Exact sequence of $R$-modules

Question

Let $0\longrightarrow N\overset{f}{\longrightarrow}M\overset{g}{\longrightarrow}L\longrightarrow0$ be a short exact sequence of $R$-modules. Prove that this chain splits iff $f(N)$ is direct summand of $M$.

Thanks a lot.

Answer 1

Suppose $\;M=f(N)\oplus K\;$ , and define $\;F:M\to N\;$ by $\;F(m) = F(f(n)+k):=n\;$ . Show that $\;F\;$ is a well defined $\;R- $ homomorphism and $\;F\circ f=\text{Id}\,_N\;$ .

The other direction: suppose there exists an $\;R-$ homom. $\;F:M\to N\;$ s.t. $\;F\circ f=\text{Id}\,_N\;$. Show that

$$\begin{align*}(1)&\;\;\ker F\cap f(N)=\{0\}\\ (2)&\;\;M=f(N)\oplus\ker F\end{align*}$$

Note: Point (2) above may be a little tricky but not too much if you understand what's going on here.