The adjoint relationship between order and topology

The categories $Ord$ and $Top$ are the two most basic categories characterized by the adjoint relationships. The two categories $Ord$ and $Top$ are in turn adjointly related to one another, so this continues the basic theme of exploring adjoint relationships in category theory.

Theorem. let $f: (A,\tau_1) \to (B,\tau_2)$ be continuous then $Ord(f) : Ord(A) \to Ord(B)$ is a monotone map of specialization preorders.

Proof. the definition of the specialization preorder yields: $$a_1 \subseteq a_2 \Leftrightarrow \forall O \in \tau_1 : a_1 \in O \Rightarrow a_2 \in O$$ We want that this would imply $f(a_1) \subseteq f(a_2)$. This will be demonstrated by using proof by contradiction. Suppose that $f(a_1) \not\subseteq f(a_2)$ then there exists $S$ such that $f(a_1) \in S$ and $f(a_2) \not\in S$. Then $f(a_1) \in S$ implies that $a_1 \in f^{-1}(S)$ and $a_2 \in f^{-1}(S)$ is logically equivalent to the condition that $f(a_2) \in S$ but we know that $f(a_2) \not\in S$ so $a_2 \not\in f^{-1}(S)$.

The inverse image of any open set is open, so $f^{-1}(S)$ is an open set, and it contains $a_1$ but not $a_2$ so it cannot be the case that $a_1 \in f^{-1}(S) \Rightarrow a_2 \in f^{-1}(S)$ which contradicts that $a_1 \subseteq a_2$. So by contradiction it cannot be the case that $f(a_1) \not= f(a_2)$ so $f(a_1) \subseteq f(a_2)$ which implies that $Ord(f) : Ord(A) \to Ord(B)$ is monotone. $\square$

Theorem. let $f: (A, \subseteq_A) \to (B, \subseteq_B)$ be a monotone map, then $f$ reflects upper sets.

Proof. let $I$ be an upper set of $B$ then consider $f^{-1}(I)$ and suppose that $a \in f^{-1}(I)$ then $f(a) \in I$ and now consider a $b$ with $a \subseteq b$. By monotonicity we have that $f(a) \subseteq f(b)$ and since $I$ is an upper set this implies that $f(b) \in I$. This in turn means that $b \in f^{-1}(I)$ so that $f^{-1}(I)$ is an upper set. $\square$

These two theorems are enough to construct an adjoint pair of functors from $Top$ to $Ord$, and these two theorems prove that these relationships are functorial.

• The specialization preorder functor: $P: Top \to Ord$ maps topologies to preorders.
• The Alexandrov topology functor: $T: Ord \to Top$ maps preorders to topologies.

Then these two functors define an adjoint relationship between order and topology. In particular, the Alexandrov topology is the largest topology with a given specialisation preorder. So the relationship $(P,\tau)$ which states that $P$ is a subpreorder of the specialisation preorder of $\tau$ is characterized by the monotone Galois connection $P(\tau) \subseteq P \Leftrightarrow \tau \subseteq T(P)$. So the relationship between preorders and topologies is governed by this adjoint pair of functors.

References:
Specialization order