TheoremStatement: AI-adaptedProof: Literature-sourcedprecheck passjudge pass (deepseek-v4-pro + claude-sonnet-5)audited 2026-08-17

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On a preorder the monads are exactly the monotone extensive maps with T(Tp) below Tp; on a poset they are exactly the closure operators

Statement

Let $P$ be a preorder, regarded as a category. A monad on $P$ is equivalently a monotone map $T : P \to P$ such that $p \leq T p$ and $T (T p) \leq T p$ for every $p$ . These inequalities force $T p$ and $T (T p)$ to be mutually comparable. If $P$ is a poset, this is equivalently a closure operator: a monotone, extensive, idempotent map.

Facts & Assumptions

Given: A preorder $P$ , regarded as a category.

[L1]

A preorder determines a category with at most one morphism between any two objects, and functors between such categories are exactly monotone maps (A preorder is a category with at most one morphism between any two objects, and its functors are exactly monotone maps).

[L2]

A monad has natural transformations $η : 1 \Rightarrow T$ and $μ : T^{2} \Rightarrow T$ satisfying the monad equations (Monad on a category).

[L3]

A partial order is a preorder satisfying antisymmetry (Partial order and partially ordered set).

Proof

technique · direct

1.1L1L2

By [L1] the endofunctor is a monotone map, and a morphism $F p \to G p$ in a preorder-category exists exactly when $F p \leq G p$ and is then the only one, so a natural transformation $F \Rightarrow G$ is precisely the family of inequalities $F p \leq G p$ . Hence by [L2] $η$ is exactly the family $p \leq T p$ , and $μ$ is exactly $T (T p) \leq T p$ . Applying monotonicity to $p \leq T p$ also gives $T p \leq T (T p)$ .

1.2L1L2

Conversely, a monotone $T$ with $p \leq T p$ and $T (T p) \leq T p$ supplies the unique transformations $η : 1 \Rightarrow T$ and $μ : T^{2} \Rightarrow T$ by [L1]. Every monad diagram commutes because a preorder has at most one arrow between any fixed source and target.

2.1L3step 1.1step 1.2∎

If $P$ is a poset, step 1.1 and antisymmetry give $T (T p) = T p$ ; conversely an ordinary closure operator is monotone and extensive and its idempotence supplies $T (T p) \leq T p$ , so step 1.2 makes it a monad.

Depends on

Used by

Dependency tree · two levels

8 results within two dependency steps of this one, each drawn at its shortest distance from it. An arrow runs from a result to what uses it, so the chart reads left to right and ends at this result, which carries a heavier outline. Every node is a link to that result. Click elsewhere on the chart to enlarge it.

Sources

E. Riehl, Category Theory in Context, 2nd ed., Example 5.1.7 (standard reference, not scraped)