There are several definitions of a projective module. For example,
1) An $R$-module $P$ is projective if for every epimorphism $f : M \rightarrow P$ there exist a homomorphism $g : P \rightarrow M$ such that $f \circ g = id_P$.
2) An $R$-module $P$ is projective if given homomorphisms $h:P \rightarrow N$, $f : M \rightarrow N$, where $f$ is an epimorphism, there exists a homomorphism $g: P \rightarrow M$ such that $f \circ g = h$.
Proving that 2) => 1) is easy: just choose $N = P$ and $h = id_P$. But how to prove the converse, i.e. 1) => 2) ?
Best Answer
Hint: Consider the fiber product $X$ of $P$ and $M$ along $N$. What can you say about this commutative square?
Alternate hint: Pick generators for $P$: we can choose a sequence $0\to K \to R^i\to P\to 0$ which is exact. Then (why?) this reduces to solving the problem for free modules.