Take the 2-minute tour ×
MathOverflow is a question and answer site for professional mathematicians. It's 100% free, no registration required.

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.

share|improve this question
1  
@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
2  
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
1  
Whoops, I am withdrawing my answer as it is obviously wrong. Thanks for pointing it out... –  Bugs Bunny Nov 23 '12 at 6:22
1  
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
4  
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
show 6 more comments

5 Answers

up vote 3 down vote accepted

My previous answer was not correct indeed, so that what remains is just the following comment.

There is a way to consider the free colimit completion in the setting of additive categories; I will essentially use construtions and results which can found in this article of B. Day and S. Lack (in the particular case of additive categories): arXiv:math/0610439. Let $C$ be a locally small additive category. I will write $\widehat C$ for the free completion of $C$ by small colimits (in the enriched sense). An explicit construction of $\widehat C$ is the following: if $Ab$ denotes the symmetric monoidal category of abelian groups, $\widehat C$ is the full subcategory of the category of additive functors $$F: C^{op}\to Ab$$ which consists of those $F$'s which are small, i.e. such that there exists a short exact sequence of the form $$F''\to F'\to F \to 0$$ with $F''$ and $F'$ isomorphic to small sums of representable presheaves. The category $\widehat C$ is always locally small and cocomplete (and the Yoneda embedding $C\to\widehat C$ is the universal additive functor from $C$ to a cocomplete additive category. I claim that, if $C$ is abelian, then $\widehat C$ is abelian as well: indeed, it has finite limits (see Prop.4.4 in Day & Lack's paper) and the Yoneda embedding $C\to\widehat C$ commutes with limits; moreover, for any object $X$ of $C$ the evaluation at $X$ functor has both a left and a right adjoint (I leave as an exercise their explicit description) and thus is exact; as these evaluation functors form a conservative family, one deduces that $\widehat C$ is abelian whenever $C$ is abelian (in fact, it is sufficient for this that $C$ is finitely complete). Note also that, whenever they exist, limits of $\widehat C$ can be computed termwise. In particular, to check that a limit is not representable in $\widehat C$, it is sufficient to check that the corresponding presheaf over $C$ is not small.

The abelian category $\widehat C$ is known to be complete if one of the following conditions is satisfied: $C$ is essentialy small, or $C$ is itself complete. Furthermore, if ever $C$ is cocomplete, then it is a refexive subcategory of $\widehat C$, so that the completeness of $\widehat C$ implies the same property for $C$. In conclusion, if ever there is an example of a cocomplete abelian category which is not complete, there must be one of the form $\widehat C$ for an additive category $C$ which is not essentially small and which is not complete. Furthermore, if there is a counter example, there must exist a small family of representable presheaves over $C$ whose product in the category of preshaves is not small. So far, I did not succeed to find such a thing.

share|improve this answer
    
Can you explain why $A^{(X)}$ is a category at all? I have tried similar examples, and the problem always was that it is not clear at all if (the set of objects in) $A^{(X)}$ can be defined within the same universe as $A$. So this is an issue of set theory. –  Martin Brandenburg Nov 26 '12 at 2:24
2  
Denis-Charles -- I had the same idea of using the category of small presheaves valued in $Ab$ of a large set, but I couldn't understand why this wasn't closed under small products. Could you not leave it as an exercise and spell out the details, please? Perhaps I am missing something simple. –  Todd Trimble Nov 26 '12 at 2:36
    
As he points out, it looks like A^(X) is just locally small, so the "set" of objects is actually a class. Meanwhile Hom-sets are honest sets. –  Dylan Wilson Nov 26 '12 at 2:38
    
I modified the definition of $A^{(X)}$ because, with the previous one, there were no room for the identity. –  Denis-Charles Cisinski Nov 26 '12 at 2:45
3  
Wait, why can there be no such epimorphism? Since any object in $C$ is "finitely supported," any cone in $C$ factors through a finite direct sum of the $M_n$ (which exists in $C$). Doesn't that imply that the desired functor admits an epimorphism from the small sum of functors represented by the finite sums of the $M_n$? Where is my mistake? –  Daniel Schäppi Nov 26 '12 at 4:32
show 8 more comments

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.

share|improve this answer
    
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
add comment

Here's an idea. Maybe somebody can either use it to construct an example or prove it can't work?

Let ${\mathcal A}$ and ${\mathcal B}$ be complete and cocomplete abelian categories, and $F:{\mathcal A}\to{\mathcal B}$ an exact functor that preserves coproducts but not products.

Then the comma categories $F\downarrow id_{\mathcal B}$ and $id_{\mathcal B}\downarrow F$ are cocomplete abelian categories (with the obvious kernels, cokernels and coproducts). They are also both complete, but for different reasons. In the first case, products are constructed using compositions of natural maps $F(\Pi X_i)\to\Pi FX_i\to\Pi Y_i$, but in the second case they are constructed by taking pullbacks of $\Pi Y_i\rightarrow \Pi FX_i\leftarrow F(\Pi X_i)$.

What if we combine the two constructions?

Let ${\mathcal C}$ be the category with objects quadruples $(X, Y, \alpha:Y\to FX, \beta:FX\to Y)$ and maps $(X,Y,\alpha,\beta)\to(X',Y',\alpha',\beta')$ consisting of pairs of maps $X\to X'$ and $Y\to Y'$ making the obvious diagrams commute.

Then ${\mathcal C}$ is a cocomplete abelian category, with the obvious kernels, cokernels and coproducts, but I don't see any obvious construction of products. Is it complete?

For an explicit example, take ${\mathcal A}$ to be the category of abelian groups, ${\mathcal B}$ the category of rational vector spaces, and $F$ the functor $-\otimes_{\mathbb Z}{\mathbb Q}$. Is the category ${\mathcal C}$ constructed from this data complete? Specifically, let $S$ be the object $({\mathbb Z},{\mathbb Q},\alpha,\beta)$, where $\alpha$ and $\beta$ are the obvious isomorphisms. Is there a product of infinitely many copies of $S$?

share|improve this answer
4  
I think what you are describing is a lax 2-limit construction. If that's true, then your category will be locally presentable if the two starting categories are. Indeed, a category is locally presentable if and only if it is cocomplete and accessible, and the 2-category of accessible categories and accessible functors is closed under lax limits (see e.g. Adamek-Rosicky, Theorem 2.77). Note that any functor which preserves colimits is accessible since it preserves filtered colimits. So, to get a counterexample from this construction, I think you would have to start with one. –  Daniel Schäppi Nov 22 '12 at 16:57
add comment

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.

share|improve this answer
6  
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
1  
THank you Todd Trimble, quite right, I find also where my answere is wrong. –  Buschi Sergio Nov 23 '12 at 14:32
add comment

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

share|improve this answer
14  
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
2  
@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
2  
-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
add comment

Your Answer

 
discard

By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.