Questions: 1. When do objects of a category form a set?
- Is there a choice function when I have a set of categories (as opposed to a set of sets)? Is there an axioma schema of separation for defining (full) subcategories?
Context: Proof of lemma 3.2 in
Martín Escardó et al. "Comparing Cartesian closed categories of (core) compactly generated spaces", (doi:10.1016/j.topol.2004.02.011)
In the argument, there is a space $X$ and a set $I$ of non-open subsets. For each $i\in I$, they choose a compact space $C_i$ with a map $p_i\colon C_i\to X$ satisfying some property. After this choice, they take a direct sum of these spaces.
The choice to each $i\in I$ is an object in a comma category $\mathcal{C}$ (of all maps from compact spaces to $X$).
First, what I mean by axiom of choice is that if I have a set $J$ of sets then there is a choice function from $J$ to the disjoint union of the sets in $J$. The question is about how to define this choice function.
I see two ways. One way is to form a set out of the objects of my category, then use the axiom of subsets to define a subset for each $i\in I$ and then use the AC as stated above.
However, the objects of a category don't usually form a set, see this answer for example. But as far as the proof above is concerned, all I need to know is that the objects in $\mathcal{C}$ form a set.
The second way to go about this is to define (full) subcategory where the objects are those that have such and such a property and then use a version of AC for when I have a set of categories. But this feels like I'm forming a set out of the objects and defining a subset there and then "lifting" it back to my category. (I am thinking here of forming subcategories like that of all two element sets or groups with every element having finite order etc.)
The third alternative is that the proof in the paper above works in an entirely different manner.
It is not possible to answer this question except in a tautological way, and in any case, to answer the question even for specific categories depends on your choice of foundations.
(For example, if you use the setup of SGA 4, then every category has a set of objects; the price you pay for this is that every category has to be cut off by a universe parameter, and sometimes you have to juggle universes. Or if you use the setup of Categories for the working mathematician, then categories always have sets of objects, but metacategories are not so constrained.)
That said, in practice, you will not go wrong if you only assume that small categories have a set of objects.
Again, it depends on your foundations, and also your definitions. In the standard set-theoretic understanding of these words, if you genuinely have a set of categories, then those categories are necessarily small. (Do you think that $\{ \textbf{Set} \}$, where $\textbf{Set}$ is the category of all sets, is a set? If you do, and you worry about set-theoretic issues, then you should probably go consult a set theory textbook first.)
But in my experience I think mathematicians are more likely to mean that they have a family of categories parametrised/indexed by a set, rather than a literal set of categories. Such a thing is not necessarily a set (or even a class). If you have the axiom of global choice, this is no obstacle: since the individual objects of each category are elements of the universe, if the universe is well-ordered, we could certainly pick an object in each category (and then apply the axiom of replacement to get a choice function).
In principle this depends on your foundations, but in commonly used class/set theories there is an axiom schema of separation for classes. Sometimes there are limitations on the complexity of predicates that can be used: for example, in NBG, the axiom schema of separation is only assumed for predicates where all bound variables are restricted to be sets. (If you are not comfortable with thinking about the complexity of predicates, there is a finite axiomatisation of NBG where the axiom schema of separation can be derived as a theorem.)