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TheoremStatement: Literature-sourcedProof: AI-generatedprecheck passjudge pass (deepseek-v4-pro + claude-sonnet-5)audited 2026-08-17
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Whenever the endofunctor category exists, monads on a fixed category and their morphisms form a category

Statement

Let C be a category for which the functor category [C,C] exists. Monads on C as objects and monad morphisms as arrows form a category.

Facts & Assumptions

Given: Monads (T,ηT,μT), (S,ηS,μS), and (R,ηR,μR) on C.

[L1]

A monad morphism is a natural transformation preserving the unit and multiplication (Morphisms between monads on one category).

[L2]

When [C,C] exists, natural transformations between endofunctors are arrows of a category and compose vertically (Functor category [C,D]).

Proof

technique · direct
1.1L1

The identity 1T:T⇒T satisfies 1TηT=ηT and 1TμT=μT T1T (1T)T, so it is a monad morphism.

2.1L1step 1.1

If α:T⇒S and β:S⇒R are monad morphisms, then (βα)ηT=βηS=ηR, so their vertical composite preserves the unit.

3.1L1L2step 2.1

Naturality of β gives βS∘Sα=Rα∘βT; substituting the multiplication equations for α and β yields (βα)μT=μR R(βα) (βα)T, so the composite preserves multiplication.

4.1L2step 1.1step 3.1∎

By [L2], vertical composition is associative and the transformations in step 1.1 are identities. Steps 2.1 and 3.1 give closure under composition under the stated endofunctor-category size condition. Hence these objects and arrows form a category.

Depends on

Used by

Nothing in the library uses this result yet.

Dependency tree · two levels

7 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