Let A be a unital Banach algebra and a∈A. Let f∈O(Sp(a)). One has f(SpA(a)) = SpA(f(a)). Moreover, for every g∈O(SpA(f(a))), one has g∘f∈O(SpA(a)) and g(f(a))=(g∘f)(a).
This results from th. 4.
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.