Let a∈An, let (h,u1,...,un) be a sequence adapted to a, and let ω be the associated differential form.
Let w be an indefinitely differentiable mapping of Cn into A, and let i and j be distinct integers between 1 and n. Define u1′,...,un′ by
ui′=ui+(zj−aj)w,uj′=uj−(zi−ai)w
uk′=uk for k=i,j.
Then the sequence (h,u1′,...,un′) is adapted to a and is linked to the sequence (h,u1,...,un).
Let us write dz=dz1∧⋯∧dzn. Since
k=1∑n(zk−ak)uk′(z)=k=1∑n(zk−ak)uk(z)+w(z)(zj−aj)(zi−ai)
−w(z)(zi−ai)(zj−aj)=1−h(z)
for every z∈Cn, the sequence (h,u1′,...,un′) is adapted to a. Further, we have
dui′∧duj′∧dz1∧⋯∧dzn=
(dui+wdzj+(zj−aj)dw)∧(duj−wdzi−(zi−ai)dw)∧dz
=(dui∧duj−(zi−ai)dui∧dw−(zj−aj)duj∧dw)∧dz
There therefore exists ε∈{−1,1} such that ε(ω−ω′) is equal to
dui′∧duj′∧k=i,j⋀duk′∧dz−dui∧duj∧k=i,j⋀duk∧dz
=−((zi−ai)dui∧dw+(zj−aj)duj∧dw)∧k=i,j⋀duk∧dz
=−(k=1∑n(zk−ak)duk)∧dw∧k=i,j⋀duk∧dz
and, taking account of (2), this is equal to
dh∧dw∧k=i,j⋀duk∧dz=d(hdw∧k=i,j⋀duk∧dz)
whence the result.
Bourbaki sets the proof directly after the statement, and the corpus does not separate them, so what is above is the statement and its proof together.
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