Bourbaki

Theorem 4

Tag 02AQ, permanent. Cite it as [Bourbaki, Tag 02AQ].

Théories spectrales, chapter I, § 4. Calcul fonctionnel holomorphe

Let A be a commutative complex unital Banach algebra, let n⩾1n\geqslant 1 be an integer, and let a=(a1,...,an)∈An\boldsymbol{a}= (a_1, . . . , a_n)\in A^n. Let f=\boldsymbol{f}= (f1,...,fp)(f_1, . . . , f_p), where f1,...,fpf_1, . . . , f_p are elements of O\mathscr{O}(Spn(a)^n(\boldsymbol{a})). The image of Spn(a)^n(\boldsymbol{a}) under the mapping z↦f(z)=(f1(z),...,fp(z))\boldsymbol{z}\mapsto \boldsymbol{f}(\boldsymbol{z}) = (f_1(\boldsymbol{z}), . . . , f_p(\boldsymbol{z})) is equal to Spp(f(a))^p(\boldsymbol{f}(\boldsymbol{a})).

For every g∈Og\in \mathscr{O}(Spp(f(a));A^p(\boldsymbol{f}(\boldsymbol{a})); A), the composite germ g∘fg\circ \boldsymbol{f} is an element of O\mathscr{O}(Spn(a);A^n(\boldsymbol{a}); A) and one has g(f(a))=(g∘f)(a)g(\boldsymbol{f}(\boldsymbol{a})) = (g\circ \boldsymbol{f})(\boldsymbol{a}).

The first assertion concerning the image of Spn(a)^n(\boldsymbol{a}) follows from cor. 2 of I, p. 67. To prove the second, we shall use the following lemma.

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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Machine translation of the English, by gpt-5-6-mini, gpt-5-mini, gpt-5.4, against glossary version 34. Not checked by a person. That model is a cut down version of the one the rest of this language was written by.