Supremum of Ideals is Upper Adjoint

Theorem
Let $L = \struct {S, \vee, \preceq}$ be a bounded below continuous join semilattice.

Let $\map {\mathit {Ids} } L$ be the set of all ideals in $L$.

Let $P = \struct {\map {\mathit {Ids} } L, \precsim}$ be an ordered set where $\mathord \precsim = \subseteq \restriction_{\map {\mathit {Ids} } L \times \map {\mathit {Ids} } L}$

Let $f: \map {\mathit {Ids} } L \to S$ be a mapping such that
 * $\forall I \in \map {\mathit {Ids} } L: \map f I = \sup I$

Then $f$ is an upper adjoint of Galois connection.

Proof
Define $d: S \to \map {\mathit {Ids} } L$
 * $\forall t \in S: \map d t = \map \inf {f^{-1} \sqbrk {t^\succeq} }$

where
 * $t^\succeq$ denotes the upper closure of $t$,
 * $f^{-1} \sqbrk {t^\succeq} $ denotes the image of $t^\succeq$ over $f^{-1}$.

We will prove that
 * $\forall t \in S: \map d t = \map \min {f^{-1} \sqbrk {t^\succeq} }$

Let $t \in S$.

By Continuous iff For Every Element There Exists Ideal Element Precedes Supremum:
 * there exists an ideal $J$ in $L$ such that
 * $t \preceq \sup J$ and for every ideal $K$ in $L$: $t \preceq \sup K \implies J \subseteq K$

We will prove that
 * $\forall K \in \map {\mathit {Ids} } L: K$ is lower bound for $f^{-1} \sqbrk {t^\succeq} \implies K \precsim J$

Let $K \in \map {\mathit {Ids} } L$ such that
 * $K$ is a lower bound for $f^{-1} \sqbrk {t^\succeq}$

By definition of $f$:
 * $t \preceq \map f J$

By definition of upper closure of element:
 * $\map f J \in t^\succeq$

By definition of image of set:
 * $J \in f^{-1} \sqbrk {t^\succeq}$

Thus by definition of lower bound:
 * $K \precsim J$

We will prove that:
 * $J$ is a lower bound for $f^{-1} \sqbrk {t^\succeq}$

Let $K \in \map {\mathit {Ids} } L$ such that:
 * $K \in f^{-1} \sqbrk {t^\succeq}$

By definition of image of set:
 * $\map f K \in t^\succeq$

By definition of upper closure of element:
 * $t \preceq \map f K$

By definition of $f$:
 * $t \preceq \sup K$

Then
 * $J \subseteq K$

Thus by definition of $\precsim$:
 * $J \precsim K$

By definition of supremum:
 * $t \preceq \map \sup {\map \inf {f^{-1} \sqbrk {t^\succeq} } }$

By definition of $f$:
 * $t \preceq \map f {\map \inf {f^{-1} \sqbrk {t^\succeq} } }$

By definition of upper closure of element:
 * $\map f {\map \inf {f^{-1} \sqbrk {t^\succeq} } } \in t^\succeq$

By definition of image of set:
 * $\map \inf {f^{-1} \sqbrk {t^\succeq} } \in f^{-1} \sqbrk {t^\succeq}$

Thus by definition of smallest element:
 * $\map d t = \map \min {f^{-1} \sqbrk {t^\succeq} }$

By Supremum of Ideals is Increasing:
 * $f$ is an increasing mapping.

By Galois Connection is Expressed by Minimum:
 * $\struct {f, d}$ is a Galois connection.

Hence $f$ is an upper adjoint of Galois connection.