Long exact sequence into short exact sequences This question is the categorical version of this question about splitting up long exact sequences of modules into short exact sequence of modules.
I want to understand the general mechanism for abelian categories. Here's the decomposition I managed to get:

But it doesn't involve any cokernels like the module version does. The closest I get to a cokernel is via the following argument (which i'm not even sure is correct):
Since exactness is autodual, we have for any composition $g\circ f$ $$\mathrm{coim}g=\mathrm{coker}f\iff \mathrm{im}f=\mathrm{ker}g$$
This, along with the Image-Coimage isomorphism, yields the isomorphism of the objects $$\mathrm{Im}f\cong\mathrm{Coker}g$$
I'm not sure whether this is correct reasoning since it also leads to the isomorphism $\mathrm{Coker}g\cong\mathrm{Ker}g$ whenever $(f,g)$ is exact. So:


*

*What is the general decomposition of a long exact sequence into
short exact sequences in an abelian category?

*Is the argument I gave for $\mathrm{Im}f\cong\mathrm{Coker}g$ correct?


You are right, it finally involves cokernels (=coimages of the next arrows).
Your first line is correct: taking cokernel of both sides of $\def\im{{\rm im}\,} \im f=\ker g$, we'll get $\def\coker{{\rm coker}\,} \def\coim{{\rm coim}\,} \coker f=\coim g$. For the converse, take kernel of both sides.
Then, for the objects, your second line should read
$${\rm Im\,}g\cong {{\rm Coker}\,f}\,.$$
The statement ${\rm Coker\,}g\cong{\rm Ker\,}g$ does not follow.
And yes, this is the way how the long exact sequence splits into short exact sequences.
