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.
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