Corollary 2

Let $f\in \mathscr{C}$ (Sp$_A(x)$).

a) We have

Sp$_A(f(x)) =f$(Sp$_A(x)$) ;

b) For every $g\in \mathscr{C}$ (Sp$_A(f(x))$), one has $(g\circ f)(x) =g(f(x))$.

Since $f(x)$ belong to the full subalgebra B of A, one has Sp$_A(f(x)) =$ Sp$_B(f(x))$ (Prop. 5 of I, p. 106). The isomorphism $f\mapsto f(x)$ of $\mathscr{C}$(Sp$_A(x)$) into B preserves the spectrum; hence one has Sp$_B(f(x)) =$ Sp$_{\mathscr{C}(Sp_A(x))}(f) =f$(Sp$_A(x)$) (Example 3 of I, p. 17). This proves assertion a).

The mapping $g\mapsto (g\circ f)(x)$ is a unital morphism of involutive algebras of $\mathscr{C}$(Sp$_A(f(x))$) into A which transforms the identical function on Sp$_A(f(x))$ into $f(x)$. By Prop. 7, one has therefore $(g\circ f)(x) =g(f(x))$ for every $g\in \mathscr{C}$ (Sp$_A(f(x))$).