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Morgan Rogers (he/him) (Sep 23 2020 at 15:03):On the nLab page for regular categories, we find the following quote:
If a regular category additionally has the property that every congruence is a kernel pair (and hence has a quotient), then it is called a (Barr-) exact category. Note that while regularity implies the existence of some coequalizers, and exactness implies the existence of more, an exact category need not have all coequalizers (only coequalizers of congruences), whereas a regular category can be cocomplete without being exact.
A similar statement can be found in section A1.3 of Johnstone's Elephant, but I have never seen an example of a parallel pair of morphisms in a regular category which did not have a coequalizer. Does anyone know of such a pathological example?
Reid Barton (Sep 23 2020 at 15:15):I think the full subcategory of coherent objects of a coherent topos can provide such an example. There is an example in the relevant section of the Elephant which involves (pre)sheaves on a category containing two objects, and .
Reid Barton (Sep 23 2020 at 15:18):Basically the issue is that the equivalence relation that you need to quotient by to form an arbitrary coequalizer might not be finitely generated, and so the quotient takes you outside of the coherent objects.
I was surprised to learn about this example but now I no longer think of it as pathological, but rather the expected situation: what requires explanation is why it sometimes happens that the coherent objects are closed under all coequalizers (and then agree with the finitely presentable ones)
Reid Barton (Sep 23 2020 at 15:20):I guess "closed under coequalizers" isn't quite the same as "has coequalizers", so there might be a different coequalizer inside the coherent objects--I'm not sure whether this happens but it seems like sort of a bad object to consider even if it does exist.
Morgan Rogers (he/him) (Sep 23 2020 at 15:25):I've found Example D3.3.9, and it is an example of a coherent category which is not closed under coequalizers in the corresponding topos. As it happens, this is probably sufficient to address the problem I was working on, even if it isn't quite what I asked for :blush:
Reid Barton (Sep 23 2020 at 15:25):Sounds like the example I was thinking of.
Morgan Rogers (he/him) (Sep 24 2020 at 14:10):The category of topological spaces and local homeomorphisms is at least a locally regular category in which some pairs of parallel morphisms don't have coequalizers (cf Elephant, discussion before Proposition A1.3.4 and Example A1.3.7); this is similarly useful to me but still not quite what I originally asked for, since any slice of is a regular category having coequalizers!
Morgan Rogers (he/him) (Sep 25 2020 at 14:46):Okay, continuing on this theme, here is part of the first paragraph of a paper of Lack from 1999; the emphasis is his:
Every kernel pair is an equivalence relation; a regular category is said to be exact if every equivalence relation is a kernel pair. Thus a regular category is a category with finite limits and coequalizers of kernel pairs, satisfying certain exactness conditions; while an exact category is a category with finite limits and coequalizers of equivalence relations, satisfying certain exactness conditions.
This strikes me as an incorrect assessment: the classic example of a regular category which is not exact is the category of torsion-free abelian groups, see here (although I do not understand the content of the commentary which begins "Herein lies a subtle trap for the unwary"...) which has all coequalizers and so in particular has coequalizers of all equivalence relations. The reason it's not exact is that some of the equivalence relations are not kernel pairs.
In particular it is not at all clear to me that Lack's construction , consisting of the closure of under coequalizers of equivalence relations in its topos of regular sheaves, produces an exact category! What am I missing here?
Morgan Rogers (he/him) (Sep 25 2020 at 15:36):I've convinced myself that Lack's construction works (I expect it would have come up at some point in the last 20 years if it didn't!) but that doesn't justify the statement of his which I quoted... unless "satisfying certain exactness conditions" means "such that the equivalence relation is the kernel pair of its coequalizer", in which case omitting that is just misleading, in light of the counterexample I mentioned :sweat_smile: