Let M be an A-module, P and P′ two projective resolutions of M; suppose that P is minimal. Let f:P→P′ and g:P′→P be two morphisms of resolutions. Then f is injective, g is surjective, and P′ is the direct sum of the subcomplexes Imf and Kerg. Moreover Kerg has zero homology.
In fact, α=g∘f is an automorphism of P (Prop. 8). Put f=f∘α−1.
We have
Imf=Imfandg∘f=1P,
which shows that P′=Imf⊕Kerg. Since the sequence
0⟶Kerg⟶P′gP⟶0
is exact and g is a homomorphism, Kerg has zero homology.
Bourbaki puts the proof directly after the statement, so what is above is the statement and its proof together.