Equivalent condition for split exact sequence

Question

Let $M,N,\&\ L$ be any $R$-modules. Then, for any short exact sequence $$0\longrightarrow N\overset{f}{\longrightarrow} M\overset{g}\longrightarrow L \longrightarrow 0$$ where $f$ is an injective $R$-homomorphism, $g$ is surjective , and $\mathrm{im}(f) = \ker(g)$, the following two conditions are equivalent :

1. There exists an $R$-homomorphism $\psi:M\rightarrow N$ such that $\psi\circ f=I_N$,

2. There exists an $R$-isomorphism $\Phi:M\rightarrow N\oplus L$ such that $\pi_2\circ \Phi=g$ 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...

Answer 1

$\Phi$ 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 $\Phi(m_1'+m_2)=(n,l)$.

Answer 2

Variant proof

$f\circ\psi\colon M\to M$ 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 and $\operatorname{Im}f\simeq L$.

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 $$M=\operatorname{Im}(f\circ\psi)\bigoplus \ker(f\circ\psi)\simeq L\bigoplus N. $$