Long exact sequence into short exact sequences

Question

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:

1. What is the general decomposition of a long exact sequence into short exact sequences in an abelian category?
2. Is the argument I gave for $\mathrm{Im}f\cong\mathrm{Coker}g$ correct?

Answer 1

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.