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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
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Lemma. Let Ab be the category of abelian groups and Tor its full subcategory of torsion ab.gr. 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
    
It seems that the argument above proves that any hereditary torsion subcategory of a Grothendieck category is bicomplete... so that is not a good source for counterexamples... –  Simone Nov 15 '12 at 23:41
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The question will be continued on mathoverflow: mathoverflow.net/questions/112574/… –  Martin Brandenburg Nov 17 '12 at 14:40

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