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TheoremStatement: AI-adaptedProof: Literature-sourcedSession-authored (Fable 5 assisted)precheck 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:CD:U be an adjunction with counit ε whose induced monad is a fixed monad T on C on the nose, and write FTUT for the Kleisli adjunction with counit εT. There is exactly one functor J:CTD satisfying

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

namely

J(A)=F(A),J(f:ATB)=εFBF(f):FAFB.

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

Facts & Assumptions

Given: An adjunction FU 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
1.1

A Kleisli arrow f:ATB=UFB has adjoint transpose εFBF(f):FAFB; define J by this formula and by J(A)=FA.

given
2.1

For an identity ηA, the first triangle identity gives J(ηA)=εFAF(ηA)=1FA. For f:ATB and g:BTC, substitution of μ=UεF, naturality of ε, and the triangle identities gives J(gf)=J(g)J(f), so J is a functor.

step 1.1given
3.1

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:AUFB to be its adjoint transpose, so they force the formula of step 1.1 on every arrow and prove uniqueness.

step 1.1step 2.1

Depends on

Used by

Dependency tree · next 3 levels

Direct dependencies and their dependencies through the next three levels: 21 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