How can I verify the following statement?

The image of a morphism $f : A \longrightarrow B$ is defined as $\text{Im}(f) = \text{Ke}r(\text{coke}r f)$ whenever it exists (e.g., in every abelian category). The morphism $f : A \longrightarrow B$ factors uniquely through $\text{Im} f \longrightarrow B$ whenever $\text{Im} f$ exists, and $A \longrightarrow \text{Im} f$ is an epimorphism and a cokernel of ker $f \longrightarrow A$ in every abelian category.

  • $\begingroup$ The statement is misleading since $\text{Im}(f)$ exists in any regular category but kernels and cokernels might not (for example the category of sets is such a category). Furthermore, there are pointed regular categories where in general $\text{Im}(f) \neq \text{Ker}(\text{coker}(f))$ (for example the category of groups). $\endgroup$ – Nex Aug 7 '16 at 5:48

Consider the exact sequence $$0 \to \operatorname{Ker}(f) \xrightarrow{\kappa} A \xrightarrow{f} B \xrightarrow{\gamma} \operatorname{Coker} (f) \to 0.$$

We have $\gamma f = 0$ by definition of the cokernel, so the universal property of the kernel $\operatorname{Ker}(\gamma) = \operatorname{Im}(f)$ yields a unique morphism $\alpha:A \to \operatorname{Im}(f)$ such that $f = \iota \alpha$, where $\iota:\operatorname{Im}(f) \to B$ is the monomorphism that comes with the kernel. By definition of a kernel, $$0 = f\kappa = \iota \alpha \kappa, $$ and therefore $0 = \alpha \kappa$ as $\iota$ is monic. Then the universal property of the cokernel $\operatorname{Coker}(\kappa) = \operatorname{Coim}(f)$ yields a morphism $\tilde{f}:\operatorname{Coim}(f) \to \operatorname{Im}(f)$.

Now assume the category is Abelian. This implies that the canonical morphism $\tilde{f}:\operatorname{Coim}(f) \to \operatorname{Im}(f)$ defined above is an isomorphism. This in turn implies that $\alpha:A \to \operatorname{Im}(f)$ is epic, as it is the composition $$\alpha: (A \to \operatorname{Coim}(f) \xrightarrow{\tilde{f}} \operatorname{Im}(f))$$ where the first map is the epimorphism given by the cokernel. It is also a cokernel of the kernel $\kappa$ by construction, as $\operatorname{Coim}(f) = \operatorname{Coker}(\kappa)$.

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    $\begingroup$ The proof, why $\tilde{f}$ is actually an isomorphism, see here $\endgroup$ – ctst Aug 26 '16 at 9:46
  • $\begingroup$ hey alex, why is $\iota$ monic? are kernels always monic in an additive category? $\endgroup$ – Tim kinsella Jun 30 '18 at 20:12
  • $\begingroup$ answer: yes. :) $\endgroup$ – Tim kinsella Jun 30 '18 at 20:28
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    $\begingroup$ @Timkinsella This holds in every category! See e.g. math.stackexchange.com/questions/645709/every-kernel-is-monic $\endgroup$ – Alex Provost Jun 30 '18 at 20:43

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