This is a duplicate of the following question to which I did not receive any answer: <https://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. 



Cocomplete but not complete abelian category
I think I have an example.


Fix a chain of fields $k\_\alpha$ indexed by ordinals $\alpha$, where $k\_\alpha\subset k\_\beta$ is an infinite field extension for all pairs $\alpha<\beta$ of ordinals.


First I'll define an "abelian category" which has large Hom-sets.


An object $V$ will consist of a $k\_\alpha$-vector space $V(\alpha)$ for each ordinal $\alpha$, together with a $k\_\alpha$-linear map $v\_{\alpha,\beta}:V(\alpha)\to V(\beta)$ for each pair $\alpha<\beta$ of ordinals, such that $v\_{\beta,\gamma}\circ v\_{\alpha,\beta}=v\_{\alpha,\gamma}$ whenever $\alpha<\beta<\gamma$.


A morphism $\theta:V\to W$ will consist of a $k\_\alpha$-linear map $\theta\_\alpha:V(\alpha)\to W(\alpha)$ for each $\alpha$, such that $w\_{\alpha,\beta}\circ\theta\_\alpha=\theta\_\beta\circ v\_{\alpha,\beta}$ for all $\alpha<\beta$.


Now let's say that an object $V$ is "$\alpha$-good" if, for every $\beta>\alpha$, $V(\beta)$ is generated as a $k\_\beta$-vector space by the image of $v\_{\alpha,\beta}$, and that $V$ is "good" if it is $\alpha$-good for some $\alpha$. If $V$ is $\alpha$-good, then any morphism $\theta:V\to W$ is determined by $\theta\_\gamma$ for $\gamma\leq\alpha$, so the full subcategory $\mathfrak{C}$ of good objects has small Hom-sets.


It's straightforward to check that $\mathfrak{C}$ is an abelian category, and it has small coproducts in the obvious way, where $\left(\coprod\_{i\in I} V\_i\right)(\alpha)=\coprod\_{i\in I} V\_i(\alpha)$.


I claim that $\mathfrak{C}$ does not have all small products.


For any $\alpha$, let $P\_{\alpha}$ be the ($\alpha$-good) object with
$$P\_\alpha(\beta)=\begin{cases}0&\mbox{if }\beta<\alpha\\k\_\beta&\mbox{if }\alpha\leq\beta\end{cases}$$
and obvious inclusion maps. Then for any object $W$, $\operatorname{Hom}\_{\mathfrak{C}}(P\_\alpha,W)$ is naturally isomorphic to $W(\alpha)$ (i.e., $P\_\alpha$ represents the functor $W\mapsto W(\alpha)$ from $\mathfrak{C}$ to $k\_\alpha$-vector spaces), and if $\alpha<\beta$ then the map
$$W(\alpha)=\operatorname{Hom}\_{\mathfrak{C}}(P\_\alpha,W)\to\operatorname{Hom}\_{\mathfrak{C}}(P\_\beta,W)=W(\beta),$$
induced by the obvious inclusion $P\_\beta\to P\_\alpha$, is just $w\_{\alpha,\beta}$.


Suppose $W$ were the product in $\mathfrak{C}$ of of a countable number of copies of $P\_0$. Since it's an object of $\mathfrak{C}$, $W$ must be $\alpha$-good for some $\alpha$.


Then 
$$W(\alpha)=\operatorname{Hom}\_{\mathfrak{C}}(P\_\alpha,W)=\prod\_{i\in\mathbb{N}}k\_\alpha$$
and for $\beta>\alpha$
$$W(\beta)=\operatorname{Hom}\_{\mathfrak{C}}(P\_\beta,W)=\prod\_{i\in\mathbb{N}}k\_\beta.$$


But then $W(\beta)$ is not generated as a $k\_\beta$-vector space by the image of the natural map $w\_{\alpha,\beta}:W(\alpha)\to W(\beta)$, since $k\_\beta$ is an infinite extension of $k\_\alpha$, contradicting the $\alpha$-goodness of $W$.


