1
$\begingroup$

As a beginner to learn homological algebra, I have just learned about the direct system and its direct limit.As R-mod categories have arbitrary coproducts indexed by a set, any direct system in R-mod indexed by J has direct limit which is defined as the quotient mod of the direct product. Then,there comes a further question: Since R-mod category is a special abelian category with arbitrary direct limit and not all of abelian categories have this property, what is the necessary and sufficient condition ? And a different question is: when does a direct system in an abelian category has direct limit? A direct system in category C indexed by a quasi-ordered set J is a covariant functor from J to C.
It seems that an abelian category has arbitrary direct limit indexed by a set iff it has arbitrary coproducts. But I can't prove it. If this is right, could you give me some hints?

$\endgroup$
  • 1
    $\begingroup$ It would be nice to restrict one question per post $\endgroup$ – Shailesh Oct 17 '15 at 2:31
  • $\begingroup$ Well, My major point is to decide the necessary and sufficient condition. The latter one is less distressing since every finite set J has its corresponding direct limit. Owing to my region(PRC),it's hard for me to search for English texts. While ,I still get the information that an abelian category with AB3* has colimit without any proof of it. Do I need to define the so-called "quotient category"? $\endgroup$ – user12580 Oct 17 '15 at 2:46
  • $\begingroup$ $R$-mod does not have arbitrary co-products, it has co-products indexed over a set. One can prove that if a category has arbitrary co-products then it has at most one morphism between any pair of objects. $\endgroup$ – Nex Oct 17 '15 at 3:05
  • $\begingroup$ To be definite,you are right. I missed the part"indexed over a set". The direct system is indexed by a quasi-ordered set, and I guess it's right that an abelian category with arbitrary coproducts indexed by a set has direct limit.If so,could you give me some hints? $\endgroup$ – user12580 Oct 17 '15 at 3:12
  • $\begingroup$ This question is related: mathoverflow.net/questions/112574/… $\endgroup$ – Nex Oct 17 '15 at 4:33
4
$\begingroup$

You are correct that every abelian category has coequalizers given by cokernels. So if the category has all small coproducts, it has all colimits (aka directed limits).

Another approach is the following: an abelian category has filtered colimits (aka filtered directed limits) iff it has all colimits.

Clearly, if it has all colimits, then it also has filtered colimits. Now let us prove the converse.

An abelian category has finite coproducts, given by direct sums. Then given a set of objects $\{A_i\}_{i\in I}$ of objects in our category, we have the coproduct as the filtered colimit of all finite coproducts of the $A_i$. Therefore, we have coproducts and coequalizers, hence all colimits.

The more interesting question is when a filtered colimit of exact sequences remains exact. This condition is part of the definition of Grothendieck categories (which also includes the existence of a generator).

$\endgroup$
  • $\begingroup$ My definition of Grothendieck category includes the assumption of a generator. $\endgroup$ – Zhen Lin Oct 17 '15 at 9:15
  • $\begingroup$ Thank you! I've got the solution to this problem: using coequalizers to get colimit. It's shown in this way:functor F:J-->C, with u:j-->k denoting an arrow of the index set J.The coproducts of {Fk} (with u:j-->k) and the coproducts of {Fj} (j∈J) exist by assumption.Then use these two coproducts to acquire the colimit. Is it the right way? $\endgroup$ – user12580 Oct 17 '15 at 10:23
  • $\begingroup$ @Zhen Lin Yes, you are right, changed the wording. $\endgroup$ – Rachmaninoff Oct 18 '15 at 22:54

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

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