# Does $\cdots \to G_1\overset f\to G_2 \overset g\to G_3\to \cdots$ exact imply $0\to \ker(g) \to G_2 \to \operatorname{coker}(f)\to 0$ exact?

Given a (part of a) long exact sequence of abelian groups (or modules over some commutative ring) $$\cdots \to G_1\overset f\to G_2 \overset g\to G_3 \to \cdots$$ we have the short exact sequence $$0 \to \ker(g) \to G_2 \to \operatorname{coker}(f)\to 0$$ which may be verified by simple diagram chasing. Does the same hold in a general abelian category (where diagram chasing doesn't make sense)? If not, does it more specifically hold in the category of $\mathcal O_X$-modules over a scheme $(X, \mathcal O_X)$?

I came across this problem because I need to do diagram chasing on the global sections of a diagram of sheaves of $\mathcal{O}_X$-modules with exact rows and columns. While the global section functor is not right-exact, I only need exactness at $\Gamma(X, G_2)$ in the short sequence to do the chasing that I want, which I have, since $\Gamma$ is a left-exact functor.

## 1 Answer

The Freyd-Mitchell theorem guarantees that you can perform diagram chasing in any abelian category (provided that your diagram only involves a set of objects).

In any case, diagram chasing is unnecessary. In any abelian category, any morphism $g : G_2 \to G_3$ gives rise to an exact sequence

$$0 \to \text{ker}(g) \to G_2 \to \text{im}(g) \to 0$$

and it's furthermore true that if $G_1 \xrightarrow{f} G_2 \xrightarrow{g} G_3$ is exact, then $\text{im}(g) = \text{coker}(f)$; this is the categorical dual of the more familiar version of exactness that $\text{im}(f) = \text{ker}(g)$.

If you prefer the dual argument, in any abelian category, any morphism $f : G_1 \to G_2$ gives rise to an exact sequence

$$0 \to \text{im}(f) \to G_2 \to \text{coker}(f) \to 0$$

and it's furthermore true by exactness that $\text{im}(f) = \text{ker}(g)$.

• Here "$=$" indicates that there's a natural map between the two things and it's an isomorphism. – Qiaochu Yuan Dec 22 '15 at 7:25
• I did not know about that theorem. That kind of renders my whole question moot (I also suddenly saw that there was a problem with taking cokernels not necessarily commuting with taking sections, which the FM-thm also avoids neatly). Thank you. – Arthur Dec 22 '15 at 7:26
• I mean, the Freyd-Mitchell theorem is really not an easy result to prove. I think it's worth avoiding if you can find cleaner categorical proofs of things instead of diagram chasing. – Qiaochu Yuan Dec 22 '15 at 7:27
• This is more of an "I need to know for my own sake that it's true", rather than "I need to write a full proof of this fact" (the short exact sequence was just an idea I has to solve a different problem, after all, and as I said in the first comment, might not even be sufficient all), so it will do just fine. But you're right, it would feel cleaner and neater without. – Arthur Dec 22 '15 at 7:30
• The embedding theorem also has its limitations. For instance, the embedding is guaranteed to be exact, but it does not necessarily preserve limits or colimits in general. – Zhen Lin Dec 23 '15 at 22:38