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Let $0 \rightarrow A \xrightarrow{\psi} B \xrightarrow{\sigma} C \rightarrow 0$ be a short exact sequence of $R$-modules.

If there is a homomorphism $\rho :C \longrightarrow B$ such that $\sigma \circ \rho$ is the identity in $C$, how do you prove that $B=\psi(A)\bigoplus \rho(B)$?

It is straightforward to check that $\psi(A)\cap \rho(B)=0$, I don't know how to show that $\psi(A) + \rho(B) = B$ ?

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    Going back to the group case, are you familiar with the fact that a short exact sequence with a section $\iff$ semidirect product?2012-03-19

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Hint. $b - \rho(\sigma(b))\in\mathrm{Ker}(\sigma) = \mathrm{Im}(\psi)$.

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    oh I see it now.Thanks.2012-03-19