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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.

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marked as duplicate by Arnaud D., Javi, Klaus, user3417, Parcly Taxel Aug 8 at 17:04

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    $\begingroup$ The construction of the canonical map is explained there : math.stackexchange.com/questions/3025415/… $\endgroup$ – Arnaud D. Aug 8 at 10:28
  • $\begingroup$ I've shown the existence of the map. How do I show it is an epimorphism and a monomorphism? Do I have to show it is the kernel and cokernel of some map? And then use the fact that we are in an abelian category $\endgroup$ – Kamil Aug 8 at 11:58