kappa ([\mathcal{I}, \mathcal{C}]) \to [\mathcal{I}, \textbf{Ind}_\kappa (\mathcal{C})]$$ is fully faithful and essentially surjective on objects.

(The key point is to show that every diagram of shape $\mathcal{I}$ can be written as a $\kappa$-filtered colimit of diagrams of $\kappa$-presentable objects of the same shape $\mathcal{I}$. Actually, the statement of the proposition is equivalent to this fact, and this is what we need for the next step.)

Proposition. Let $\mathcal{C}$ be a category with $\kappa$-small colimits, let $\mathcal{E}$ be a category with $\kappa$-filtered colimits, let $F : \mathcal{C} \to \mathcal{E}$ be a functor, and let $\tilde{F} : \textbf{Ind}_\kappa (\mathcal{C}) \to \mathcal{E}$ be the extension. Then $\tilde{F}$ preserves colimits if and only if $F$ preserves $\kappa$-small colimits.

(We already know $\tilde{F}$ preserves $\kappa$-filtered colimits, so it is enough to check whether $\tilde{F}$ preserves $\kappa$-small colimits. The "only if" direction is easy. The "if" direction is proved using the previously mentioned decomposition of $\kappa$-small diagrams.)

I don't have a good answer for (1), but perhaps this will be enough for your purposes:

Proposition. Let $\mathcal{C}$ and $\mathcal{D}$ be small categories with $\kappa$-small colimits. A functor $F : \mathcal{C} \to \mathcal{D}$ has a left adjoint if and only if $\textbf{Ind}_\kappa (F) : \textbf{Ind}_\kappa (\mathcal{C}) \to \textbf{Ind}_\kappa (\mathcal{D})$ has a left adjoint.

(The "only if" direction is easy: after all, $\textbf{Ind}_\kappa$ is a pseudofunctor so it preserves adjunctions. The "if" direction amounts to saying that the left adjoint of a $\kappa$-accessible functor between locally $\kappa$-presentable categories preserves $\kappa$-presentable objects, which is straightforward to check.)

Improve this answer
$\endgroup$
5
  • $\begingroup$ Off-topic, but the url listed on your user page is broken. It's good to see you here again, btw! $\endgroup$
    – David Roberts ♦
    Commented Sep 13, 2020 at 1:17
  • 2
    $\begingroup$ Thanks! I updated the link. (Nothing new there though.) $\endgroup$
    – Zhen Lin
    Commented Sep 13, 2020 at 2:19
  • $\begingroup$ Thank you, this is helpful! Is there reason to believe there may not be a nice characterisation for (1)? $\endgroup$
    – varkor
    Commented Sep 13, 2020 at 12:20
  • $\begingroup$ Well, I wouldn't expect anything as nice as (2). For example you might consider $\textbf{CRing}^X$ where $X$ is an infinite set. This is a locally finitely presentable category for general reasons, but the terminal object – the limit of the empty diagram! – is not finitely presentable. So one has to do more work to figure out how a finitely accessible functor defined on finitely presentable objects is going to act on the terminal object. $\endgroup$
    – Zhen Lin
    Commented Sep 13, 2020 at 14:14
  • $\begingroup$ Thanks, this is a nice example. I think simply knowing that there isn't such an elegant characterisation for (1) is enough to answer that part of my question. I had hoped there would be a condition not much more complex than that for (2), but suspected it might not be the case. $\endgroup$
    – varkor
    Commented Sep 14, 2020 at 1:25

You must log in to answer this question.

Not the answer you're looking for? Browse other questions tagged .