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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 $g: \operatorname{im}(f)\rightarrow C$ ?

Bernard
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Mr. Cooperman
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    Maybe it's helpful: https://math.stackexchange.com/questions/1884741/a-longrightarrow-textim-f-is-an-epimorphism-and-a-cokernel-of-ker-f-long?rq=1 – Ben West May 07 '18 at 01:55
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    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. – Kevin Carlson May 07 '18 at 02:57
  • 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. – Mr. Cooperman May 07 '18 at 10:26

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