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I know that a consequence of the Gabriel-Popescu theorem (i.e., every Grothendieck category is a torsion-theoretic localization of a full category of modules) is that any Grothendieck category (which by definition is cocomplete) is complete. I guess that this is not true for general abelian categories, so here is the question:

is it true that a cocomplete abelian category is complete? If no (as I suppose) is there some canonical counterexample?

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I don't see any obvious reason for it to be true, but at the same time it is difficult to come up with a counterexample because the cocomplete abelian categories I know of are Grothendieck categories... – Zhen Lin Nov 15 '12 at 22:10
Well, a natural way to find abelian non-Grothendieck category is to take full subcategories of some Grothendieck category. For example the category of torsion abelian groups is not grothendieck (there is no progenerator). It is cocomplete but it seems to me that the torsion part of the product in the category of ab. groups is a product in the category of the tor. ab. groups... so even this seems not to be a counterexample. – Simone Nov 15 '12 at 22:29
Lemma. Let Ab be the category of abelian groups and Tor its full subcategory of torsion Given a family {$A_i:i\in I$} we have that $t(\prod_IA_i)$ is a product in tor. Proof. Take a group $P$ and maps $p_i:P\to A_i$. By the universal property of the product in Ab, there is aunique morphism $\phi:P \to \prod_IA_i$ with the wanted compatibilities. As $P$ is torsion $\phi(P)\subseteq t(\prod_IA_i)$ and so restricting the codomain of $\phi$ we have verified the universal property of the product in tor.\\\ – Simone Nov 15 '12 at 23:35
The question will be continued on mathoverflow:… – Martin Brandenburg Nov 17 '12 at 14:40
The question was answered on mathoverflow, giving an example of a cocomplete but not complete Abelian category. – Simone Oct 23 '14 at 15:51
up vote 5 down vote accepted

I think it's a good idea to take questions off the unanswered list whenever possible. Here is a direct link to the MO answer:

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