Free modules lift across every surjection.Suggest a slogan
Let \(R\) be a ring. Every free \(R\)-module is projective.
Let \(F\) be free with basis \((e_i)_{i\in I}\), let \(g\colon N\to M\) be a surjective \(R\)-linear map, and let \(f\colon F\to M\) be \(R\)-linear. For each \(i\in I\) choose \(n_i\in N\) with \(g(n_i)=f(e_i)\). There is a unique \(R\)-linear map \(\tilde f\colon F\to N\) with \(\tilde f(e_i)=n_i\) for all \(i\), and \(g\circ\tilde f\) agrees with \(f\) on the basis, so \(g\circ\tilde f=f\).