# On relationship of two categorical characterization of finitely generated objects.

I've encountered The following categorical characterization of finitely generated modules:

A $R$-module $M$ is finitely generated iff it satifies one of the following properties:

a): for any family of $R$-module $\{U_i\}_{i\in\mathcal I}$ and any epimorphism $f:\bigoplus_{i\in\mathcal I} U_i\twoheadrightarrow M$, there exists a finite subset $\mathcal F$ of $\mathcal I$ such that the restriction of $f$ on $\bigoplus_{i\in\mathcal F} U_i$ is also epic.

b): for any category $\mathscr I$ representing a directed partially ordered set and any functor $F:\mathscr I\to R\text -\mathsf{Mod}$ such that every arrow of $\mathscr I$ is mapped to an injection via $F$, then the canonical homomorphism $\varphi:\varinjlim\text{Hom}(M,-)\circ F\to\text{Hom}(M,\varinjlim F)$ is epic in $\mathsf{Ab}$.

My question is:

If $R\text -\mathsf{Mod}$ is replaced by an arbitrary cocomplete abelian category, are characterization a) and b) remain equivalent?

I've already proved that b) always implies a).

• This is at least very close to true: you can embed your abelian category in a category of modules, possibly by passing to a higher universe, and use the argument you already have, there. This embedding is exact, so a) reads the same in both categories. However, the embedding is generally not cocontinuous, so b) might change. I don't know whether there are some strange counterexamples where this actually separates the concepts. – Kevin Carlson Dec 13 '16 at 10:53
• @KevinCarlson I think there is no guarantee that such an embedding exists which preserves infinite colimit. – Censi LI Dec 13 '16 at 10:58
• Yes, that's what I said-"cocontinuous" means "preserving all colimits." – Kevin Carlson Dec 13 '16 at 11:43
• @KevinCarlson then a) might change, too, for it involves infinite coproduct. – Censi LI Dec 13 '16 at 11:54

The two are not equivalent in the opposite category of the category of vector spaces over a field $k$, since a one-dimensional vector space $k$ satisfies (a) but not (b).
• Excellent! In fact I was considering $\mathsf{Ab}^\text{op}$ and was stuck there. I never thought of trying opposite category of the simplest module category! – Censi LI Dec 13 '16 at 14:25