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In an abelian category, every map $f:B \to C$ factors as $B \xrightarrow{e} \text{im}(f) \xrightarrow{m} C$ with $m = \ker(\text{coker}f)$ monic and $e $ epi.

How can $\text{im}(f)$ be defined in an abelian category?

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  • $\begingroup$ The answer is in these notes. Get the most recent version and look in the section on abelian categories in the chapter on category theory: math.stanford.edu/~vakil/216blog $\endgroup$ – user4571 Dec 21 '15 at 21:31
  • $\begingroup$ Thanks @Patrick, found it in the notes, but how do they identify $\text{im} f$ with an object? $\endgroup$ – BananaCats Category Theory App Dec 21 '15 at 21:38
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    $\begingroup$ The image in an abelian category for a morphism $f:X\rightarrow Y$ is a morphism $\operatorname{im}(f)\rightarrow Y$, right? So just take the object named $\operatorname{im}(f)$. $\endgroup$ – user4571 Dec 21 '15 at 22:01
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The question is essentially answered in the comments, so I'm writing this just so it has an official answer.

Depending on how you define abelian categories, it is either a theorem or by definition that every arrow factors (essentially uniquely) as an epi followed by a mono. The epi is called the coimage and the mono is called the image. The object you are asking about is the domain of this monomorphism. As usual in category theory, it is only identified up to isomorphism.

If you're asking how to realize this image morphism - i.e how to construct it using more familiar creatures -the answer is that the image of an arrow may be realized as the kernel of its cokernel. To verify this you can check both have the same universal property. In familair territories like the category of abelian groups, you will be able to realize the image of a morphism as the usual set theoretic image of a function with extra structure.

The image can't really be identified with an object: for example, the very fact every arrow factors as the coimage followed by the image means that image and coimage objects are isomorphic, so from looking at the object alone you cannot tell "which role it's playing". Usually, by slight abuse of notation, one identifies $\mathrm{Im}f$ with the subobject of $Y$ that it represents as a monomorphism.

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