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Let $\mathcal C$ be a finitely complete, finitely cocomplete category. Then the following are equivalent:

  1. $\mathcal C$ is regularly well-powered (i.e. every $C \in \mathcal C$ has a small set of regular subobjects);

  2. $\mathcal C$ is regularly co-well-powered (i.e. $\mathcal C^{op}$ is regularly well-powered).

Proof:

Freyd showed in his paper Concreteness that (1) is equivalent to

  1. $\mathcal C$ is concrete (i.e. there exists a faithful functor $\mathcal C \to Set$);

which is in turn equivalent (because there exist faithful functors $Set^{op} \to Set$, e.g. the contravariant powerset functor) to

  1. $\mathcal C^{op}$ is concrete;

which by the dual of Freyd's result, is equivalent to (2).


Question:

  1. Is there an elementary proof that $(1) \Leftrightarrow (2)$?

  2. Can the hypotheses be relaxed? For instance, does $(1) \Leftrightarrow (2)$ hold when $\mathcal C$ is assumed to have equalizers and coequalizers, but not necessarily all finite limits and colimits?

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    $\begingroup$ By well-powered do you mean "connected to a continuous energy source"? $\endgroup$
    – Asaf Karagila
    Feb 13, 2021 at 12:45
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    $\begingroup$ @AsafKaragila Exactly. Unfortuately, "co-well-powered" means that it's also connected to a continuous energy drain. So the above theorem that $(1) \Leftrightarrow (2)$ is really a weak form of the 0th law of thermodynamics. $\endgroup$
    – Tim Campion
    Feb 13, 2021 at 12:51
  • $\begingroup$ One way to think about this could be to study the presheaf $\mathsf{RSub}$ of regular subobject and the copresheaf $\mathsf{RQuo}$ of regular quotient. Now one could try to say that there must be an injective map $\mathsf{RQuo} \to 2^\mathsf{RSub}$ (notice that the variance now match). Similar tricks are used by Freyd in Sec 5 of "On the concreteness of certain categories", for very different purposes. $\endgroup$ Feb 13, 2021 at 12:59
  • $\begingroup$ @IvanDiLiberti That sounds promising. The usual ways I know of relating subobjects and quotients -- taking kernels / cokernels, taking kernel pairs / quotienting by a congruence etc. seem to require stronger exactness properties than I'm assuming here. It would be interesting if there were something like this which didn't require any sort of exactness. $\endgroup$
    – Tim Campion
    Feb 13, 2021 at 13:00
  • $\begingroup$ That was my idea. It's hard to say without spending time on a whiteboard. $\endgroup$ Feb 13, 2021 at 13:01

1 Answer 1

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Your first question is handled in Freyd’s paper, which shows that the notion of “a binary relation controlled by a set” is richer than you would think. Freyd says a relation $\square$ from $L$ to $R$ between large sets is resolvable if there is a small subset $L’\subseteq L$ such that every $l\in L$ has the same $\square$-relatives as some $l’\in L’$.

Remarkably, resolvability of a relation implies that of its opposite! This is the key elementary foundation for everything in this paper. The proof is easier than thinking of the statement: for each subset $A$ of $L’$ for which this is possible, pick $r_A$ such that $A$ is the set of all $r_A$’s $\square$-relatives in $L’$. Then it’s easy to check that the (small!) set of all $r_A$s resolves $\square^{\mathrm{op}}$.

The categorical application of this notion is to say that a span $A\leftarrow X\to B$ $\square$-relates to a cospan $A\to Y\leftarrow B$ if and only if the square commutes. Isbell proved that a concrete category has $\square$ resolvable for every $A,B$, and Freyd proved the converse. This is saying that a category is concrete if and only if it doesn’t have too many distinguishable maps into or equivalently out of any pair of objects, which is a hyper-general (co)wellpoweredness condition.

If now the category has binary products, respectively coproducts, then it is equivalent to look at the analogous relations involving forks, respectively coforks, centered at a single object $A$, and one proves this directly. It’s actually the first lemma toward proving the concreteness theorem. Resolvability of these relations is what Freyd calls generalized regular (co)-well-poweredness, so we do know that these are equivalent for categories with binary products and coproducts; adding equalizers and coequalizers gets you the result you mentioned, and again it’s easy to prove directly that resolvability of the generalized regular subobject relation is equivalent to regular wellpoweredness when there are equalizers. So that handles your (1).

Toward your (2), I’m pretty convinced you can show these results are optimal by fiddling with Isbell’s original non-concrete category, which is a union of a large number of commutative squares spanning and cospanning two fixed objects $A,B$. This is (generalized) regular well- and co-wellpowered because there aren’t any nontrivial forks or coforks. Now freely adjoin a product $A\times B$. You haven’t added any coforks but $A\times B$ is at the middle of a large number of distinguishable forks, so this is a guy which is not generalized regular well powered but is (generalized) regular cowellpowered. Now freely adjoin equalizers and then coequalizers. Since we’ve never created any new coforks and every nontrivial fork is centered on $A\times B$, I claim that all this adds is an initial object and a terminal object plus a large family of distinct regular subobjects of $A\times B$. Now this has equalizers and coequalizers, is not regular well powered but is regular well copowered.

So I think the moral is that forks are less meaningful that squares in the absence of products.

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  • $\begingroup$ (Since I think you're still proofreading, I'll point out that Freyd's name is misspelled in the first sentence.) $\endgroup$
    – Tim Campion
    Feb 13, 2021 at 14:42
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    $\begingroup$ Ah, a Freydian slip! $\endgroup$ Feb 13, 2021 at 14:43
  • $\begingroup$ Thanks! I suppose what you're pointing out is that in order to get $(1) \Leftrightarrow (2)$ from Freyd's paper, I really just need to read the first section, which is much more elementary than the subsequent sections, which start doing weird things to embed non-finitely-complete categories into abelian categories. That does seem like a win! I'm now wondering if there are any other reasonable ways one might try to weaken the hypotheses... (I was planning to delete my previous comment after you saw it, but then your response wouldn't make sense any more, which would be a shame!) $\endgroup$
    – Tim Campion
    Feb 13, 2021 at 14:49
  • $\begingroup$ Er... hm... I think I'm still bothered, because Freyd's proof that $(1) \Rightarrow (3)$ still does require one to go through a bunch of rigamarole about embedding abelian categories into $Ab$, and that part doesn't really simplify when we assume that $\mathcal C$ is finitely-complete. I think in order to have an "elementary" proof of any of these implications, we should at least stipulate that at no point in the proof should one need to know what an abelian category is! $\endgroup$
    – Tim Campion
    Feb 13, 2021 at 15:17
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    $\begingroup$ Ok ok -- Freyd doesn't necessarily explicitly state that $(1) \Leftrightarrow (2)$, but your argument shows how his observations in the first part of the paper get us there without delving into the later part of the paper or concreteness at all. Thanks! I'll probably let this sit a few days before accepting in hopes of maybe seeing something in the way of positive results for (2), but I think this looks fairly definitive. $\endgroup$
    – Tim Campion
    Feb 13, 2021 at 15:26

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