Let $P$ be an $R-$module and let $F$ be a free module (say $F$ isomorphic to $R^n$). If $P$ is a summand of $F$, clearly there is a presentation (namely a surjective $R-$homomorphism) $p:F\to P$. Prove that $p$ admits a splitting, namely a right-inverse $\sigma:P\to F$ ($p\sigma=id_P)$ that is also an $R-$homomorphism.
Clearly since $p$ is surjective, it admits a right-inverse but I can't prove that it is an $R-$homomorphism, so probably the way to get the thesis is different. Thank you
You can write $F=P\oplus Q$, and define $p(u\oplus v)=u$, $u\in P,v\in Q$, clearly $\sigma:P\rightarrow F=P\oplus Q$ defined by $\sigma(u) =u$ is a splitting.