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In chapter 4 of Handbook of Categorical Algebra, vol 1, the author defines a "subobject of $A$" as "an equivalence class of monomorphisms with codomain $A$" (for a suitable notion of equivalence). He then defines what it means for a category to be well-powered: "$\mathcal{A}$ is well-powered when the subobjects of every object constitute a set". Thus, for instance, the category of sets is well-powered.

I'm having trouble understanding exactly what it means to have such a set of subobjects. As far as I can tell, each element of a set should also be a set, but an equivalence class of monomorphisms could be a proper class: for instance, the class of singleton sets is not a set. On the other hand, it seems that one can cheat by defining a subobject to be a class containing one representative of each equivalence class of monomorphisms, even though this is not, strictly speaking, what's stated in the book.

How can one solve this problem? Is there a "normal" set theory where such a set of subobjects can contain proper classes? Or does one need to cheat like suggested above?

Arthur Azevedo De Amorim
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    Elements of a set must be a set so there is no set which has a proper class as element. You need a another word to denote `family of proper classes'... – Hanul Jeon Nov 05 '14 at 14:59
  • That's what I suspected indeed, but how to make sense of the definition of "well-powered", then? – Arthur Azevedo De Amorim Nov 05 '14 at 15:20
  • If classes are objects, then it seems to make just as much sense to have a set of them as it does to have a set of cats. We need a slightly different theory of sets, see http://en.wikipedia.org/wiki/Urelement, though. –  Nov 05 '14 at 15:22
  • This just makes sense with a global choice. Maybe he's assuming locally small categories... – user40276 Nov 13 '14 at 21:10
  • @user40276 Yes, he is assuming locally small categories, but even with this assumption it seems a bit strange, because the monomorphisms in the definition need not have the same domain. Each hom set is indeed a set, but when you add up hom sets for lots of different objects, you can get something that is not one. – Arthur Azevedo De Amorim Nov 14 '14 at 14:57

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No, proper classes are not elements of other sets (or other classes, for that matters).

But this is the great power of universes. Classes of one universes are just sets in a larger universe. So when you want to talk about collections of classes, you move to a larger universe, where you can treat them formally as sets.

Another, more complicated method of solving this issue, is to talk about schema of definitions when it comes to equivalence, and then the collections are represented by one of the classes (and you can sometimes prove that this representative doesn't matter). Then the whole thing becomes a much more technical and involved from a formal point of view, which is not a bad thing. But I think that if you're interested in category theory and want to talk about larger and larger categories, then universes are probably the way to go.

Asaf Karagila
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  • Why aren't proper classes elements of sets? –  Nov 05 '14 at 15:29
  • @GME: Because proper classes are collections which do not exist (from a semantic point of view), and elements are objects in the universe and therefore exist. – Asaf Karagila Nov 05 '14 at 15:32
  • I'm not sure I'm following. What's a collection which doesn't exist when it's at home? –  Nov 05 '14 at 15:36
  • (BTW: I'm not trying to be difficult. I genuinely don't understand what a "collection which doesn't exist" is nor what the semantic point of view is) –  Nov 05 '14 at 15:40
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    If $M$ is a model of set theory, then $M$ thinks that $\varphi(y)$ defines a set if and only if $M\models\exists x\forall y(y\in x\leftrightarrow \varphi(y))$. Do you agree with that? – Asaf Karagila Nov 05 '14 at 15:53
  • Certainly, Asaf! –  Nov 05 '14 at 15:53
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    Good. So what does it mean when $\varphi$ defines a proper class? It means that $M$ does not satisfy the statement "There exists $x$ ...", so proper classes are collections of objects in $M$ which do not exist in $M$. But being elements, from the point of view of $M$, requires first to exist, because the $\in$ is only defined on the objects inside $M$. Ergo, proper classes are collections of objects of $M$ which do not exist in $M$. So they cannot be elements of other sets, since in this context we limit "set" to mean "an object inside $M$". – Asaf Karagila Nov 05 '14 at 15:58
  • Right, so if we ignore things which aren't sets then there aren't proper classes. Much like, if we ignore things which aren't sets then there aren't cats. But that doesn't tell us whether there are proper classes or cats, and it doesn't tell us whether such things can be in sets. No? –  Nov 05 '14 at 16:03
  • Sure it does. Any member of the universe is a set, and a proper class is by definition not a set. So a set can't have a member that is a proper class because members of sets are also members of the universe! – Zhen Lin Nov 05 '14 at 16:05
  • What do you mean by "the universe"? Doesn't it follow by the same reasoning that a set can't have a member that is a cat? –  Nov 05 '14 at 16:07
  • @GME: Again, $\in$ is only defined on objects in the universe of $M$. Proper classes are not objects in the universe of $M$. So proper classes cannot be members of objects of the universe of $M$. Remember that $(M,\in)$ is a structure for a first-order language. – Asaf Karagila Nov 05 '14 at 16:08
  • Yep. Cats aren't elements of sets in $M$ either. That doesn't establish that there's no set of the cats. BTW: we can build models of versions of ZFC with urelements where all proper classes of the pure sets exist in the model and are elements of impure sets. –  Nov 05 '14 at 16:11
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    @GME: No, it means exactly that. If categories are not sets, they don't exist in the universe. I really don't know how to make this any clearer than that. As for your second remark, that might depend on your axioms, but in the usual axiomatization of $\sf ZFA+AC$ you can prove that every set is in bijection with a pure set (more specifically, an ordinal). So this remark is certainly false there. – Asaf Karagila Nov 05 '14 at 16:16
  • @AsafKaragila I see how universes can be sort of a solution to this problem, but then I'm not entirely sure about how to interpret the sentence "there is a set of subobjets". To take the example of Set, if our category of sets is restricted to sets in some universe $U$, then subobjects can still be outside of this set, meaning that if the "power set" constitutes indeed a set, it'll be a set outside of $U$, implying that this "power set" wouldn't be an object of Set. Wouldn't that be a problem? Besides that, is there a good reference of "schemas of definitions"? – Arthur Azevedo De Amorim Nov 05 '14 at 16:17
  • Sorry, so you are claiming that there's no set of cats? Re. my second remark: I said "versions". –  Nov 05 '14 at 16:19
  • @Arthur: I'm afraid that this is beyond my understanding in the usual abuse of terminology in category theory. But perhaps Zhen Lin could shed some light on that. Schema of definitions simply means that we have to step "up" to the meta-theory and prove there that there is some "pattern" of sentences where you can plug in any formula and be able to prove the sentence you get. I'm not sure where you can really read about it. It's the sort of thing you "learn on the job". – Asaf Karagila Nov 05 '14 at 16:20
  • @GME: Yes, that is what I'm claiming. To your second remark, you also said "ZFC+Urelements", and I don't know any variant which allows proper classes of pure sets to be realized as impure sets. – Asaf Karagila Nov 05 '14 at 16:21
  • Oh, I see! Good. I didn't look like your answer depended on such a controversial claim. On the other point: Replacement fails in general. Everything else works as usual. –  Nov 05 '14 at 16:25
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    @GME: You can't say that something is "a version of ZFC" if replacement fails. – Asaf Karagila Nov 05 '14 at 16:26
  • It holds for the pure sets. –  Nov 05 '14 at 16:26
  • @Zhen Lin: See Arthur's comment to my answer please. I think that you might be able to be more helpful regarding that. – Asaf Karagila Nov 05 '14 at 17:13
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    @ArthurAzevedoDeAmorim "Power set" has a well-established meaning in ZF which does not run into these problems. In general, yes, a subobject in the sense of Borceux will be a proper class. I agree with Asaf in that (Grothendieck) universes are the most convenient way of avoiding trivial set-theoretic difficulties of this kind. – Zhen Lin Nov 05 '14 at 17:50
  • @Zhen Lin: I should probably point out that it is I who is in agreement with you, rather than the other way around. :-) – Asaf Karagila Nov 05 '14 at 18:53