Proposition 10

Let A and B be commutative unital Banach algebras, $h: A\rightarrow B$ a unital algebra morphism, not necessarily continuous. If $\mathsf{X}(h)$ is surjective, then $h(A)$ is a full subalgebra of B.

Let $x\in A$ be such that $h(x)$ is invertible in B. For every $\chi \in \mathsf{X}(A)$, there exists $\xi \in \mathsf{X}(B)$ such that $\chi =\mathsf{X}(h)(\xi )$, hence $\chi (x) =\xi (h(x))\not= 0$. Prop. 6 of I, p. 37 then shows that $x$ is invertible in A, and hence that $h(x)$ is invertible in $h(A)$.

The necessary condition of the proposition is not sufficient, even if $h$ is isometric (I, p. 168, exerc. 14). One has nevertheless the following result: