Take the 2-minute tour ×
Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

Why do we say "This functor is left exact, but not right exact" instead of "This functor preserves limits, but not colimits". It seems more natural to base the theory of derived functors on the second statement instead of the first. I've never really understood what it meant to say that the global sections functor is left exact. From looking at the definition on the wikipedia page, it says given an exact sequence $$0 \to A \to B \to C \to 0,$$ the functor is left exact when $$0 \to F(A) \to F(B) \to F(C)$$ is exact. However, if you consider the category of open sets on a space $X$, do exact sequences make sense in this case? $0$ is supposed to be a zero object, which means it is both initial and terminal. I don't think the category of open sets has that.

Also, are the following equivalent?

  1. $F$ preserves limits/colimits
  2. $F$ is left exact/right exact
  3. $F$ is right adjoint/left adjoint

Edit Number One: I think I have a better understanding now. I think has more to do with preserving arbitrary limits/colimits vs finite limits/colimits. If a functor is right adjoint, then it preserves arbitrary limits. From Goldbatt's Topoi book, I see that the definition of left exact is for the functor to preserve finite limits. So, do we lose flavor of the theory of Derived Functors when we say the functor preserves arbitrary limits/colimits instead of the restricted finite limits/colimits? What if we say that the functor is left adjoint/right adjoint?

Edit Number Two: Should the wiki article on Exact Functors be edited? Comparing to the article given over at nlab, I think the wiki article does a poor job. It seems to me that the wiki article gives the definition via short exact sequences, and then derives the "preserving limits/colimits" fact. This approach however assumes that the category has a zero object, and the source of some of my confusion. The nlab approach is better; it gives the definition of left exact functor as preserving finite limits.

share|improve this question
    
Yes, they are very near to each other, under certain circumstances, they are equivalent. –  Berci Nov 23 '12 at 23:58
    
Is the right adjoint / left adjoint situation the most general? Is it okay if I base my intuition of derived functors on this fact? –  six Nov 24 '12 at 0:11
1  
I think I have a better understanding now. I think has more to do with preserving arbitrary limits/colimits vs finite limits/colimits. –  six Nov 24 '12 at 0:26
1  
The utilitarian point of view: to check left exactness you only have to check a single shape of diagrams. Looks like an easier task than arbitrary finite limits although the conditions are equivalent for abelian categories. Moreover, when doing basic homological algebra you will need to talk about exactness in any case, while arbitrary limits need not necessarily enter the picture explicitly. –  commenter Nov 24 '12 at 2:03
1  
The title should be edited; this question isn't about derived functors but about exactness. You can talk about derived functors for an arbitrary (not necessarily left or right exact) additive functor. –  Qiaochu Yuan Nov 24 '12 at 4:53

1 Answer 1

up vote 3 down vote accepted

The basic point here is that there are various ways of defining both exact sequences and left / right exact functors. These definitions are equivalent in abelian categories but inequivalent in general, so which one you want to take outside of the setting of abelian categories depends on what theorems you want to still be true.

As for your specific question: in an abelian category, "preserves finite limits" is equivalent to the usual definition of left exactness, and dually "preserves finite colimits" is equivalent to the usual definition of right exactness. Preserving finite limits is a strictly weaker condition than having a left adjoint, which requires at least that a functor preserves arbitrary limits.

share|improve this answer

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.