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 $\psi(A) + \rho(B) = B$ ?