Direct summand of modules proof I'm have problems trying to show this, any help would be appreciated.
Show $i: N \subset M$ is a direct summand iff $\exists$ a module map $r: M \to N$ s.t $ri = 1_{N}$, and that any complement of $N$ is isomorphic to $M/N$
I'm using it as a build up to prove Maschke's Theorem.
Thanks
 A: Let $A$ be a commutative ring with identity and
$$
0 \rightarrow N \xrightarrow{\alpha} M \xrightarrow{\beta} P \rightarrow 0
$$
be an exact sequence of $A$-modules and homomorphisms. Then the followings are equivalent:
(i) There is an isomorphism of sequences with the sequence
$$
0 \rightarrow N \xrightarrow{i} N\oplus P \xrightarrow{\pi} P \rightarrow 0
$$
where the maps $i$ and $\pi$ are the natural injection and projection respectively.
(ii) There is a $A$-morphism $P \xrightarrow{f} M$ such that $\beta f = 1_P.$
(iii) There is a $A$-morphism $M \xrightarrow{g} N$ such that $g\alpha = 1_N.$
(i)$\Rightarrow$(ii) and (i)$\Rightarrow$(iii) are obvious.
(ii)$\Rightarrow$(i): $M =$ Im$f$ + Ker$\beta$, because any $m \in M$ can be written as $m = m - f\beta(m) + f\beta(m)$ and $m - f\beta(m) \in$ Ker$\beta$. Also $m \in$ Im$f \cap$ ker$\beta \Rightarrow f(n) = m, \beta(m) = 0$ for some $n \in P$ $ \Rightarrow n = \beta f(n) = \beta(m) = 0 \Rightarrow m = f(n) = 0.$
(iii)$\Rightarrow$(i): $M =$ Im$\alpha$ + ker$g$ because any $m \in M$ can be written as $m = m - \alpha g(m) + \alpha g(m)$ and $m - \alpha g(m) \in$ Ker$g$. Also $m \in$ Im$\alpha \cap$ ker$g \Rightarrow \alpha(n) = m, g(m) = 0$ for some $n \in N$ $ \Rightarrow n = g \alpha(n) = g(m) = 0 \Rightarrow m = \alpha(n) = 0.$
A: One direction is trivial: if $M=N\oplus N'$, then by definition any element $m\in M$ can be written in one and only one way as $m=x+y$, with $x\in N$ and $y\in N'$; verifying that $r\colon m\mapsto x$ is the homomorphism you need is very easy.
Conversely, let $N'=\ker r$. Try to prove that


*

*$M=N+N'$; hint: write $m=r(m)+(m-r(m))$.

*$N\cap N'=\{0\}$.
If $M=N\oplus N'$, then the homomorphism theorem says
$$
M/N=(N+N')/N\cong N/(N\cap N')
$$
and…
