TheoremStatement: Literature-sourcedProof: AI-generatedSession-authored (Fable 5 assisted)precheck passaudited 2026-08-17

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The ultrafilter endofunctor with principal unit and flattening multiplication is a monad

Statement

The ultrafilter endofunctor $β$ together with the principal unit $η$ and flattening multiplication $μ$ is a monad on $S e t$ .

Facts & Assumptions

Given: The functor $β$ and the natural transformations $η$ and $μ$ from The ultrafilter endofunctor with principal unit and flattening multiplication.

[L1]

For $A \subseteq X$ , write $\hat{A} = {U \in β X : A \in U}$ ; then $A \in μ_{X} (W)$ exactly when $\hat{A} \in W$ (The ultrafilter endofunctor with principal unit and flattening multiplication).

[L2]

The data in [L1] are already well-defined and natural (The ultrafilter endofunctor with principal unit and flattening multiplication).

Proof

technique · direct

1.1

For $U \in β X$ , one has $A \in μ_{X} (η_{β X} (U))$ iff $\hat{A} \in η_{β X} (U)$ iff $U \in \hat{A}$ iff $A \in U$ . Thus $μ η β = 1_{β}$ .

1.2

Likewise $A \in μ_{X} (β η_{X} (U))$ iff $(η_{X})^{- 1} [\hat{A}] \in U$ . This inverse image is $A$ , so $μ β η = 1_{β}$ .

L1L2

2.1

For $Z \in β^{3} X$ , expanding membership in $A$ along either $μ_{X} \circ β μ_{X}$ or $μ_{X} \circ μ_{β X}$ gives the same condition $\hat{\hat{A}} \in Z$ . Hence multiplication is associative, and steps 1.1–1.2 give the unit laws.

L1L2step 1.1step 1.2∎

Depends on

Used by

On finite sets the ultrafilter monad is naturally isomorphic to the identity; assuming the ultrafilter lemma, its unit is not invertible on the natural numbers Example

Dependency tree · next 3 levels

Direct dependencies and their dependencies through the next three levels: 19 results over 10 levels. An arrow runs from a result to what uses it, and this result sits at the bottom with a heavier outline. Click the chart to enlarge it.

Sources

E. Riehl, Category Theory in Context, 2nd ed., Example 5.1.4(v) and Exercise 5.1.ii (standard reference, not scraped)