We say that $( \mathfrak{A} , \psi )$ is *tight* if $\psi$ is a homeomorphism.

We will always identify $\mathbb{O} ( \mathrm{Prim} ( \mathfrak{A} ) )$ and $\mathbb{I} ( \mathfrak{A} )$ using the canonical lattice isomorphism $U \mapsto \bigcap_{ \mathfrak{p} \in \mathrm{Prim} ( \mathfrak{A} ) \setminus U } \mathfrak{p}$. Let $( \mathfrak{A} , \psi )$ be a $C^{*}$-algebra over $X$. Then we get a map ${  \psi^{*}  \colon  \mathbb{O} ( X )  \rightarrow  \mathbb{O} ( \mathrm{Prim} ( \mathfrak{A} ) )  }$ defined by $U \mapsto {\ensuremath{\left\{  \mathfrak{p} \in \mathrm{Prim} ( \mathfrak{A} )  \: : \:  \psi ( \mathfrak{p} ) \in U  \right\}}}$. Using the lattice isomorphism $\mathbb{O} ( \mathrm{Prim} ( \mathfrak{A} ) ) \cong \mathbb{I} ( \mathfrak{A} )$, we get a map, which we again denote by $\psi^{*}$, from $\mathbb{O}(X)$ to $\mathbb{I}(\mathfrak{A} )$ by $$\begin{aligned}
U \mapsto \bigcap {\ensuremath{\left\{  \mathfrak{p} \in \mathrm{Prim} ( \mathfrak{A} )  \: : \:  \psi ( \mathfrak{p} ) \notin U  \right\}}}.\end{aligned}$$ Denote this ideal by $\mathfrak{A}(U)$. For $Y = U \setminus V \in \mathbb{LC} ( X )$, set $\mathfrak{A}(Y) = \mathfrak{A} (U) / \mathfrak{A}(V)$. By [@rmrn:bootstrap Lemma 2.15], $\mathfrak{A} ( Y)$ (up to a canonical choice of isomorphism) does not depend on $U$ and $V$.

By 