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TheoremStatement: AI-adaptedProof: Literature-sourcedprecheck passjudge pass (deepseek-v4-pro + claude-sonnet-5)audited 2026-08-17
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The Kleisli factorisation functor for an adjunction inducing a monad exists and is unique

Statement

Let F:C⇄D:U be an adjunction with counit ε whose induced monad is a fixed monad T on C on the nose, and write FT⊣UT for the Kleisli adjunction with counit εT. There is exactly one functor J:CT→D satisfying

JFT=F,UJ=UT,J(εBT)=εFB for every object B,

namely

J(A)=F(A),J(f:A→TB)=εFB∘F(f):FA→FB.

These three equalities are what it means for J to be a morphism of adjunctions from the Kleisli adjunction to F⊣U.

Facts & Assumptions

Given: An adjunction F⊣U as in Adjunction by unit, counit, and the triangle identities, inducing T=UF with μ=UεF as in Every adjunction induces a monad on the domain of its left adjoint, and the Kleisli adjunction of The Kleisli adjunction induces the given monad.

Proof

technique · direct
1.1given

A Kleisli arrow f:A→TB=UFB has adjoint transpose εFB∘F(f):FA→FB; define J by this formula and by J(A)=FA.

2.1step 1.1given

For an identity ηA, the first triangle identity gives J(ηA)=εFA∘F(ηA)=1FA. For f:A→TB and g:B→TC, substitution of μ=UεF, naturality of ε, and the triangle identities gives J(g⋆f)=J(g)∘J(f), so J is a functor.

3.1step 1.1step 2.1∎

The formulas give JFT=F and UJ=UT on objects and arrows, and J sends the Kleisli counit to ε. Conversely, those equalities force the image of f:A→UFB to be its adjoint transpose, so they force the formula of step 1.1 on every arrow and prove uniqueness.

Depends on

Used by

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Sources