4
$\begingroup$

Two short questions:

  • Is there any work classifying the lattice of subcategories of an arbitrary (sufficiently small) category $\mathcal{C}$, similar to the way that the set of subsets of set $\mathcal{S}$ is isomorphic to the functions $\mathcal{S}\to\mathbf{2}$, where $\mathbf{2}$ is a two point set?

  • Is there standard notation denoting the lattice of subcategories of some category?

The definitions found in nLab are phrased in terms of functors going into $\mathcal{C}$, but the definition for sets talks about functions out of the set $\mathcal{S}$. Why are things done differently? That is, rather than characterising subcategories in terms of functors into $\mathcal{C}$, why not characterise them in terms of functors out of $\mathcal{C}$? Something like:

The lattice of subcategories of $\mathcal{C}$ is isomorphic to the functor category $\mathcal{SO}^\mathcal{C}$, for some "subobject classifier" $\mathcal{SO}$.

$\endgroup$
2
  • 1
    $\begingroup$ If you have the good fortune to have a theorem like Birkhoff's HSP theorem, your lattice will be dual to some lattice of something like equational theories are for varieties. Is there even a Galois correspondence between the categories and something else that gives you a similar lattice to study? Gerhard "Ask Me About System Design" Paseman, 2010.06.19 $\endgroup$ Commented Jun 19, 2010 at 16:19
  • 1
    $\begingroup$ Be careful. The notion of "subcategory" is deeply problematic. See the nLab page at ncatlab.org/nlab/show/subcategory $\endgroup$ Commented Jun 19, 2010 at 20:36

3 Answers 3

7
$\begingroup$

In his comment on Finn Lawler's response, Mike Shulman points out that Cat has no subobject classifier in the 2-categorical sense. It also fails to have a subobject classifier in the 1-categorical sense. To see this, note that if Cat had a subobject classifier, then it would be a topos, as it is cartesian closed. However, Cat is not a topos since, e.g., it is not locally cartesian closed (cf. Johnstone's "Elephant", A1.5).

$\endgroup$
3
$\begingroup$

Remember that $\mathbf{2}$ is (classically) the subobject classifier in Set, so that there is a natural bijection $\mathrm{Sub}(X) \cong \hom(X,\mathbf{2})$ given by pulling back along $\mathrm{true} \colon \mathbf{1} \hookrightarrow \mathbf{2}$. So you can think of a subset of X as either an isomorphism class of monos into X or a function $X \to \mathbf{2}$.

Mike Shulman's pages about 2-toposes say that there is no subobject classifier in Cat. I don't know of a reference, but you could drop a query box there asking. Such a thing would classify functors that are faithful and injective on objects (i.e. the monos in Cat).

Edit: I misinterpreted Mike's remark at loc. cit., but see Michael Warren's answer.

The closest thing I know of (see loc. cit.) is to classify discrete opfibrations with an equivalence between functors $A \to \mathrm{Set}$ and discrete opfibrations over A, given by pullback along the forgetful functor $\mathrm{Set}_\bullet \to \mathrm{Set}$ out of the category of pointed sets (see here).

These don't look very similar, but in fact the Cat example is related to the comprehensive factorization system (initial, discrete opfibration) in Cat (see Street--Walters, The comprehensive factorization of a functor, Bull. AMS, (5)79, 1973), while the corresponding system in Set is (epi, mono). The latter also arises as the image factorization system, while in Cat, as far as I can tell, the two notions don't coincide.

$\endgroup$
2
  • 1
    $\begingroup$ What I meant on that page is that the 2-category Cat has no subobject classifier which classifies its 2-categorical subobjects (probably meaning fully faithful functors) 2-categorically, i.e. in terms of hom-categories rather than homsets. I'm not sure if the 1-category Cat has a subobject classifier in the 1-categorical sense. $\endgroup$ Commented Jun 21, 2010 at 0:10
  • $\begingroup$ Oh, OK. My mistake. But as Michael Warren points out, if Cat had a subobject classifier then it would be a topos and so locally cartesian closed, but it isn't so it doesn't. $\endgroup$ Commented Jun 21, 2010 at 16:09
0
$\begingroup$

Special cases being the submonoids of a monoid, and suborders of a partial order. What would one expect to be a common generalisation of those two, that was of interest? All one really asks for is a closure condition on morphisms (modulo talk about identity maps).

$\endgroup$
1
  • $\begingroup$ So, for the case where the "subcategories" are "ordered by inclusion" then asking if it is a lattice makes sense. But otherwise ... ??? $\endgroup$ Commented Jun 20, 2010 at 11:33

You must log in to answer this question.

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