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.
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 $\text{im}(f) = \text{ker}(g)$.