# What is the definition of a $\mathcal{U}$-category?

Let $$\mathcal{U}$$ be a universe. The adaptation of the concept of a locally small category to universes is a $$\mathcal{U}$$-category.

There are two definitions of $$\mathcal{U}$$ category I've met.

$$(1)$$ A category $$\mathsf{C}$$ is a $$\mathcal{U}$$-category if $$\forall X,Y \in \mathsf{C}, \mathsf{Hom_C}(X,Y) \in \mathcal{U}$$.

$$(2)$$ A category $$\mathsf{C}$$ is a $$\mathcal{U}$$-category if

• $$\mathsf{Ob(C)} \subseteq \mathcal{U}$$,

• $$\forall X,Y \in \mathsf{C}, \mathsf{Hom_C}(X,Y) \in \mathcal{U}$$.

(SGA takes a different approach because they define a $$\mathcal{U}$$-small set not as an element of $$\mathcal{U}$$, but as a set which is equinumerous to some element of $$\mathcal{U}$$, but this is not something I would like to discuss here)

As you can see, these two definitions differ in whether we require the set of objects of a $$\mathcal{U}$$-category to be a subset of $$\mathcal{U}$$. If I'm not mistaken, the straightforward adaptation of the concept of a locally small category seems to be the second approach: with universes, we treat elements of a universe $$\mathcal{U}$$ as "sets" and subsets of a universes $$\mathcal{U}$$ as "classes".

What I want to know is which definition is more useful? In particular, is there any reason to restrict the definition of a $$\mathcal{U}$$-category by requiring the set of object to be a subset of $$\mathcal{U}$$? It appears to create some problems with functor categories (for example, given a $$\mathcal{U}$$-small category $$\mathsf{C}$$ and a $$\mathcal{U}$$-category $$\mathsf{D}$$, $$[\mathsf{C,D}]$$ is not a $$\mathcal{U}$$-category).

• Definition (2) seems more reasonable and more in tune with we'd like to call $\mathcal{U}$-category. The problem with functor categories is an important one, and it makes sense that in general $[C,D]$ should not be a $\mathcal{U}$-category – Maxime Ramzi Nov 1 '18 at 20:43

I asked a similar question (note that there I called "$$\mathcal{U}$$-locally small categories" what you call "$$\mathcal{U}$$-categories"). I still don't have any strong opinion about this, but here are a few relevant points.

1. One generally wants to put some kind of bound on the objects of a locally small for the same reason one normally uses universes in category theory: in order to make the collection of "all" locally small categories into a legitimate 2-category, for example, so that one can talk about its relationship to other 2-categories.

The question is then: what bound? A natural alternative is to require that the collection of objects belongs to the successor universe of $$\mathcal{U}$$; or one could parameterize the definition on two universes.

2. A minor advantage to requiring the collection of objects of a category $$C$$ to be a subset of $$\mathcal{U}$$ is that, if you write "set" for "element of $$\mathcal{U}$$", then a "set of objects" of $$C$$ automatically has the correct meaning for settings in which the distinction between sets and classes of objects is important, like the theory of locally presentable categories. If the objects of $$C$$ don't have to belong to $$\mathcal{U}$$, then technically $$C$$ might not have any nonempty "sets" of elements in the above sense; in that case you may find yourself needing a word for "a set which is equinumerous to some element of $$\mathcal{U}$$"...

3. I used to believe something like your final parenthetical, but I think it is actually false. If $$x$$, $$y \in \mathcal{U}$$, then $$(x, y) \in \mathcal{U}$$; and then the key point is that if $$A \in \mathcal{U}$$ and $$B \subset \mathcal{U}$$, then each function $$f : A \to B$$ is actually an element of $$\mathcal{U}$$ under the standard encoding as a set of ordered pairs, because this set has the same cardinality as $$A$$ and each member $$(a, f(a))$$ belongs to $$\mathcal{U}$$. Then, functors from a $$\mathcal{U}$$-small category $$C$$ to a $$\mathcal{U}$$-category $$D$$ are also elements of $$\mathcal{U}$$, and so $$[C, D]$$ has as set of objects a subset of $$\mathcal{U}$$.

This argument is somewhat dependent on the encodings of ordered pairs and functions and you may not care for it much. In type-theoretic foundations, an argument like this may not even be possible; that is perhaps one philosophical reason to prefer the requirement that the collection of objects belongs to the successor universe of $$\mathcal{U}$$.

• But isn't the approach where one requires a set of objects to belong to the successor universe is essentially the same where one requires nothing about it? Indeed, let $\mathsf{C}$ be a category where for any $X,Y \in \mathsf{C}$ we know that $\mathsf{Hom_C}(X,Y) \in \mathcal{U}$? Then by the axiom of universes we have a universe $\mathcal{V}$ which contains the set $\{\mathsf{Ob(C)}, \mathcal{U} \}$, that is, which contains $\mathsf{Ob(C)}$ and which is a successor of $\mathcal{U}$. – Jxt921 Nov 2 '18 at 10:38
• I'm not really convinced by your #2, because I think very rarely in category theory does one want to talk about a "set of objects" of some large category anyway. What we actually talk about all the time are set-indexed families of objects, and those should work equally well regardless of what choice is made here. The only case I can think of where we talk about a "set of objects" is when the set is a generating set, but even in such cases it is always just as good to talk about an indexed family. – Mike Shulman Nov 2 '18 at 17:59