Adjoint functor theorem for preorders
Suppose \(Q\) is a preorder that has all meets and \(P\) is any preorder.
A monotone map \(Q \xrightarrow{g} P\) preserves meets iff it is a right adjoint.
Likewise, suppose \(P\) has all joins and \(Q\) is any preorder:
A monotone map \(P \xrightarrow{f} Q\) preserves joins iff it is a left adjoint.
Proved the reverse direction in Proposition 1.91.
Given a right adjoint, construct the left adjoint by:
\(f(p) := \bigwedge\{q \in Q\ |\ p \leq g(q)\}\)
First need to show this is monotone:
If \(p \leq p'\), the relationship between the joined sets of \(f(p)\) and \(f(p')\) is that the latter is a subset of the former.
By Proposition 1.91 we infer that \(f(p) \leq f(p')\).
Then need to show that it is satisfies the left adjoint property:
Show that \(p_0 \leq g(f(p_0))\)
\(p_0 \leq \bigwedge \{g(q)\ |\ p_0 \leq g(q)\} \cong g(\bigwedge\{q\ |\ p_0 \leq g(q)\}) = g(f(p_0))\)
The first inequality comes from the fact that the meet of the set (of which \(p_0\) is a lower bound) is a greatest lower bound.
The congruence comes from the fact that right adjoints preserve meets.
Show that \(f(g(q_0)) \leq q_0\)
\(f(g(q_0)) = \bigwedge\{q\ |\ g(q_0) \leq g(q)\} \leq \bigwedge \{q_0\} = q_0\)
The first inequality comes from Proposition 1.91 since \(\{q_0\}\) is a subset of the first set.
The second equality is a property of the meet of single element sets.
Proof of a left adjoint construction (assuming it preserves joins) is dual.