Cocomplete but not complete abelian category

Question

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.

Answer 1

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 $W$.

Answer 2

The following does't give an example as required, but eliminates some candidates. If $\scr A$ is an additive (not necessary abelian) category which is complete, well powered and has a cogenerator then $\scr A$ has coproducts. Indeed, $\scr A$ satisfies the hypotheses of Fred's Special Adjoint Functor Theorem (see Adamek Rosicky, Locally Presentable and Accessible Categories, Section 0.7). Thus every functor which preserves limits is a right adjoint. Let $X_i\in\scr A$, $i\in I$ be a set of objects. The functor $F=\prod_{i\in I} \scr {A}(X_i,-):{\scr A}\to \mathfrak {A}\mathfrak{b}$ preserves limits, so it has a left adjoint $G$. Straightforwardly $F$ is represented by the object $G(\mathbb Z)$ which has to be isomorphic to $\coprod_{i\in I}X_i$. The same is also true if we work with non-additive categories, but replacing the category of abelian groups with the category of sets.

In conclusion, if $\scr A$ has in addition push-outs (or equivalently coequalizers) then $\scr A$ is cocomplete. Note that an abelian category has always cokernels, therefore coequalizers.

Answer 3

According to Weibel, the category of torsion abelian groups is cocomplete but not complete. No proof is offered. pg. 426 of that book of his

Answer 4

I'm no totally sure (as ever), If no I hope could suggest some ideas..

Let $fAb$ the abelian category of finite abelian groups, and let $\mathcal{C}:= Ind(fAb)$ its ind-category, this is the full category of presheaves on $fAb$ isomorphic to a filtred diagram of representable. Now form usual literature (e.g. Artin MAzur "Etale Homotopy" appendix) $\mathcal{C}$ is abelian, (then has finite sums), and has filtered colimits then has (small) sums, then is cocomplete.

Consider a countable numeration of finite cyclic groups $(C_n)_{n\in \mathbb{N}}$, and suppose that exist the product $P:= \prod_n h_{C_n}$ in $\mathcal{C}$ of associate representable of the $C_n$'s, let $P\cong \varinjlim_{i\in I} h_{G_i}$ for some direct diagrams of finite abelian groups $G_i$. We have a split monomorphisms $\delta: \sum_n h_{C_n}\to P$, i claim the the family of maps $h_{C_n}\to \sum_n h_{C_n}\to P$ is epimorphic, this follow because the the family $G_i\to P$ is epimorphic, and any $G_j$ is a (finite) sums of cyclic groups. But then $\delta$ is a epimorphism, then a isomorphism. Now fix a cyclic group $C_{m}\neq 0$, and consider $(h_{C_m}, \sum_n h_{C_n})\cong \bigoplus_n fAb(C_m, C_n)$ (the sum is a direct colimits of finite sums, and finite sums are representable by a biproduct) and each elemts of this sum as all $0$'s but finite components, but this isnt true for $(h_{C_m}, P)\cong \prod_n fAb(C_m, C_n)$, and considering that $1_{C_n}:= \pi_i\circ \delta\circ \epsilon_i : h_{C_n}\to \sum_n h_{C_n}\to P \to h_{C_n} $ we get a absurd condition.