Projective modules are retracts of free modules

Proof. Suppose first that \(P\) is projective. We can choose a map of left \(R\)-modules \(p\colon N \to P\), where \(N\) is free, such that the induced map \(\pi _0(p)\colon \pi _0(N) \to \pi _0(P)\) is surjective. (For instance, we can take \(N\) to be the free module generated by the set \(\pi _0(P)\).) Letting \(N'\) be the fiber of \(p\), it follows from the long exact sequence on homotopy groups that \(N'\) is connective as well. The exact sequence \(N' \to N \to P\) of connective \(R\)-modules induces by Observation 6.4.10 a sequence \[ \Ext ^0_R(P,N') \to \Ext ^0_R(P,N) \to \Ext ^0_R(P,P) \to \Ext ^1_R(P,N'). \] Since \(P\) is projective, the last term vanishes by criterion (3) of Observation 8.5.3. In particular, there exists a map \(s\colon P \to N\) such that \(p \circ s = \id _P\), exhibiting \(P\) as a retract of the free \(R\)-module \(N\).

For the converse, we observe that the collection of projective left \(R\)-modules is stable under retracts. It will therefore suffice to show that every free left \(R\)-module is projective. We check criterion (2) from Observation 8.5.3. Since \(\Ext ^i_R(\bigoplus _I R, Q) \cong \prod _{I} \Ext ^i_R(R,Q)\), we may assume \(P = R\) is free on a single generator. But then \[ \Ext ^i_R(R,Q) = \pi _{0} \Hom _{\LMod _R}(R,Q[i]) \simeq \pi _0 \Hom _{\Sp }(\S ,Q[i]) = \pi _{-i}(Q). \] Since \(Q\) is connective, this is zero for \(i > 0\), as desired. □