Let A, B be commutative rings, φ a homomorphism of A to B, E1,E2,E3 three A-modules, f:E1×E→E3 an A-bilinear mapping and
fB:E1(B)×E2(B)→E3(B)
its extension. Consider a submodule F2 of E2, a submodule F3 of E3, and denote by T the submodule of E1 consisting of those x1∈E1 such that f(x1,x2)∈F, for all x2∈F. Suppose that B is a flat A-module and that F2 is finitely generated. Then T(B) is the set of those x1′∈E1(B) such that fB(x1′,x2′)∈F3(B) for all x2′∈F2(B).
Let p be the canonical surjection E3→E3/F3; with each x1∈E1 we associate the A-linear mapping x2↦p(f(x1,x2)) of F2 to E3/F3, which we denote by g(x1); then g is an A-homomorphism of E1 to HomA(F2,E3/F3) and the kernel of g is precisely T. Since B is a flat A-module, we have the exact sequence
0⟶T(B)⟶E1(B)1⊗g(HomA(F2,E3/F3))(B)
(no. 3, Proposition 1). By Proposition 11 the canonical homomorphism
is injective. On the other hand, as B is a flat A-module, (E3/F3)(B) is canonically identified with E3(B)/F3(B); taking the composition of ω with 1⊗g, we obtain a homomorphism u for which the sequence!
0⟶T(B)⟶E1(B)uHomB(F2(B),E3(B)/F3(B))
is exact. It follows immediately from the definitions that u(x1′), where
x1′=1⊗x1∈E1(B),
is the linear mapping which maps each x2′∈F2(B) to the class mod. F3(B) of fB(x1′,x2′); by linearity this is also true for all x1′∈E1(B); since the kernel of u is T(B), the proposition is proved.
Bourbaki puts the proof directly after the statement, so what is above is the statement and its proof together.