# Equivalent condition for split exact sequence

the question is in Module Theory,

Let $M,N,\&\ L$ be any R-modules. Then , for any short exact sequence $0\longrightarrow N\overset{f}{\longrightarrow} M\overset{g}\longrightarrow L \longrightarrow 0$ ($f$ is injective R-homo, $g$ is surjective , $Im(f)=ker(g)$)

The following two conditions are equivalent :

1) there exist an R-homo $\psi:M\rightarrow N$ s.t. $\psi\circ f=I_N$

2) there exist an R-isomo. $\Phi:M\rightarrow N\oplus L$ s.t. $\pi_2\circ \Phi=g$ and $\Phi \circ f=i_1$ (where $\pi_2$ is projection on second coordinate and $i_1$ is the inclusion map on first coordinate )

I proved $(2)$ implies $(1)$. But , I am stuck in defining an $R$-isomorphism from $M\rightarrow N\oplus L$ or vice versa.

I tried to define $\Phi(m)=(\psi(m),g(m))$. But, I'm stuck in proving it's isomorphism. I don't think it's true...

• Have you managed to show that $\Phi$ is injective? Or that it is surjective? If you haven't managed, what were the problems you encountered? – Daniel Fischer Jul 5 '15 at 20:42
• It is also equivalent to; 3) There exists a $R$-homomorphism $s\colon L\to M$ such that $\;g\circ s=\operatorname{id}_L$. – Bernard Jul 5 '15 at 21:01
• Yes Bernard. I have showed that $2$ equivalent to the condition you wrote – Leonardo Jul 5 '15 at 21:04

$f\circ\psi\colon M\to M$ is a projector. Indeed, $(f\circ\psi)\circ(f\circ\psi)=f\circ(\psi\circ f)\circ\psi=f\circ\psi$.
There results a direct sum decomposition $\;M=\operatorname{Im}(f\circ\psi)\bigoplus\ker(f\circ\psi)$.
Now, from $\psi\circ f=\operatorname{id}_L$, we deduce $\psi$ is surjective, hence =\operatorname{Im}f\simeq L$. The restriction of$g$to the direct summand$P=\ker(f\circ\psi)$is injective. Indeed, if$g(m)=0,\ m\in P$we have$m=f(n),\ n\in N$, by exactness. Since$m\in P$,$0=(f\circ\psi)(m)=f\circ(\psi\circ f)(n)=f(n)=m$. Moreover, since$\operatorname{Im}(f\circ\psi)\subset\ker g$and$g$is surjective, the retriction of$g$to$P$is also surjective, hence$P\simeq N$, so that finally $$M=\operatorname{Im}(f\circ\psi)\bigoplus \ker(f\circ\psi)\simeq L\bigoplus N.$$$\Phi$is indeed an isomorphism. For injectivity, suppose $$\Phi(m)=0$$ Then$\psi(m)=0$and$g(m)=0$.$g(m)=0$means that$m$is in the image of$f$. Thus there is an$n\in N$such that$f(n)=m$. Then$\psi(f(n))=n=0$, so we must have had that$n=0$, hence$m$was$0$in the first place. For surjectivity, given$(n,l)$we want to find$m$such that$\psi(m)=n$and$g(m)=l$. Take$m_1\in M$so that$g(m_1)=l$. If we have$\psi(m_1)=n'$, take$m_1'=m_1-f(n')$; then$g(m_1')$is still equal to$l$because$f(n')\in\mathrm{Ker}(g)$, and$\psi(m_1-f(n'))=0$. Finally take$m_2\in\mathrm{Ker}(g)$such that$\psi(m_2)=n$. Then$\Phi(m_1'+m_2)=(n,l)$. • For surjectivity,$let (n,l)\in N\oplus L$, by surjectivity of$g$we can find$m\in M$s.t.$g(m)=L$. But does this$m$satisfies$\psi(m)=n$??? i need to find$m\$?? I couldn't in fact.. – Leonardo Jul 5 '15 at 20:57