Let n be an integer > 0, G an n-dimensional Lie group, U a symmetric open neighbourhood of e in G and ψ:U2→Kn a chart of G such that ψ(e)=0. If (x1,…,xn) are the coordinates of x∈ψ(U) and (y1,…,yn) the coordinates of y∈ψ(U), we denote by
m1(x1,…,xn,y1,…,yn),…,mn(x1,…,xn,y1,…,yn)
the coordinates of ψ(ψ−1(x)−1ψ−1(y)). Then, if we write, for 1≤k≤n,
ϖk(x1,…,xn)=Dn+1mk(x1,…,xn,x1,…,xn)dx1+⋯+D2nmk(x1,…,xn,x1,…,xn)dxn,
the differential forms ϖk on ψ(U) are derived through ψ from left invariant differential forms on G and are such that ϖk(0,…,0)=dxk.
We apply Lemma 2 with E=K, taking ϕ(g) to be the coordinate of ψ(g) of index k. We obtain a differential form ωk; let ϖk be its transform under ψ. The value of ϖk at (x1,…,xn) is the differential at (x1,…,xn) of the function y↦mk(x1,…,xn,y1,…,yn); this value is therefore given by formula (32). It then suffices to use the conclusion to Lemma 2.
Bourbaki puts the proof directly after the statement, so what is above is the statement and its proof together.