# Showing there is a natural map between coim$(f)$ and im$(f)$ in an abelian category [duplicate]

We defined an abelian category as one who is (i) additive, (ii) every morphism has a kernel and a cokernel, (iii) and every monomorphism is a kernel and every epimorphism is a cokernel.

I'm given a morphism $$f: A \rightarrow B$$ is an abelian category $$\mathcal{C}$$.

I need to show that there is a natural (isomorphism) map $$\overline{f}$$ between the image and the coimage of $$f$$.

Attempt: I know the image of $$f$$ is the kernel of its cokernel, and the coimage of $$f$$ is the cokernel of its kernel.

So I have arrows $$\ker(f) \xrightarrow{k} A \xrightarrow{f} B \xrightarrow{q} coker(f).$$ I also have maps $$c: A \rightarrow coim(f) = coker(k)$$ and $$m: im(f) = ker(q) \rightarrow B.$$

Now we know that $$f \circ k = 0$$, since $$k$$ is the kernel of $$f$$. Since $$c$$ is the cokernel of $$k$$, by def. of cokernel there exists a unique factorization such that $$f = \psi \circ c.$$

Since $$q$$ is the cokernel of $$f$$, we have $$q \circ f = 0$$. Since $$m$$ is the kernel of $$q$$, we have a unique factorization $$f = m \circ \phi$$.

Now I'm stuck. I have these two unique morphisms $$\psi: coim(f) = coker(k) \rightarrow B$$ and $$\phi : A \rightarrow im(f) = ker(q)$$.

But how do I construct a morphism $$\overline{f} : coim(f) \rightarrow im(f)$$ ?? And how do I show it is an isomorphism?

Thank you for any help.