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My textbook, when discussing exact sequences of modules, makes the following claim.

If the short exact sequence $0\to M'\xrightarrow{u} M\xrightarrow{v} M''\to 0$ is exact and there is a morphism $u':M'\to M$ so that $uu'=\text{id}_{M'}$, it is easy to prove that $M\cong\text{Ker}(u')\oplus\text{Im}(u)$, and this means that this exact sequence splits.

However, the proof of this isn't obvious to me. Could anyone lend any hints? Many thanks, it's probably just a silly question.

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  • $\begingroup$ The composition $uu'$ doesn't make sense. $\endgroup$
    – Kenta S
    Commented Mar 28, 2021 at 2:09
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    $\begingroup$ Your $u’$ is wrong. Either it should go from $M$ to $M’$ and the composition you want is $u’u$; or else it should go from $M’’$ to $M$ and then you want the composition to be $u’v=\mathrm{Id}_{M’’}$. Also, you should state your definition of “the exact sequence splits” (there are several equivalent ways of defining it, so it’s important to know which one you are thinking about...) $\endgroup$ Commented Mar 28, 2021 at 2:11
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    $\begingroup$ (Correction: you want $uu’$ even if $u\colon M\to M’$... too late to edit the comment) $\endgroup$ Commented Mar 28, 2021 at 2:18
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    $\begingroup$ As a general principle with these sorts of questions, first ask yourself if there are any "obvious" maps between the things that are supposed to be isomorphic, in this case $M \to M' \oplus M''$ or the other way around. Then try to see if you can show that map is actually bijective. $\endgroup$ Commented Mar 28, 2021 at 3:54

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Take the sequence $0\rightarrow A \xrightarrow{\alpha} B \xrightarrow{\beta} C \rightarrow 0 $ an exact sequence.

And take $u:B\rightarrow A$ the inverse of $\alpha$, then the sequence splits. To prove this, you can take the morphism \begin{equation} \psi: B \rightarrow A \oplus C\ \end{equation} defined by $\psi(b)=u(b)+\beta (b)$. Now it's easy to check that this is an isomorphism and this is the (correct) isomorphism

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  • $\begingroup$ It is incorrect to refer to $u$ as “the inverse of $\alpha$“. If $\alpha$ is not surjective, then it has no inverse. When the sequence splits, $\alpha$ has a left inverse. $\endgroup$ Commented Mar 28, 2021 at 17:41
  • $\begingroup$ You are right, I meant that $\endgroup$
    – LuckyS
    Commented Mar 28, 2021 at 18:03

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