— Let B be a perfect ring of characteristic p and A a subring of B. Write Ap−∞=⋃f≥0Ap−f and denote by j the canonical injection of A in Ap−∞. Then Ap−∞ is the smallest perfect subring of B containing A and (Ap−∞,j) is a perfect closure of A.
For each integer f∈Z denote by πf the automorphism b↦bpf of B. The sequence of subrings π−f(A) of B (for f≥0) is ascending and its union Ap−∞ is thus a subring of B. We have π1(Ap−∞)=⋃f≥0π−(f−1)(A)=Ap−m, hence Ap−∞ is a perfect subring of B. Finally let B0 be a perfect subring of B containing A; for every integer f≥0 we have π−f(A)⊂π−f(B0)=B0, whence Ap−∞⊂B0.
If v is a homomorphism of A into a perfect ring B′ of characteristic p, then for every integer f≥0 we can define a homomorphism hf of π−f(A) into B′ by hf(π−f(a))=v(a)p−f for all a∈A. We see at once that hf+1 agrees with hf on π−f(A); thus there exists a homomorphism h of Ap−∞ into B′ which induces hf on π−f(A) for all f≥0 and in particular, h extends h0=v. If h′ is another extension of v to a homomorphism of Ap−∞ into B, the equality h′=h may be established as in the proof of Th. 3.
Bourbaki puts the proof directly after the statement, so what is above is the statement and its proof together.