Proposition 11

Let A and B be commutative unital Banach algebras, let $a$ be an element of A and let $h: A\rightarrow B$ be an injective unital morphism (not necessarily continuous). Suppose that the full closed subalgebra of A generated by $a$ is equal to A. The following conditions are equivalent:

(i) $\mathsf{X}(h)$ is surjective;

(ii) $h(A)$ is a full subalgebra of B;

(iii) Sp$_A(a) =$ Sp$_B(h(a))$.

(i) $=\Rightarrow$ (ii) follows from prop. 10.

(ii) $=\Rightarrow$ (iii) results from the formula Sp$_A(a) =$ Sp$_{h(A)}(h(a)) =$ Sp$_B(h(a))$, valid since $h(A)$ is a full subalgebra of B.

(iii) $=\Rightarrow$ (i) by the formula (4) of No.$^o6$ of I, p. 6, we have the commutative diagram

Sp$_B(h(a))\rightarrow$ Sp$_A(a)$

where the vertical arrows denote surjective mappings (I, p. 37, prop. 6) and $i$ is the canonical inclusion. Assumption (iii) means that $i$ is bijective. Moreover the surjective mapping $\mathscr{G}_A(a) :\mathsf{X}(A)\rightarrow$ Sp$_A(a)$ is bijective: in fact, for every characters $\chi_1$ and $\chi_2$ of A, the set $A_{\chi_1,\chi_2}=\{x\in A|\chi_1(x) =\chi_2(x)\}$ is a closed full subalgebra of A. By hypothesis, therefore, one has $A_{\chi_1,\chi_2}$ = A if $a\in A_{\chi_1,\chi_2}$, that is to say $\chi_1=\chi_2$ if $\mathscr{G}_A(a)\chi_1=\mathscr{G}_A(a)\chi_2$. The diagram then implies that the mapping $\mathsf{X}(h)$ is surjective.