Let $R$ be a ring. I know that there is a proof for the statement that "every free $R$-module is a projective $R$-module". And, of course, $R$ is a free $R$-module. I was wondering whether there is a direct proof of the projectiveness of $R$.
Suppose $e: A \to B$ is an epimorphism, and $f: R \to B$ is a homomorphism. Is there a way to `know' the homomorphism $g: R \to A$ that has $eg = f$?
Yes. The morphism $f$ is given by sending 1 to an element $b\in B$. Because e is onto, there is $a\in A$ that covers it. Define $g$ that sends 1 to $a$. Note that this is the proof free modules are projective: you lift the basis elements.