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...

  • $\begingroup$ 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? $\endgroup$ – Daniel Fischer Jul 5 '15 at 20:42
  • $\begingroup$ It is also equivalent to; 3) There exists a $R$-homomorphism $s\colon L\to M$ such that $\;g\circ s=\operatorname{id}_L$. $\endgroup$ – Bernard Jul 5 '15 at 21:01
  • $\begingroup$ Yes Bernard. I have showed that $2$ equivalent to the condition you wrote $\endgroup$ – Leonardo Jul 5 '15 at 21:04

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

  • $\begingroup$ 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.. $\endgroup$ – Leonardo Jul 5 '15 at 20:57
  • $\begingroup$ @Leonardo Realized surjectivity wasn't as easy as I thought, please see my edit. $\endgroup$ – Matt Samuel Jul 5 '15 at 21:01

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