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The Kleisli factorisation functor for an adjunction inducing a monad exists and is unique
Statement
Let be an adjunction with counit whose induced monad is a fixed monad on on the nose, and write for the Kleisli adjunction with counit . There is exactly one functor satisfying
namely
These three equalities are what it means for to be a morphism of adjunctions from the Kleisli adjunction to .
Facts & Assumptions
Given: An adjunction as in Adjunction by unit, counit, and the triangle identities, inducing with 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
A Kleisli arrow has adjoint transpose ; define by this formula and by .
For an identity , the first triangle identity gives . For and , substitution of , naturality of , and the triangle identities gives , so is a functor.
The formulas give and on objects and arrows, and sends the Kleisli counit to . Conversely, those equalities force the image of to be its adjoint transpose, so they force the formula of step 1.1 on every arrow and prove uniqueness.
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
- E. Riehl, Category Theory in Context, 2nd ed., Proposition 5.2.13 (standard reference, not scraped)
- B. Richter, From Categories to Homotopy Theory, Theorem 6.3.10 (standard reference, not scraped)