Exercise 15, § 3

Let A and B be commutative unital Banach algebras, $\varphi$ a unital morphism from A into B and $(x_{\lambda})_{\lambda\in\Lambda}$ a family of elements of A such that the full closed subalgebra of A generated by the $x_{\lambda}$ is equal to A. In order that $\mathsf{X}(\varphi )$ be surjective, it is necessary and sufficient that Sp$^{\Lambda}_A((x_{\lambda})) =$ Sp$^{\Lambda}_B((\varphi (x_{\lambda})))$. (Use the diagram (1) of No.$^o6$ of I, p. 41, where the right arrow is bijective according to prop. 12 of I, p. 43.)