In Algebra: Chapter 0, the author proves the first part of the splitting lemma with the following: Proposition:
And the proof:
However, why do we have that coker $\phi$ $\cong$ ker $\psi$? I see that this would hold for when $\phi$ is surjective, as then the ker $\psi$ would be the trivial group and so would the cokernal $\phi$, however, I'm not sure what an isomorphism would be between ker $\psi$ $\subset$ N and coker $\phi$ would be in general.
Thanks in advance
Note that the author shows that $\theta\colon N\to M\oplus \ker(\psi):n\mapsto (\psi(n),n-\phi\psi(n))$ is an isomorphism.
Recall that kernels, cokernels, etc..., are defined up to isomorphism. Thus the cokernels of the maps $\phi$ and $\theta\circ \phi$ are isomorphic, i.e. $\text{coker}(\theta\circ \phi)\cong \text{coker}(\phi)$.
Now $$\theta\circ \phi\colon M\to M\oplus \ker(\psi):m\mapsto (\psi\phi(m), \phi(m)-\phi\psi\phi(m))=(m,\phi(m)-\phi(m))=(m,0).$$
As the cokernel of a module map $f\colon X\to Y$ is isomorphic to $Y/\text{im}(f)$, we find that $\text{coker}(\theta\circ \phi)\cong (M\oplus \ker(\psi))/M\cong \ker(\psi)$.