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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: enter image description here

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?
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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.

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    $\begingroup$ Here is the updated decomposition. Can anything additional be said about the arrows? In particular, how should one intuitively interpret the coimage arrows? $\endgroup$ – user153312 Jan 17 '15 at 12:29

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