Let $\mathcal{C}$ is a grothendiect category and consider all of what follows in $\mathcal{C}$.

Let $$\{\varepsilon_i: 0\to A_i \to B_i \to C_i\to 0\ ,\ \phi_i^j\}$$ be a direct system of short exact sequences.

Is it true to say that $\varepsilon: 0\to \lim A_i \to \lim B_i \to \lim C_i \to 0$ (where $lim$ denotes the direct limit) is the direct limit $\lim \varepsilon_i$.

Actually we know that there exists a morphism $\varepsilon_i\to \varepsilon$ for each $i$.

What does occur when $\varepsilon_i$ is a split short exact sequence for each $i$? In fact why in this fact the direct limit is a pure short exact sequence?

In a locally finitely presented category, an exact sequence is called pure if it is a direct limit of split short exact sequences. In a locally presentable category the notion of $\alpha$-purity arises in a similar way (In fact pure exact sequences are $\aleph$-pure exact sequences).

For example in the category of modules over a ring, an exact sequence is pure if and only if it is a direct limit of split exact sequences.

  • 1
    $\begingroup$ Your question has a large number of typos and confusing phrasing. The answer to your first question "is it true" is yes. For your second question on splittings I see no reason why splittings at each $i$ would not assemble to a splitting for $\epsilon$, but I could be wrong here. For your last question I don't understand what you're asking. You'll get better answers if you take the time to make your question clearer $\endgroup$ May 18, 2013 at 10:04
  • $\begingroup$ @David White: There is no reason that one should be able to choose the splittings to be compatible with the direct system so as to yield a splitting in the limit. For example, the submodule $\bigoplus_{\mathbb{N}} \mathbb{Z}$ of $\prod_{\mathbb{N}} \mathbb{Z}$ is pure, but it is not a direct summand. $\endgroup$
    – Martin
    May 19, 2013 at 4:08

1 Answer 1


I am going to concentrate on the underlying categorical issue here and leave the homological algebra to the experts in that subject.

A direct limit is also known as a colimit. Yours, I take it, is over a sequence $N$, and as such is called directed or filtered.

Being an short exact sequence $0\to A\to B\to C\to 0$ amounts three things:

  • $A\to B$ is a monomorphism or $0$ is its kernel, which are kinds of limit (or projective limit in old terminology), but finitary ones;

  • $B\to C$ is an epimorphism, or $C$ is its image, which are other kinds of colimit property, this time finitary ones.

  • the image of $A\to B$ is the kernel of $B\to C$, which combines properties of both kinds.

Now, limits commute with limits and colimits commute with colimits.

The question is whether filtered colimits commute with finite limits.

Indeed, they do if the category described is by a finitary algebraic theory.

I presume that a Grothendieck category is like this, though Zhen Lin says otherwise.

As David White says, the interest in filtered colimits arises from the fact that they are the ones that commute with finitary stuff such as limits in $\bf Set$.

However, this is not a theorem of categories in general.

As Zhen Lin points out, ${\bf Set}^{op}$ does not have this property. Limits and colimits in $\bf Set$ behave quite differently, as I emphasise in Chapter V of my book, Practical Foundations of Mathematics.

A preorder in which this happens is called meet-continuous. See Counterexamples O 4.5 of A Compendium of Continuous Lattices for complete lattices that are not meet-continuous.

(I had difficulty with my Internet connection when I originally posted this, so some of the intended text got deleted by mistake.)

I repeat that I am not an expert on homological algebra, but I note that the Wikipedia article on Grothendieck categories says explicitly that "direct limits (a.k.a. filtered colimits) of exact sequences are exact".

  • $\begingroup$ I thought filtered colimits always commute with finite limits. Isn't that basically what filtered colimits are good for? There's an old mathoverflow question on this subject $\endgroup$ May 18, 2013 at 11:15
  • $\begingroup$ A Grothendieck category has a lot of smallness (I think it's locally presentable for example) so that means Hom commutes with colimits, maybe with some hypothesis on the kind of colimit or its length. It seems your argument might hold more generally and not really need this smallness hypothesis $\endgroup$ May 18, 2013 at 11:17
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    $\begingroup$ @DavidWhite Filtered colimits commute with finite limits in any finitely accessible category (hence Paul's emphasis on finitary algebraic theory). In particular it will be true in a locally finitely presentable abelian category, but Grothendieck categories need not be locally finitely presentable. Moreover it is not true in general that filtered colimits commmute with finite limits: take $\mathbf{Set}^\mathrm{op}$, for example. $\endgroup$
    – Zhen Lin
    May 18, 2013 at 12:36
  • $\begingroup$ To be clear, filtered colimits do commute with finite limits in a Grothendieck category – but this is basically part of the definition! For an example of a Grothendieck category that is not locally finitely presentable, one just has to look for a topos that is not locally finitely presentable. $\endgroup$
    – Zhen Lin
    May 18, 2013 at 13:34
  • $\begingroup$ I thought that might be the case, but maybe for completeness you could either say explicitly what a Grothendieck category is or provide a link to a definition. $\endgroup$ May 18, 2013 at 14:29

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