If $f'\circ f=\operatorname{id}_N$, then $f(N)$ is a direct summand

Question

(This is a modified version of my previous question)

Suppose there is an injective $R$-module homomorphism $f:N\to M$ and a homomorphism $f':M\to N$ such that $f'\circ f=\operatorname{id}_N$. I need to show that $f(N)$ is a direct summand of $M$.

My idea is to consider the short exact sequence $$0\to f(N)\to M\to M/f(N)\to 0$$ where the first map is the inclusion $\iota$ and the second map is the projection $\pi$, and to define a map $h:M\to f(N)$ such that $h(\iota(x))=x$ for all $x\in f(N)$. This will imply that $M\simeq f(N)\oplus M/f(N)$.

The condition $h(\iota(x))=x$ is equivalent to $h(x)=x$, this suggests to define $h$ to be the identity. But the identity makes sense only on $f(N)$, and not on the whole $M$. How to define it on the whole $M$? I guess I need to use $f'$ somehow...

Answer 1

Let $U=ker(f')$, show that $U\oplus f(N)=M$. Let $x\in M, x=x-f(f'(x))+f(f'(x))$, $f'(x-f(f'(x))=f'(x)-f'(f(f'(x)))=0$. If $x\in f(N)\cap U$, $x=f(y), y\in N$, $f(f'(x))=f(f'(f(y))=f(y)=x$, and $f(f'(x))=0$ since $x\in ker f'$.