The image of the mapping X(h) is called the simultaneous spectrum of x, and denoted by SpAΛ(x) or SpΛ(x).
The simultaneous spectrum of x is a compact subset of CΛ. An element c=(cλ)∈CΛ belong to SpAΛ(x) if and only if the elements xλ−cλ belong to one and the same maximal ideal of A, in other words if the family (xλ−cλ)λ∈Λ does not generate the algebra A.
If Λ contains only one element, so that the family x reduces to a single element x∈A, we have SpAΛ(x)= SpA(x) (I, p. 37, prop. 6, b)). If Λ′⊂Λ, then SpAΛ′((xλ)λ∈Λ′) is the image of SpAΛ((xλ)λ∈Λ) under the canonical projection mapping of CΛ onto CΛ′. In particular, we have
SpAΛ(x)⊂∏λ∈Λ SpA(xλ).
Let zλ, for λ∈Λ, denote the coordinate functions on CΛ. If χ∈ X(A), the value at χ of zλ∘X(h) is χ(xλ), hence zλ∘X(h)=G(xλ).
Let A and B be commutative unital Banach algebras, φ a unital morphism from A into B, and x=(xλ)λ∈Λ a family of elements of A. Let φ(x) denote the family (φ(xλ))λ∈Λ of elements of B. We have, for every χ∈X(B), and every λ∈Λ
χ(φ(xλ))=(X(φ)(χ))(xλ)
hence SpBΛ(φ(x))⊂ SpAΛ(x). The diagram
←X(φ)
X(B)→X(A)
→←→←(1)
Λ←iΛ
SpB(φ(x))→ SpA(x)
where i denotes the inclusion, and where the vertical arrows denote the mappings defined by the families φ(x) and x, is therefore commutative.
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.
Nothing.
Nothing.