Let I be a right directed preordered set, (Sα)α∈I an increasing family of multiplicative subsets of a ring A and S=⋃α∈ISα. We write ρβα=iASβ,SαSβ,Sα for α≤β, ρα=iSα,Sα. Then (Sα−1A,ρβα) is a direct system of Zings and, if, for all α∈I, ρα′ is the canonical mapping of Sα−1A to lim→Sα−1A, there exists a unique isomorphism j from lim→Sα−1A to S−1A such that j∘ρα′=ρα for all α∈I.
For α≤β≤γ, ργα=ργβ∘ρβα (no. 1, Corollary 3 to Proposition 2), hence (Sα−1A,ρβα) is a direct system. We write A′=lim→Sα−1A; as ρα=ρβ∘ρβα for α≤β (no. 1, Corollary 3 to Proposition 2), (ρα) is a direct system of homomorphisms and j=lim→ρα is the unique homomorphism from A′ to S−1A such that j∘ρα′=ρα for all a∈I. The homomorphisms ρα′∘iAαS:A→A′ are all equal, for ρβα∘iAαS=iAβS for a≤β; let u be their common value. Clearly the elements of u(S) are invertible in A′, which shows that there exists a homomorphism h:S−1A→A′ such that h∘iAS=u (no. 1, Proposition 1). Then
j∘h∘iSA=j∘u=j∘ρα′∘iSαAα=ρα∘iAαS=iAS
for all α∈I and consequently j∘h is the identity automorphism of S−1A. On the other hand, for all α∈I,
h∘j∘ρα′∘iAαS=h∘ρα∘iAαS=h∘iAS=u=ρα′∘iAαS,
whence h∘j∘ρα′=ρα′ for all α∈I; it follows that h∘j is the identity automorphism of A′ and consequently j is an isomorphism.
Bourbaki puts the proof directly after the statement, so what is above is the statement and its proof together.