If we have an exact sequence of modules $O\rightarrow A\stackrel{f}{\rightarrow}B\stackrel{g}{\rightarrow}C$, then from the third isomorphism theorem we can conclude that $ \operatorname{Im} g\cong \operatorname{Coker} f$.

I think that this is also true in every abelian category.

We have that

\begin{align} \operatorname{Im}g&=\operatorname{Ker}(\operatorname{Coker}g)\cong \operatorname{Coker}(\operatorname{Ker}g)\cong \operatorname{Coker}( \operatorname{Im} f)\\ &= \operatorname{Coker}(\operatorname{Coker}(\operatorname{Ker} f))\cong \operatorname{Coker}(f). \end{align}

First isomorphism holds by the axiom(s) of abelian categories, second holds by exactness of the sequence $O\rightarrow A\stackrel{f}{\rightarrow}B\stackrel{g}{\rightarrow}C$ at $B$ and the last holds because $f$ is a monomorphism.

Is my reasoning correct?


This is correct. Note that if you just want to verify whether a statement like this is true (rather than finding a proof directly from the axioms for an abelian category), there is a very quick and simple method: invoke the Freyd-Mitchell embedding theorem, which says that any small abelian category has an exact fully faithful functor into the category of modules over some ring. So any statement about a small collection of objects in an abelian category and properties preserved by an exact fully faithful functor which is true in the category of modules over a ring is true in any abelian category. In particular, for instance, an exact functor preserves exact sequences, images, and cokernels, and so you can deduce your statement for general abelian categories from the statement for modules.

(Of course, this is not actually logically simpler than your argument, since the proof of the Freyd-Mitchell embedding theorem is hard. But it's a really handy tool for quickly understanding what statements are true in arbitrary abelian categories.)


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