# All Galois Connections are Essentially Uninteresting
$\DeclareMathOperator{\id}{id}$ $\DeclareMathOperator{\im}{im}$ $\newcommand{\sh}{^\sharp}$ $\newcommand{\nat}{^\natural}$

**Definition.** A *(monotone) Galois connection* $f \dashv g$ from a poset $P$ to a poset $Q$ is a pair of monotone maps $f \colon P \to Q$ and $g \colon Q \to P$ such that $\forall~ x \in P$, $y \in Q$,
\begin{equation*}
f(x) \le y \iff x \le g(y).
\end{equation*}

**Lemma 1.** The pair of monotone maps $f \colon P \to Q$, $g \colon Q \to P$ is a Galois connection if and only if it satisfies the following two conditions.
\begin{align*}
\id_P & \le g \circ f \\
f \circ g & \le \id_Q.
\end{align*}

**Lemma 2.** If $f \dashv g$, then
\begin{align*}
f \circ g \circ f & = f\\
g \circ f \circ g & = g.
\end{align*}

**Definition.** A Galois connection $f \dashv g$ is *uninteresting* if it is a Galois correspondence -- i.e., if $f$ and $g$ are poset isomorphisms.

**Definition.** A Galois connection $f \dashv g$ is *essentially uninteresting* if the restrictions of $f$ and $g$ to $\im g$ and $\im f$ respectively form an uninteresting Galois connection.

**Theorem.** All Galois connections are essentially uninteresting.
*Proof.* Let $f \dashv g$ be a Galois connection from $P$ to $Q$.
Let $P\sh = \im g$ and $Q\sh = \im f$, and let $f\sh \colon P\sh \to Q\sh$ and $g\sh \colon Q\sh \to P\sh$ be the obvious restrictions of $f$ and $g$ respectively.
We first prove the following:
1. $P\sh = \{\, x \in P \mid g(f(x)) = x \,\}$.
2. $Q\sh = \{\, y \in Q \mid f(g(y)) = y \,\}$.

Suppose $x \in P\sh = \im g$. Then $x = g(y)$, $\exists~ y \in Q \implies$ $g(f(x)) = g(f(g(y)) = g(y) = x$, by Lemma 2.
On the other hand, any $x \in P$ such that $g(f(x)) = x$ is clearly in $\im g = P\sh$. Thus, 1. holds.
Similary 2. holds.

Now, we show that $f\sh$ and $g\sh$ are mutually inverse isomorphisms. Being restrictions of monotone maps, they are themselves monotone, so it is sufficient to show that they are mutually inverse bijections.

By 1., $x \in P\sh \iff g(f(x)) = x \iff g\sh(f\sh(x)) = x \implies g\sh \circ f\sh = \id_{P\sh}$.
By 2., $y \in Q\sh \iff f(g(x)) = y \iff f\sh(g\sh(y)) = y \implies f\sh \circ g\sh = \id_{Q\sh}$.

**Theorem.** Let $P$ and $Q$ be posets, and $f \colon P \to Q$ and $g \colon Q \to P$ be a pair of mutually inverse poset isomorphisms between them. If $P\nat$ and $Q\nat$ are complete lattices containing $P$ and $Q$, respectively, as subsets, then there is a unique Galois connection $f\nat \dashv g\nat$ from $P$ to $Q$ such that $f\nat|_P = f$ and $g\nat|_Q = g$.
*Proof.* For $x \in P\nat$, define $U_P(x) = \{\, x_P \in P \mid x \le x_P \,\}$.
For $y \in Q\nat$, define $D_Q(y) = \{\, y_Q \in Q \mid y_Q \le y \,\}$.
Now, define $f\nat$ and $g\nat$ as follows:
For all $x \in P\nat$, and $y \in Q\nat$, let
\begin{align*}
f\nat(x) = \bigwedge_{x_P \in U_P(x)} f(x_P) = \bigwedge f(U_p(x)), &&
g\nat(y) = \bigvee_{y_Q \in D_Q(y)} g(y_Q) = \bigvee g(D_Q(y)).
\end{align*}

To see that these maps are monotone, let $x_1, x_2 \in P\nat$. Then $x_1 \le x_2 \implies U_P(x_2) \subseteq U_P(x_1) \implies$ $f(U_P(x_2)) \subseteq f(U_P(x_1)) \implies \bigwedge f(U_P(x_1)) \le \bigwedge f(U_P(x_2))$.
By duality, $g$ is also monotone.

To see that $f\nat$ and $g\nat$ extend $f$ and $g$, let $x \in P$. Then $\forall~ x_P \in U_P(x)$, $x \le x_P \implies f(x) \le f(x_P) \implies f\nat(x) = \bigwedge f(U_P(x)) = f(x)$. Hence, $f\nat$ extends $f$, and dually, $g\nat$ extends $g$.

Now, we show that $f\nat \dashv g\nat$. Let $x \in P\nat$, $y \in Q\nat$.
First, suppose that $f\nat(x) \le y$.

x₁    x₂   f(x₁)   f(x₂)
 \   /        \  y  /
  \ /          \ | /
   x        f(x₁)∧f(x₂)

$x \le g(f(x_1)),\, g(f(x_2))$
$x \le g(f(x_1)) \wedge g(f(x_2))$
$g\nat(f\nat(x)) \le g(f(x_1)) \wedge g(f(x_2))$

a ∧ b ≤ a, b
g(a ∧ b) ≤ g(a), g(b)
g(a ∧ b) ≤ g(a) ∧ g(b)