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This is an exercise in Ravil Vakil's Algebraic Geometry notes labeled hard

Let $f: A\rightarrow B$ be a morphism in an abelian category. Then $f$ factors through a morphism $f':A\rightarrow \operatorname{im}(f)$ (recall that $\operatorname{im}(f)$ is the (domain of the) kernel of the cokernel of $f$).

Why is $f'$ an epimorphism? That is why does $g\circ f'=0$ imply $g=0$ for any $g: \operatorname{im}(f)\rightarrow C$ ?

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marked as duplicate by Tim kinsella, Community May 7 '18 at 10:29

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    $\begingroup$ Maybe it's helpful: math.stackexchange.com/questions/1884741/… $\endgroup$ – Ben West May 7 '18 at 1:55
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    $\begingroup$ The difficulty of this depends on the assumptions you're using. The point is that the image is isomorphic to the coimage, and $A$ maps epimorphically on the coimage by definition. Sometimes the fact that the image is the coimage is included as one of the axioms of an abelian category, but I guess not in Vakil. $\endgroup$ – Kevin Carlson May 7 '18 at 2:57
  • $\begingroup$ Thanks Ben and Kevin. So it seems to be a duplicate, and yes the hard part definitely seems to be showing the natural map coimage to image is an isomorphism. $\endgroup$ – Tim kinsella May 7 '18 at 10:26