# All Infima Preserving Mapping is Upper Adjoint of Galois Connection

## Theorem

Let $\struct {S, \preceq}$ be a complete lattice.

Let $\struct {T, \precsim}$ be an ordered set.

Let $g: S \to T$ be all infima preserving mapping.

Then there exists a mapping $d: T \to S$ such that $\tuple {g, d}$ is Galois connection and:

$\forall t \in T: \map d t = \map \min {g^{-1} \sqbrk {t^\succsim} }$

where:

$\min$ denotes the minimum
$g^{-1} \sqbrk {t^\succsim}$ denotes the image of $t^\succsim$ under relation $g^{-1}$
$t^\succsim$ denotes the upper closure of $t$

## Proof

Define a mapping $d: T \to S$:

$\forall t \in T: \map d t := \map \inf {g^{-1} \sqbrk {t^\succsim} }$

We will prove as lemma $1$ that

$g$ is an increasing mapping.

Let $x, y \in S$ such that

$x \preceq y$
$y^\succeq \subseteq x^\succeq$
$\map \inf {x^\succeq} = x$ and $\map \inf {y^\succeq} = y$

By definition of mapping preserves all infima:

$g$ preserves the infimum on $x^\succeq$ and $g$ preserves the infimum on $y^\succeq$

By definition of mapping preserves the infimum of set:

$\map \inf {\map {g^\to} {x^\succeq} } = \map g x$ and $\map \inf {\map {g^\to} {y^\succeq} } = \map g y$
$\map {g^\to} {y^\succeq} \subseteq \map {g^\to} {x^\succeq}$

Thus by Infimum of Subset:

$\map g x \precsim \map g y$

This ends the proof of lemma $1$.

We will prove as lemma $2$ that

$\forall t \in T: \map d t = \map \min {g^{-1} \sqbrk {t^\succsim} }$

Let $t \in T$.

By definition of $d$:

$\map d t = \map \inf {g^{-1} \sqbrk {t^\succsim} }$
$g \sqbrk {g^{-1} \sqbrk {t^\succsim} } \subseteq t^\succsim$
$t = \map \inf {t^\succsim} \precsim \map \inf {g \sqbrk {g^{-1} \sqbrk {t^\succsim} } }$

By definition of upper closure of element:

$\map \inf {g \sqbrk {g^{-1} \sqbrk {t^\succsim} } } \in t^\succsim$

By definition of complete lattice:

$g^{-1} \sqbrk {t^\succsim}$ admits an infimum

By definitions mapping preserves the infimum:

$\map \inf {g \sqbrk {g^{-1} \sqbrk {t^\succsim} } } = \map g {\map d t}$

Thus by definition of image of set:

$\map d t \in g^{-1} \sqbrk {t^\succsim}$

Thus

$\map d t = \map \min {g^{-1} \sqbrk {t^\succsim} }$

This ends the proof of lemma $2$.

$\tuple {g, d}$ is a Galois connection.

Thus by lemma $2$:

$\forall t \in T: \map d t = \map \min {g^{-1} \sqbrk {t^\succsim} }$

$\blacksquare$