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.
Best Answer
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 $W$.