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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    $\begingroup$ They're not equivalent in general. It's true in an abelian category, but that's basically one of the axioms. $\endgroup$ – Zhen Lin Sep 27 '13 at 21:24
  • $\begingroup$ How do you define kernel and cokernel in a general category? $\endgroup$ – Thomas Andrews Sep 27 '13 at 21:25
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    $\begingroup$ @ThomasAndrews The non-additive version is "equaliser of the cokernel pair". $\endgroup$ – Zhen Lin Sep 27 '13 at 21:27
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    $\begingroup$ 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. $\endgroup$ – Johannes Kloos Sep 27 '13 at 21:28

The two concept aren't always equivalent. For example, in category of groups this is no longer true. In particular, the image of a group homomorphism is the kernel of its cokernel if and only if the image is a normal subgroup of the codomain.

You can prove easily that a kernel is always the kernel of its cokernel.


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