Supremum of Ideals is Upper Adjoint

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

Let $\mathit{Ids}\left({L}\right)$ be the set of all ideals in $L$.

Let $P = \left({\mathit{Ids}\left({L}\right), \precsim}\right)$ be an ordered set where $\mathord \precsim = \subseteq\restriction_{\mathit{Ids}\left({L}\right)\times \mathit{Ids}\left({L}\right)}$

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

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

Proof
Define $d: S \to \mathit{Ids}\left({L}\right)$
 * $\forall t \in S: d\left({t}\right) = \inf \left({f^{-1}\left[{t^\succeq}\right]}\right)$

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

We will prove that
 * $\forall t \in S: d\left({t}\right) = \min \left({f^{-1}\left[{t^\succeq}\right]}\right)$

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
 * $\forall K \in \mathit{Ids}\left({L}\right): K$ is lower bound for $f^{-1}\left[{t^\succeq}\right] \implies K \precsim J$

Let $K \in \mathit{Ids}\left({L}\right)$ such that
 * $K$ is lower bound for $f^{-1}\left[{t^\succeq}\right]$

By definition of $f$:
 * $t \preceq f\left({J}\right)$

By definition of upper closure of element:
 * $f\left({J}\right) \in t^\succeq$

By definition of image of set:
 * $J \in f^{-1}\left[{t^\succeq}\right]$

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

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

By Galois Connection is Expressed by Minimum:
 * $\left({f, d}\right)$ is a Galois connection.

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