If A is a Gorenstein ring, for example a regular ring, one can apply Example 3 by taking Ω=A and d=dim(A) (§ 3, No. 7, Prop. 11).
For every integer n⩾0, set In=T(A/mAn). For m⩾n, denote by pmn:A/mAm⟶A/mAn the canonical surjection and by imn:T(A/mAn)⟶T(A/mAm) the A-linear mapping T(pmn). It is injective by FD 4), and one has imn∘inp=imp for m⩾n⩾p by FD 3). Let I=limT(A/mAn) be the inductive limit A-module of the system ((In),(imn)). For n⩾0, the canonical mapping In→I is injective; we identify In with its image in I, so that I is the increasing union of the In.
Let M be an A-module of finite length, and n an integer ⩾0 such that mAnM=0. For x∈M, let us denote by φM,xn the a-linear mapping of A/mAn into M which maps the class of 1 to x. The mapping T(φM,xn):T(M)→In is a-linear, and one has T(φM,axn)=aT(φM,xn) for a∈A by FD 1). Hence the mapping (x,u)↦T(φM,xn)(u) of M×T(M) into I is A-bilinear. It does not depend on the choice of the integer n: in fact, for every integer q⩾n and every element x of M, one has φM,xq=φM,xn∘pqn, whence by FD 3) T(φM,xq)=iqn∘T(φM,xn). We deduce an a-linear mapping
θM:T(M)⟶HomA(M,I)
satisfying θM(u)(x)=T(φM,x)(u) for u∈T(M),x∈M.
Bourbaki puts the proof directly after the statement, so what is above is the statement and its proof together.