This is a duplicate of the following question to which I did not receive any answer: http://math.stackexchange.com/questions/238247/complete-but-not-cocomplete-category

Let $\mathfrak C$ be an abelian, cocomplete category. If $\mathfrak C$ has a generator and colimits are exact (i.e., $\mathfrak C$ is Grothendieck) then $\mathfrak C$ is the torsion-theoretic localization of a full category of modules (by the Gabriel-Popescu Theorem) and so it is also complete. Anyway I'm not aware of any counter-example showing that a cocomplete abelian category may not be complete. So my question is: could you provide such example or a reference to a proof of the bicompleteness of cocomplete abelian categories?

My first idea was to look for counterexamples in non-Grothendieck subcategories of a Grothendieck category. After some attempt I realized the following

**Lemma.** *Let $\mathfrak C$ be a Grothendieck category and $\mathcal T$ a full hereditary torsion subcategory (i.e. $\mathcal T$ is closed under taking sub-objects, quotient objects, extensions and coproducts). Then $\mathcal T$ is bicomplete.*

*Proof.* Let $T:\mathfrak C\to \mathcal T$ be the hereditary torsion functor associated to $\mathcal T$. Now, given a family {$C_i:i\in I$} of objects in $\mathcal T$ we can take the product $(P,\pi_i:P\to C_i)$ of this family in $\mathfrak C$. We claim that $(T(P), T(\pi_i))$ is a product in $\mathcal T$. Indeed, let $X\in \mathcal T$ and choose maps $\phi_i:X\to C_i$. By the universal property of products in $\mathfrak C$, there exists a unique morphism $\phi:X\to P$ such that $\pi_i\phi=\phi_i$ for all $i\in I$. Now, since $X\in\mathcal T$, there is an induced map $T(\phi):X\to T(P)$ which is clearly the unique possible map satisfying $T(\pi_i)T(\phi)=T(\phi_i)=\phi_i$. \\\

Thus there are lots of non-Grothendieck bicomplete abelian categories.

EDIT: notice that in the lemma we never use the hypothesis that the subcategory $\mathcal T$ is closed under taking extensions or subobjects. In fact, if $\mathcal T$ is just closed under taking coproducts and quotients, one defines the functor $T:\mathfrak C\to \mathcal T$ such that, for all object $X\in\mathfrak C$, $T(X)\in \mathcal T$ is the direct union of all the subobjects belonging to $\mathcal T$ (image (which is a quotient) of the coproduct of all the subobject of $X$ belonging to $\mathcal T$ under the universal map induced by the inclusions of the subobjects in $X$). Clearly $T(X)$ is fully invariant as a subobject of $X$ (by the closure of $\mathcal T$ under taking quotients and the construction of $T$) and so $T$ can be defined on morphisms by restriction. It is also clear that $T(X)=X$ if $X\in\mathcal T$ so the proof of the lemma can be easily adapted to this case.

REMARK: the new relaxed hypotheses of the lemma allow us to exclude other "exotic" examples... in particular, if you want to take the abelian subcategory of all the semisimple objects in a given Grothendieck category, this is closed under coproducts and quotients.