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Adjoining one of two fixed points defines commuting closure-operator monads whose distributive law yields their composite
Example
On ordered by inclusion, define
These closure-operator monads commute, and their equality gives a distributive law whose composite adjoins both and .
Facts & Assumptions
Given: The inclusion poset and the displayed maps .
A monotone, extensive, idempotent map on a poset defines a monad (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).
A distributive law must satisfy the unit and multiplication compatibility diagrams (Distributive law between two monads).
Such a distributive law gives a monad structure on (A distributive law makes the composite endofunctor a monad).
Verification
Union with a fixed subset is monotone, extensive, and idempotent. Thus and are closure-operator monads by [L1].
For every , . This equality gives ; all diagrams in [L2] commute because is thin and the parallel arrows have the displayed common endpoints.
On , both composites respectively give . This checks the formula at every object, including the empty set.
By [L3], the distributive law makes a monad; its unit is the inclusion and its multiplication is the idempotence equality for the closure operator adjoining both fixed points.
Depends on
- Distributive law between two monads
- A distributive law makes the composite endofunctor a monad
- 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
- The power set $\mathcal{P}(x) = \{\, z : z \subseteq x \,\}$
- Partial order and partially ordered set
Used by
Nothing in the library uses this result yet.
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