Let $\phi: M \to N$ be an $R$-module homomorphism having a left-inverse $\psi$. Prove that $\ker (\psi) \cong \operatorname{coker} (\phi)$ Let $\phi: M \to N$ be an $R$-module homomorphism having a left-inverse $\psi$. 
I need to prove that $\ker ( \psi )\cong \operatorname{coker}  \phi $.
How to approach the problem? Any hints will be appreciated.
 A: Verify that the sequence
$$0\longrightarrow \ker \phi\stackrel{}{\longrightarrow}M\stackrel{\phi}{\longrightarrow}N\stackrel{(Id-\phi \circ \psi)|^{\ker \psi}}{\longrightarrow}\ker \psi ~ {\longrightarrow} ~0$$
is well-defined and exact.

Clarification:
The sequence 
$$0\longrightarrow \ker \phi\stackrel{}{\longrightarrow}M\stackrel{\phi}{\longrightarrow}N\stackrel{\pi}{\longrightarrow}\text{coker} \phi ~ {\longrightarrow} ~0$$
is clearly exact. It follows that $\text{coker} \phi \cong \ker \psi$. In general, if two sequences
$$0\longrightarrow A \stackrel{}{\longrightarrow}B\stackrel{f}{\longrightarrow}C\stackrel{h}{\longrightarrow}D ~ {\longrightarrow} ~0$$
and 
$$0\longrightarrow A \stackrel{}{\longrightarrow}B\stackrel{f}{\longrightarrow}C\stackrel{h'}{\longrightarrow}D' ~ {\longrightarrow} ~0$$
are exact, then $D \cong D'$. One can see this, for example, by defining the map $h(x) \mapsto h'(x)$. This is well defined, since $h$ is surjective and $h(x)=h(y) \implies x-y \in \ker h \implies x-y \in \text{Im}~ f \implies x-y \in \ker h' $
$\implies h'(x-y)=0 \implies h'(x)=h'(y).$ It is also a homomorphism, since $h$ and $h'$ are homomorphisms. And clearly it is an isomorphism.
A: You know that $\operatorname{coker}\phi=N/\phi(M)$. Let's define a homomorphism $f\colon N\to N$ by
$$
f(x)=x-\phi\psi(x)
$$


*

*Prove that $f(N)=\ker\psi$

*Prove that $\ker f=\phi(M)$

*Deduce that $\ker\psi=f(N)\cong N/\ker f=N/\phi(M)$.
