[@rmrn:bootstrap Lemma 2.25] it follows that if $X$ is a sober space (which is no actual restriction on $X$) then any $C^\ast$-algebra $\mathfrak A $ together with a map $\psi^\ast\colon \mathbb O(X) \to \mathbb I(A)$, which respects arbitrary suprema, finite infima and such that $\psi^\ast(\emptyset)=0,\psi^\ast(X)=\mathfrak A$, gives rise to a continuous map $\phi \colon \mathrm{Prim} ( \mathfrak A ) \to X$ such that $\psi^\ast=\phi^\ast$.

Let $(\mathfrak{A} , \psi )$ be a $C^{*}$-algebra over $X$. We say that $( \mathfrak{A} , \psi )$ is *continuous* if $\psi^{*}$ respects arbitrary infima, i.e., for any collection of open subsets $\{ U_{\lambda} \}$ of $X$, then $$\begin{aligned}
\mathfrak{A} (U) = \bigcap_{\lambda} \mathfrak{A}(U_{\lambda})\end{aligned}$$ where $U$ is the interior of $\bigcap_{\lambda}U_{\lambda}$.

We should remark, that in the case that $X$ is a locally compact Hausdorff space, a $C^\ast$-algebra over $X$ is the same as a $C_{0}(X)$-algebra by [@rmrn:bootstrap Proposition 2.11]. Combining [@rmrn:bootstrap Lemma 2.9] with [@mn:bundles Corollary 2.2] one gets that continuous $C^\ast$-algebras over $X$ correspond exactly to continuous $C_{0}(X)$-algebras.

Let $\mathfrak{A}$ and $\mathfrak{B}$ be $C^{*}$-algebras over $X$. A $*$-homomorphism ${  \phi  \colon  \mathfrak{A}  \rightarrow  \mathfrak{B}  }$ is *$X$-equivariant* if $\phi ( \mathfrak{A} (U) ) \subseteq \mathfrak{B} ( U )$ for all $U \in \mathbb{O}(X)$. Hence, 