I was wondering if the following idea is a notion that has been studied. I thought of it while thinking of functorial invariants (such as (co)homology functors) and wondering what kind of category has a 'compactness condition' for a collection of functors that are able to distinguish non-isomorphic elements in a category. I tried searching for some likely terminology with no luck.

Definition: Let $\mathcal{C}$ be a category, and $\mathfrak{F} = \{ F_i \}_{i \in I}$ be a family of functors $F_i: \mathcal{C} \to \mathcal{D}_i$ for all $i \in I$ for some index set $I$ (where $I$ may be infinite). Then we say that $\mathfrak{F}$ distinguishes objects of $\mathcal{C}$ if for all $a, b \in \mathcal{C}$, with $a$ not isomorphic to $b$, there exists $F_j \in \mathcal{F}$ such that $F_j(a) \neq F_j(b)$.

Definition: $\mathcal{C}$ is functorially compact if for all families of functors $\mathfrak{F}$ that distinguish objects of $\mathcal{C}$ there exists a finite subset $\mathfrak{G} \subseteq \mathfrak{F}$ that distinguishes objects of $\mathcal{C}$.

Question: What sort of category can be 'functorially compact'?

Thoughts: The identity functor always distinguishes objects of $\mathcal{C}$, so we know that every category has at least one family of functors that distinguishes objects in it. I believe categories with a finite number of objects would always be functorially compact as well. I imagine that this is a rather strict property that wouldn't allow sophisticated categories, but I'd like to be proven otherwise. But what about categories with an infinite number of objects (small or large)?

Note: These definitions may not be perfect - maybe some restriction must be put on what kind of functor is allowed to make things interesting, but I don't know enough about category theory to know where to begin with that.

Edit: I made corrections as per Qiaochu's comments below.

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    $\begingroup$ When you say $\neq$ here do you mean "not isomorphic"? $\endgroup$ – Qiaochu Yuan Jun 28 '12 at 21:00
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    $\begingroup$ The motivating question in your first paragraph doesn't seem very closely related to what you've actually asked. Of course as you observe the identity functor always distinguishes objects, so every category only needs a finite number of functorial invariants to distinguish all objects in it. What makes cohomology useful isn't (just) that it's functorial but that it's extremely computable in practice. $\endgroup$ – Qiaochu Yuan Jun 28 '12 at 21:02
  • $\begingroup$ Yes, I meant not isomorphic. And you are correct that I did not write my first paragraph carefully; I will make these changes. Thanks for pointing them out. $\endgroup$ – Jon Paprocki Jun 28 '12 at 23:04

I think that a category is "functorially compact" iff it is finite.

It is easy to see that finite $\Rightarrow$ functorially compact: for any family of functors $\mathfrak{F}$, we construct a finite sub-family by picking one distinguishing functor from $\mathfrak{F}$ for each pair of objects in $\mathcal{C}$.

To show that functorially compact $\Rightarrow$ finite, consider the family $\mathfrak{F} = \{F_A\}_{A \in Obj(\mathcal{C})}$ indexed by the objects of $\mathcal{C}$, where each $F_A : \mathcal{C} \rightarrow \mathbf{2}$. We take $\mathbf{2}$ to be the category consisting of two objects $X$ and $Y$, and two (non-identity) morphisms $f : X \rightarrow Y$ and $g : Y \rightarrow X$. We define $F_A$ to be the functor that sends $A$ to $X$ and every other object to $Y$ (this map uniquely determines where the functor maps the morphisms of $\mathcal{C}$). $\mathfrak{F}$ distinguishes objects of $\mathcal{C}$, but there can be no finite subfamily that does the same, unless $\mathfrak{F}$ was finite to begin with (if there exist objects $A$ and $B$ such that neither $F_A$ nor $F_B$ belong to this subfamily, then $A$ and $B$ are not distinguished).

  • $\begingroup$ This can't be right. You need at least to assume that $C$ is skeletal (otherwise an infinite collection of objects all of which are isomorphic is a counterexample). Even then, if $C$ has two objects $X, Y$ and two morphisms $f : X \to Y, g : Y \to X$ such that $fg = \text{id}_Y$ then the functor you want doesn't exist because the corresponding relation is false in $2$. $\endgroup$ – Qiaochu Yuan Jun 28 '12 at 21:08
  • $\begingroup$ How is that a counterexample? The definition of "functorial compactness" requires us to distinguish isomorphic objects as well. $\endgroup$ – Zak Jun 28 '12 at 21:12
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    $\begingroup$ A category all of whose objects are isomorphic is functorially compact (since there is nothing to distinguish) but it need not have finitely many objects. In other words, being functorially compact is invariant under equivalence but having finitely many objects isn't. $\endgroup$ – Qiaochu Yuan Jun 28 '12 at 21:13
  • $\begingroup$ I think our problem is that I read $\neq$ as "not identical objects" whereas you read it as "not isomorphic objects". Regarding your second objection, $fg = id_Y$ does hold in 2 (since there is only one morphism from Y to Y in my definition of 2, $fg$ must be the identity) $\endgroup$ – Zak Jun 28 '12 at 21:25
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    $\begingroup$ if there's only one morphism $Y \to Y$ and presumably only one morphism $X \to X$ then $X$ and $Y$ are isomorphic. You are right that I read $\neq$ as "not isomorphic." The other question is evil (ncatlab.org/nlab/show/evil). $\endgroup$ – Qiaochu Yuan Jun 28 '12 at 21:31

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