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TheoremStatement: Literature-sourcedProof: AI-adaptedSession-authored (Fable 5 assisted)precheck passjudge pass (deepseek-v4-pro + claude-sonnet-5)audited 2026-08-17
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A monad morphism induces restriction of algebras and a natural comparison of free algebras

Statement

A monad morphism α:TS on C induces a functor α:CSCT over C, defined by restriction of algebra structure. Its components also define a natural transformation from the free T-algebra functor to the free S-algebra functor followed by α.

Facts & Assumptions

Given: A monad morphism α:(T,ηT,μT)(S,ηS,μS) on C.

[L1]

The equations for α are αηT=ηS and αμT=μSSααT (Morphisms between monads on one category).

[L2]

An S-algebra (A,a) satisfies aηAS=1A and aμAS=aS(a), and its homomorphisms satisfy fa=bS(f) (Algebra and algebra homomorphism for a monad).

[L3]

The free T-algebra on A is (TA,μAT) (Free algebra for a monad).

Proof

technique · direct
1.1

For an S-algebra (A,a), put aT=aαA:TAA. By [L1]–[L2], aTηAT=aηAS=1A, while naturality of α and the two multiplication equations give aTμAT=aTT(aT). Thus (A,aT) is a T-algebra.

L1L2
2.1

If f:(A,a)(B,b) is an S-algebra homomorphism, then faαA=bS(f)αA=bαBT(f) by naturality of α. Hence the unchanged underlying arrow is a T-algebra homomorphism, and unchanged identities and composites define a functor α over C.

L1L2step 1.1
3.1

For every A, the multiplication equation in [L1] says precisely that αA:(TA,μAT)(SA,μASαSA) is a T-algebra homomorphism. Naturality of α makes these maps natural in A, giving the claimed comparison of free algebras.

L1L3step 2.1

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