Let \(P\) be a preorder. Monotone maps \(P \rightarrow \mathbb{B}\) are in one-to-one correspondance with upper sets of \(P\).
Let \(P \xrightarrow{f} \mathbb{B}\) be a monotone map. The subset \(f^{-1}(true)\) is an upper set.
Suppose \(p \in f^{-1}(true)\) and \(p \leq q\).
Then \(true = f(p) \leq f(q)\) because \(f\) is monotonic.
But there is nothing strictly greater than \(true\) in \(\mathbb{B}\), so \(f(q) = true\) and therefore \(q \in f^{-1}(true)\) too.
Suppose \(U \in U(P)\), and define \(P\xrightarrow{f_U}\mathbb{B}\) such that \(f_U=true \iff p \in U\)
Given there are only two values in \(B\) and an arbitrary \(p\leq q\), the only way for this to not be monotone is for \(f_U(p) \land \neg f_U(q)\) but this defies the definition of an upper set.
The two constructions are mutually inverse.