# “Uniformly (co)well-powered” categories?

Let $C$ be a category and for each object $X$ let us denote by ${\sf Mono}(X)$ the category of all monomorphisms in $C$ going to $X$ (i.e. monomorphisms which have $X$ as range -- I hope it is clear how morphisms in ${\sf Mono}(X)$ are defined).

$C$ is said to be well-powered (see MacLane, or nLab) if for each object $X$ the category ${\sf Mono}(X)$ has a "small" skeleton $S$ (i.e. a skeleton, which is a set, not a proper class).

Is this property equivalent to the following one:

• there is a map $X\to S_X$ which assigns to each object $X$ a "small" skeleton $S_X$ of the category ${\sf Mono}(X)$.
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If you assume global choice, then this is equivalent to being well-powered, so you're not really going to be able to exhibit an example that you want in (2) (since global choice cannot be disproven from the usual axioms). The best you can hope for is to construct a model of set theory in which you have a counterexample. Or do you want to demand some sort of compatibility between the choices of $S_X$ for different objects X? – Eric Wofsey Jul 20 '13 at 16:40
@Eric Wolfsey: this must be a misunderstanding... I don't see how this follows from global choice. – Sergei Akbarov Jul 20 '13 at 16:58
@SergeiAkbarov In the presence of the other usual axioms of set theory, including especially the axiom of regularity, the axiom of global choice gives you a well-ordering of the class of all sets. Given such a well-ordering, you can define $S_X$ to contain just the first (with respect to the well-ordering) representative of each isomorphism class of monomorphisms into $X$. – Andreas Blass Jul 20 '13 at 17:27
More explicitly, using regularity you can get the map you ask for by sending $X$ to the set of all skeleta of ${\sf Mono}(X)$ which have minimal rank. – Eric Wofsey Jul 20 '13 at 17:39
@SergeiAkbarov OK, I've copied my comment into an answer. What Eric mentioned is often known as "Scott's trick" (named after Dana Scott). The idea is that, if you have a (definable) class (definably) partitioned into subclesses, then there is a (defniable) way to shrink each of those subclasses to a nonempty subset of itself, namely the set of those members of the subclass that have the smallest rank (in the sense of the cumulative hierarchy of sets) among the members of that subclass. In your situation, you'd have the class of monomorphisms partitioned into isomorphism classes. – Andreas Blass Jul 21 '13 at 4:33

In the presence of the other usual axioms of set theory, including especially the axiom of regularity, the axiom of global choice gives you a well-ordering of the class of all sets. Given such a well-ordering, you can define $\mathcal S_X$ to contain just the first (with respect to the well-ordering) representative of each isomorphism class of monomorphisms into $X$.