Proposition 6.4.8.label Let $A$ be a unital $C^{*}$-algebra and $x\in A$ be normal. Let $B$ be a $C^{*}$-subalgebra of $A$ generated by $x$ and $1$. Then $\exists!$ isomorphism
\begin{align*}\p:C(\spec{A}{x})\to B&&\p(1)=1,\p(\indi{\spec{A}{x}})=x\end{align*}
where $\indi{\spec{A}{x}}$ is the identity on $\spec{A}{x}$.
Proof. The normality of $x$ implies that $B$ is abelian. Since the polynomials of $x$ and $x^{*}$ form a dense subalgebra of $B$, the map $\psi:\om\in\OM(B)\mto\om(x)\in \spec{A}{x}$ is a homeomorphism. By Gelfand Representation 6.3.11. $\psi$ is well-defined and continuous. By Proposition 6.3.8. $\psi$ is bijective. By Propositionn 6.3.9, $\psi\inv$ is continuous. By Gelfand-Naimark 6.4.5, the composed map $\p:f\in C(\spec{A}{x})\mto\scr{F}\inv(f\circ\psi)\in B$ is the desired isomorphism.$\square$