# Cocomplete but not complete abelian category

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.

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@Andrej Bauer: It is what I wrote in the question: an abelian category is Grothedieck if it is cocomplete, colimits are exact and is has a set of generators. The canonical example is that of a category of modules. A deep result of Gabriel and Popescu characterizes Grothendieck categories as the torsion-theoretic localizations of module categories. (see also en.wikipedia.org/wiki/Grothendieck_category) –  Simone Virili Nov 16 '12 at 14:02
Your lemma is a special case of the general observation that coreflective subcategories of complete categories are complete. –  Martin Brandenburg Nov 17 '12 at 14:46
Wrong answers are also hepful. They prevent us from falling in the same mistakes again and again. –  Fernando Muro Nov 25 '12 at 23:37
This question is quite challenging, despite of being quite basic at first sight. I have played around with a lots of cocomplete abelian categories, but somehow they always turn out to be complete. Meanwhile I suspect that counterexamples will be quite strange ... –  Martin Brandenburg Nov 26 '12 at 2:04
This question is still open (although an answer was accepted by the bounty rule)! –  Martin Brandenburg Nov 26 '12 at 20:22

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 $W$.

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I'm not seeing a thing wrong with this, and it's carefully written. Congratulations on this ingenious answer. Just a small note that $k_\beta \otimes_{k_\alpha} -$ preserves arbitrary products iff $k_\beta$ is a finite extension of $k_\alpha$ (being canonically isomorphic to $\hom_{k_\alpha-\text{Vect}}(k_\beta, -)$), in which case the canonical map $k_\beta \otimes_{k_\alpha} W(\alpha) \to W(\beta)$ would be an isomorphism. That's why we needed $k_\beta$ to be an infinite extension of $k_\alpha$. –  Todd Trimble Oct 15 '14 at 14:05
@ToddTrimble Yes, exactly. In fact, it would be enough for $k_\beta$ to be an infinite extension of $k_\alpha$ for some $\beta>\alpha$. –  Jeremy Rickard Oct 15 '14 at 14:54
This looks like the direct limit of an $\mathbf{On}$-indexed sequence of cocomplete abelian categories along colimit-preserving exact functors, so it should be just abstract nonsense that it is a cocomplete abelian category. –  Zhen Lin Oct 15 '14 at 15:14
@ZhenLin I think that's a nice way of thinking about what's happening. The abelian categories in the $\mathbf{On}$-indexed sequence are all complete and cocomplete. The functors between them preserve coproducts, and so the coproduct of any set of objects can be constructed in any of the categories that contains all the objects, and survives in the direct limit. But the functors don't preserve products, and the product of a set of objects gets bigger as we go through the ordinals and gets pushed out of the top of the category (technical term) as we take the direct limit. –  Jeremy Rickard Oct 15 '14 at 15:30
This is an example from the book! –  Fernando Muro Oct 15 '14 at 16:22

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.

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Actually I think the last hypothesis that $\scr A$ has push-outs is superfluous; it is enough to replace products with arbitrary limits in the definition of the functor F above, in order to derive directly the existence of colimits. –  George C. Modoi Mar 29 '13 at 16:07
The settings in which I worked are rather dual as those of the question. Dualizing again we obtain: A cocomplete well copowered additive category with a generator is also complete. In particular this gives another proof for the fact that Grothendieck categories are complete. –  George C. Modoi Mar 29 '13 at 16:35
@ Dominic Michaelis: Why are you editing my answer by replacing a capital $X$ with a script $\scr X$ without any relation with the rest? –  George C. Modoi Jul 25 '13 at 6:45
This is nothing new and can be found in any treatment of category theory. It is just the usual proof that Grothendieck abelian categories are complete. –  Martin Brandenburg Nov 1 '13 at 0:21

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.

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Once again, the Ind-completion of a small finitely cocomplete category is finitely accessible and cocomplete, hence locally finitely presentable, hence complete. Every answer that has appeared thus far has failed basically for this reason. –  Todd Trimble Nov 22 '12 at 21:48
THank you Todd Trimble, quite right, I find also where my answere is wrong. –  Buschi Sergio Nov 23 '12 at 14:32

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

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People keep making this same mistake. The fact that the product of some torsion abelian groups in the category of all abelian groups is not torsion does not imply that there is no product of these same objects in the subcategory. In fact, in the product of all the groups Z/n the subgroup consisting of torsion elements is a product (in the category of torsion abelian groups). –  Tom Goodwillie Nov 26 '12 at 0:55
@William: Weibel is wrong here. To anyone who has submitted an answer which has not resolved the question: please consider deleting your answer, because this silly MO mechanism will automatically award an "accepted answer" to the highest voted answer after the bounty is over and the OP hasn't accepted anything. In this case, that would be the present answer. :-) I might also mention that I am trying to email people who might well be able to answer this. (Pity that there aren't a huge number of categorists who tune in here.) –  Todd Trimble Nov 26 '12 at 1:18
-1. The full subcategory of abelian groups consisting of torsion abelian groups is coreflective, hence complete (see also the other comments). By the same reason, many other potential examples you first think of don't work. It is a common false belief (which also gets repeated again and again on mathoverflow) that limits and colimits are reflected/preserved/created by forgetful functors, see also mathoverflow.net/questions/23478/… –  Martin Brandenburg Nov 26 '12 at 2:01
Whoops! Silly me (comment removed, as well as the fake proof) –  Dylan Wilson Nov 26 '12 at 2:33