# Equivalence of the definitions of “image” in category theory

The concept of "image" in Category Theory is, depending on source, defined in two possible ways: Either as a factorization of a morphism, or as the kernel of a cokernel.

More precisely, let some category with zero morphisms be given. If $f: A \to B$ is a morphism in that category, we can define the image of $f$ via factorization: $$A \to I \underset{\operatorname{im}(f)}\hookrightarrow B = f$$ where $I$ is the smallest object with this property (this is the definition that Wikipedia and nLab give) or as the kernel of the cokernel: $$\operatorname{im}(f) = \ker(\operatorname{coker}(f))$$

What I would like to understand is when and why these definitions are equivalent. Thus, my (two-part) question is:

• Are these definitions equivalent in general, or only for Abelian categories?
• Where can I find a simple proof of the equivalence?

I have seen a (very sketchy) proof in the Abelian case that I did not fully understand, and I could prove that the kernel-cokernel definition gives a factorization for any category with zero morphisms, but not that it is minimal.

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They're not equivalent in general. It's true in an abelian category, but that's basically one of the axioms. – Zhen Lin Sep 27 '13 at 21:24
How do you define kernel and cokernel in a general category? – Thomas Andrews Sep 27 '13 at 21:25
@ThomasAndrews The non-additive version is "equaliser of the cokernel pair". – Zhen Lin Sep 27 '13 at 21:27
I use the definition that $f \circ \ker f = 0$ and if $f \circ k = 0$, there is a unique $g$ such that $g = \ker f \circ g$, similarily for cokernels. Note that I require my category to have zero morphisms. – Johannes Kloos Sep 27 '13 at 21:28