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.