Let , $0\to A \overset{f}\rightarrow B\overset{g} \rightarrow C\to 0$ be an exact sequence of modules over a commutative ring $R$. Given $A$ and $C$ are projective modules. Prove that $B\cong A\oplus C$.
As, $C$ is projective and $g$ is onto so there exists $h:C\to B$ such that $gh=id_C$. Now we define a map $\psi :A\oplus C\to B$ by $\psi(a,c)=h(c)-f(a)$.
To show injectivity of $\psi$ , $\psi(a,c)=0$ implies $f(a)=h(c)$. Now , $c=gh(c)=g(f(a))=0$. Also, $f(a)=h(c)=h(0)=0$ , implies $a=0$, as $f$ is one-one.
I've stuck to show the surjectivity of $\psi$. Let, $b\in B$. Then we have to find an element in $(a,c)\in A\oplus C$ such that $\psi(a,c)=b$.
Can anyone help to find this element ? Or any other way to prove the isomorphism ?
Hint:
Show that, for any $b\in B$, $\;h\bigl(g(b)\bigr)-b\in\ker g$.
Some details:
Indeed, $\bigl(g\circ(h\circ g)\bigl)(b)=\bigl((g\circ h)\circ g\bigl)(b)=g(b)$.
Thus, $h\bigl(g(b)\bigr)-b\in \operatorname{Im}f$. Let $a\in A$ such that $f(a)=h\bigl(g(b)\bigr)-b$, and set $c=g(b)$. We have $$\psi(a,c)=h(c)-f(a)=h\bigl(g(b)\bigr)-h\bigl(g(b)\bigr)+b=b.$$