8
$\begingroup$

Let $\mathscr{U}$ be a universe. Call a set $X$ $\mathscr{U}$-small if there is a set $Y \in \mathscr{U}$ so that $X \cong Y$. Call a category $\mathsf{C}$ a $\mathscr{U}$-category if for any $X,Y \in \mathsf{C}$, $\mathsf{Hom_C}(X,Y)$ is $\mathscr{U}$-small.

Assume $\mathsf{ZFC}$ as our foundational system (not Bourbaki set theory).

Let $\mathsf{C}$ be a $\mathscr{U}$-category and let $\mathscr{U}\text{-}\mathsf{Set}$ be the category of all sets which belong to $\mathscr{U}$.

How do we construct a $\mathsf{Hom}$-functor $\mathsf{Hom_C}(X,-)\colon\mathsf{C}\to\mathscr{U}\text{-}\mathsf{Set}$? Note for every $Y \in \mathsf{C}$, $\mathsf{Hom_C}(X,Y)$ doesn't belong to $\mathscr{U}\text{-}\mathsf{Set}$, but rather is isomorphic to a set in there. Grothendieck in SGA uses Bourbaki set theory and $\tau$ choice operator (also axiom $\mathscr{U}$B), while in $\mathsf{ZFC}$ we don't have that.

Is it even possible to work with these definition in $\mathsf{ZFC}$?

$\endgroup$
14
  • 1
    $\begingroup$ @Harry: That is not the definition used in SGA4. $\endgroup$ Nov 23, 2018 at 22:19
  • 1
    $\begingroup$ @Harry: No, see SGA4.I.1.1.2. $\endgroup$ Nov 23, 2018 at 22:31
  • 1
    $\begingroup$ "...let $\mathscr{U}\text{-}\mathsf{Set}$ be a set of all sets which belong to $\mathscr{U}$". In ZFC, the set of all sets that belong to $\mathscr{U}$ is $\mathscr{U}$ itself, isn't it? Or you mean the category of all $\mathscr{U}$-small sets? In the latter case $\mathscr{U}\text{-}\mathsf{Set}$, though a $\mathscr{U}$-small category according to your definition, will not be a set in ZFC. $\endgroup$
    – Qfwfq
    Nov 23, 2018 at 23:23
  • 1
    $\begingroup$ @FredRohrer Dear Fred, I'm trying to understand how to correctly use universes without using Bourbaki set theory as it is quite nonconventional. There are approaches to handle $\mathscr{U}$-smallness: to say that a $\mathscr{U}$-small set an element of $\mathscr{U}$ of to say it is merely equinumerous to an element of $\mathscr{U}$. Grothendieck used the second approach, and it apparently heavily relies on Bourbaki machinery such as tau operator. I'm trying to understand if we can use this approach in conventional foundations such a $\mathsf{ZFC}$ without specifically resorting to $\endgroup$
    – Jxt921
    Nov 24, 2018 at 10:31
  • 3
    $\begingroup$ On the other hand, the OP is I think greatly overstating the differences between Bourbaki and ZFC: there’s no issue whatsoever in “adapting” SGA to ZFC. Are worst, one can assume global choice in addition to ZFC, and then use that to uniformly interpret the τ/ε operator. But hardly anything really requires that extra assumption of global choice — the τ/ε operator is used mostly just as a notational convenience, and uses of it can be replaced straightforwardly by appeals to AC on a case-by-case basis (as my answer spells out for this example of hom-functors). $\endgroup$ Nov 24, 2018 at 12:21

3 Answers 3

7
$\begingroup$

This depends on what definition of category you have taken: do you assume the objects form a set?

(I’ll avoid the terminology “small category”, since while this is sometimes used to mean “objects form a set”, it’s also used to mean “objects form a $U$-set”, for some universe $U$ whose sets have been earlier designated as “small”.)

If you’ve taken the definition where objects of a category are always assumed to form a set, then there’s no problem: just use the axiom of choice. For each $X, Y$ in $\operatorname{ob} C$, you know there exist some set in $U$ and isomorphism $C(X,Y) \cong U$; so by the axiom of choice, there’s a function which chooses, for each $X$, $Y$, some such set and isomorphism. Given this, the construction of the hom-functor is straightforward.

If you’re not assuming that the objects of a category form a set — i.e. you define a category to have a class of objects — then life gets rather subtle, because categories in this sense aren’t something you can really talk about inside ZFC. You can represent classes by predicates, and so talk about individual categories (or parametrised families of categories) schematically — so any theorem about arbitrary categories is really a theorem schema. Or you can extend your language to something like NBG set theory, where you really can talk about classes (and hence categories) directly.

In either of those setups of “large categories”, I don’t see how you can define the hom-functor for an arbitrary $U$-category without invoking something approaching the axiom of global choice, which essentially gives you choice functions on classes instead of sets. Given global choice, we can construct hom-functors essentially by re-running the argument used for the small case above.

Using the language of NBG, one can ask whether the existence of hom-functors for all such categories is (a) independent of NBG (which is conservative over ZFC), or (b) equivalent over NBG to the axiom of global choice. It’s equivalent to the statement “there is a global choice function for non-empty $U$-small sets”, if I’m not mistaken. My guess would be that this statement is independent of NBG, but strictly weaker than full global choice. I can’t substantiate either part of this, off the top of my head, but I think for someone less rusty than me on the relevant techniques, it should be fairly standard.

$\endgroup$
6
  • 1
    $\begingroup$ I've taken the definition where objects of a category are always assumes to form set as getting rid of proper classes is one of main points of introducing universes. However, there can be many choice functions and hence we can construct the $\mathsf{Hom}$-functor canonically. $\endgroup$
    – Jxt921
    Nov 24, 2018 at 10:33
  • $\begingroup$ “However, there can be many choice functions and hence we can construct the 𝖧𝗈𝗆-functor canonically” — did you mean …can’t construct… there? In which case: absolutely, this construction (like most that rely on AC) isn’t canonical. On the other hand the construction in Bourbaki set theory using the τ/ε operator isn’t really canonical either in any useful sense of the term (i.e. naturality or similar properties). If you want something meaningfully “canonical”, you probably need to define $U$-categories as carrying specified isomorphisms to sets in $U$, as discussed in other comments. $\endgroup$ Nov 24, 2018 at 10:46
  • $\begingroup$ Can you, please, elaborate on "..define $U$-categories and carrying specified isomorphisms to set sets in $U$..."? $\endgroup$
    – Jxt921
    Nov 24, 2018 at 11:17
  • 1
    $\begingroup$ I mean the same suggestion Harry Gindi made in comments on Fred Rohrer’s answer. Instead of just assuming that for each hom-set there exists some isomorphic set in $U$, assume that each hom-set is equipped with a choice of some set in $U$ and isomorphism to it. $\endgroup$ Nov 24, 2018 at 12:10
  • 3
    $\begingroup$ @Jxt921: I don’t quite understand your question — how is what I wrote any less formal than the definition you started with? Just to spell the proposal out in full, for comparison: I’m suggesting the definition “A $U$-category is a category $C$ equipped with, for each pair of objects $x, y \in \operatorname{ob} C$, an element $s_{x,y} \in U$ and an isomorphism $\varphi_{x,y} : C(x,y) \cong s_{x,y}$”. $\endgroup$ Nov 24, 2018 at 13:41
5
$\begingroup$

If we wish to do this while keeping all the properties that the construction in SGA 4.I.1.3 has, then I think it is not possible:

Given objects $X$ and $Y$ such that ${\rm Hom}(X,Y)\notin\mathscr{U}$, we have to choose canonically a set in $\mathscr{U}$ with the same cardinality as ${\rm Hom}(X,Y)$; the only candidate that comes to my mind is the cardinality of ${\rm Hom}(X,Y)$. But now we also have to canonically choose a bijection between ${\rm Hom}(X,Y)$ and its cardinality, which is impossible.

$\endgroup$
7
  • $\begingroup$ When Grothendieck demands that each Hom set is isomorphic to one in $U$, maybe we should interpret this as meaning that for each pair $X,Y$, there is an actual distinguished isomorphism as part of the data? $\endgroup$ Nov 23, 2018 at 22:31
  • 4
    $\begingroup$ @Harry: Essentially yes, see SGA4.I.1.3. $\endgroup$ Nov 23, 2018 at 22:32
  • $\begingroup$ Haha, sorry for asking all of these stupid questions! $\endgroup$ Nov 23, 2018 at 22:34
  • $\begingroup$ @FredRohrer Yeah, I have suspected something like that. If I can ask one more question, Fred. Does this mean that if we try to use universes in any other mainstream foundation other than Bourbaki set theory, we have to adopt the approach to universes where $\mathscr{U}$-categories are categories whose $\mathsf{Hom}$-sets are not merely isomorphic to an element of $\mathscr{U}$, but actually belong to $\mathscr{U}$? $\endgroup$
    – Jxt921
    Nov 24, 2018 at 10:36
  • 1
    $\begingroup$ In Proposition 1.14 of this paper by Zhen Lin Low something along this line is done, but when trying to provide rigorous foundations it seems questionable to change encodings (here: of maps) whenever you run into problems. $\endgroup$ Nov 24, 2018 at 11:01
4
$\begingroup$

I think in this approach, you also need Grothendieck's axiom for universe besides ZFC, namely you have to admit for any set $A$, there is a universe $\mathcal{V}$ with $A \in \mathcal{V}$.

Once Grothendieck's axiom on universe is valid, one can only consider small categories in the sense that both the class of the objects and class of morphisms are sets---all the usual categorical constructions, such that forming functor categories as in this question, just produces small categories in an even bigger universe.

With that being said, we can assume that there exists a big universe $\mathcal{V} $, such that the set of objects of $C$ and all him-sets of $C$ are all elements of $\mathcal{V} $. Then for any $X\in \mathcal{V} $, we can form the set $\tau_X$ of injective maps from $X$ into $\mathcal{U} $. Then you can form the product set $\prod_{X \in \mathcal{V} } \tau_X$. An element in this product set, which exists by the axiom of choice for sets, serves as your $\tau$ operator. And you can continue your constructions with the help of this element. When more and more categories are involved for whatever purpose later, you can always take even bigger universe $\mathcal{V} $ to have a global choice of bijections. This allows you to abandon the full power of the $\tau$ operator and use the axiom of choice for a big enough universe instead.

$\endgroup$
3
  • 2
    $\begingroup$ Are you sure that this is invariant under change of universe, i.e., do the Hom functors defined like this stay the same if you enlarge the universe $V$? $\endgroup$ Nov 23, 2018 at 19:14
  • $\begingroup$ It of course not invariant under change of universe. The point is, for a fixed question(e. g., the question of constructing certain functors like asked here), you can always choose a big enough universe and be done with it. $\endgroup$ Nov 23, 2018 at 19:26
  • 2
    $\begingroup$ And this method will fail spectacularly once you consider things like the category of all small categories, or any large categories (you will need the full power of the $\tau$ operator when the category is locally small for a fixed universe). That's why we consider only small categories in this approach. $\endgroup$ Nov 23, 2018 at 19:29

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

Not the answer you're looking for? Browse other questions tagged or ask your own question.