$A \longrightarrow \text{Im} f$ is an epimorphism and a cokernel of ker $f \longrightarrow A$ in every abelian category 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.

 A: 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 $\operatorname{Coim}(f) = \operatorname{Coker}(\kappa)$.
