1
$\begingroup$

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.

  • 2
    When you say $\neq$ here do you mean "not isomorphic"?2012-06-28
  • 1
    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.2012-06-28
  • 0
    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.2012-06-28

1 Answers 1