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Are filtered colimits exact in all abelian categories?

In Set, filtered colimits commute with finite limits. The proof carries over to categories sufficiently like Set (i.e. where you can chase elements round diagrams), in particular A-Mod where A is a commutative ring. This implies that filtered colimits are exact in A-Mod.

I am aware of a vague principle that things that are true in A-Mod are true for all abelian categories, but I have never seen a precise statement of this principle so I am not sure if it applies in this case.

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  • $\begingroup$ In case you are still curious about your "vague principle," you might be thinking of the Freyd-Mitchell Embedding Theorem: en.wikipedia.org/wiki/Mitchell's_embedding_theorem $\endgroup$
    – Justin Curry
    Commented Nov 14, 2012 at 15:34
  • $\begingroup$ See also Steve Lack's similar question (as yet not answered successfully) mathoverflow.net/questions/57099/… $\endgroup$
    – ziggurism
    Commented Mar 21, 2013 at 4:26
  • $\begingroup$ Filtered colimits commute with finite limits in any algebraic category (for a definition and characterization of 'algebraic categories' see for instance Borceux, Vol II, Thm 3.9.1). Thus filtered colimits are exact in any cocomplete algebraic category. $\endgroup$
    – Exterior
    Commented Feb 25, 2015 at 0:48

2 Answers 2

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Here's a dumb counterexample. If C is an abelian category, so is Cop. In Cop, filtered colimits are filtered limits in C. And, of course, there are many examples of abelian categories (such as abelian groups) where filtered limits aren't exact.

Of course, your question is really: when is an abelian category C sufficiently close to Set, so that we can ratchet up the fact that filtered colimits are exact in Set to a proof for C.

Any category of sheaves of abelian groups on a space (or on a Grothendieck topos) will have exact filtered colimits, for instance.

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  • $\begingroup$ OK thanks. What is the easiest way to see that filtered colimits are exact for sheaves of abelian groups? Look at stalks? Is there any sort of simple sufficient condition that would include this example? $\endgroup$
    – Martin Orr
    Commented Oct 24, 2009 at 10:45
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    $\begingroup$ Yes, it is apparent by looking at stalks. Or, you can do it in two steps: (1) presheaves of abelian groups have this property, and (2) prove it for sheaves, using that the sheafification commutes with all colimits and is exact. $\endgroup$
    – Charles Rezk
    Commented Oct 24, 2009 at 19:22
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    $\begingroup$ @Martin: An exercise with a hands-on hint to do this is 1.6.K in the notes at math216.wordpress.com (Jan 15 2013 version; presumably all later versions, and most earlier versions). $\endgroup$
    – Ravi Vakil
    Commented Feb 16, 2013 at 0:56
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A counterexample which is non-trivial is given in Chapter 6 of Neeman's book Triangulated Categories. The category in question is the full subcategory of additive functors Cat(S^{op}, Ab) where S satisfies some hypotheses (e.g. is an essentially small triangulated category) and we take those functors which are product preserving for small enough products (so as contravariant functors S-> Ab they send sufficiently small coproducts to products). This category is complete and cocomplete but has neither exact filtered limits or exact filtered colimits. More precisely it satisfies [AB4] and [AB4*] but neither [AB5] nor [AB5*].

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