Studying homological algebra,I got stuck into this question:(all the objects and morphisms are in a general abelian category) I have $a \to b \to c$ with morphisms $\phi , \psi$ such that $\psi\phi=0$. In this situation there are natural maps $h : Imm \phi \to Ker \psi$ and $g : Coker \phi \to Imm \psi$. I would like to demonstrate that $ker h= coker g$.
If someone uses some immersion theorem (my teacher used it to justify Snake Lemma in a general abelian category without demonstrating it), this is quite simple ,as we know how modules are explicitly done.
I'm missing the way to demonstrate it in this general setting, just with "categorical" approach,without using immersion
Consider the morphism $u\colon \ker\psi\rightarrow b\rightarrow \rm{coker}\phi$. Since $\rm{im}~\psi\cong \rm{coker}(\ker\psi\rightarrow b)$, we easily find that $\rm{im}~\psi$ is the cokernel of $u$ via $g$ by checking that it satisfies the appropriate universal property. Dually, we find that $\rm{im}~\phi$ is the kernel of $u$ via $h$.
Now, since $\rm{im}~u\cong\ker{\rm{coker}~u}\cong\rm{coker}~\ker u$ holds and since $\rm{coker}~u$ is $\rm{im}~\psi$ via $g$ and $\ker u$ is $\rm{im}~\phi$ via $h$, we find $\rm{coker}~h\cong\rm{im}~u\cong\ker g$.
The whole thing probably becomes clearer once you sketch the diagram of the involved morphisms, but unfortunately I don't know how to do that in MathJax.