I'm trying to understand the proof of proposition 25 in section 10.5 of Dummit and Foote:
The short exact sequence $$0 \to A \overset \psi \to B \overset \varphi \to C \to 0$$ of $R$-modules is split if and only if there is an $R$-module homomorphism $\mu:C \to B$ such that $\varphi \circ \mu$ is the identity map on $C$.
My question is this: The proof states that if $\mu$ is given define $C' = \mu(C)$. But how do we conclude $B \cong \psi(A) \oplus \mu(C)$?
Note that $B=\ker\phi+\operatorname{im}\mu$: every $b\in B$ can be written as $b=\underbrace{b-\mu\phi(b)}_{\in\ker\phi}+\underbrace{\mu\phi(b)}_{\in\textrm{im}\mu}$, indeed \begin{equation} \phi(\mu\phi-1_B)(b)=(\phi\mu)\phi(b)-\phi(b)=0. \end{equation} Moreover $\phi\mu(c)=0=1_C(c)=c\iff c=0$ and $\mu$ is injective (it has left inverse) i.e. $\textrm{im}\mu\cap\ker\phi=\lbrace0\rbrace$, therefore $B\simeq\psi(A)\oplus\mu(C)$ since $\psi(A)=\ker\phi$ by short exactness.
I'd like to get some sort of feedback: does my post answer to your question? If no, why? May I improve it? If yes, why haven't you accepted it yet? Thanks.