Let f:M′→M be a homomorphism of A-modules of type C, and p:P→M a bounded left resolution of P of type C0. There exist a bounded left resolution p′:P′→M′ of type C0 and a morphism of complexes u:P′→P such that p∘u=f∘p′.
Let us argue by induction on the length n of P, the assertion being trivial when the latter is <0. Consider the mapping g:M′×P0→M such that
g(x,r′)=f(x)−p0(r′)
for x∈M′,r′∈P0,
and its kernel K; the A-module K is of type C since g is surjective and M′×P0 and M are of type C. Let h:P0′→K be a surjective homomorphism, where P0′ is of type C0; let us denote by p0′:P0′→M′ (resp. u0:P0′→P0) the homomorphism composed of h and the projection K→M (resp. K→P0); the homomorphism p0′ is surjective and we have a commutative diagram
P0′p0′↓M′u0fP0↓p0M.
It is then enough to apply the induction hypothesis to the homomorphism
Kerp0′→Kerp0
deduced from u0.
Bourbaki puts the proof directly after the statement, so what is above is the statement and its proof together.