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The codensity construction satisfies the monad laws

Statement

Let $G : B \to C$ be a functor, and suppose its codensity monad $(T, η, μ)$ is supplied as in Codensity monad from a right Kan extension $ε : T G \Rightarrow G$ of $G$ along itself.

Then $(T, η, μ)$ is a monad on $C$ (Monad on a category): the two unit laws and the associativity law all hold.

Facts & Assumptions

Given: A right Kan extension $ε : T G \Rightarrow G$ of $G$ along itself, and the induced natural transformations $η : 1_{C} \Rightarrow T$ and $μ : T^{2} \Rightarrow T$ .

[F1]

In the codensity construction, $1_{G} = ε \circ (η G)$ and $ε \circ (μ G) = ε \circ (T ε)$ (Codensity monad).

[F2]

A right Kan extension $(T, ε)$ is terminal among natural transformations $S G \Rightarrow G$ (Left and right Kan extensions).

[F3]

A monad is an endofunctor with unit and multiplication satisfying $μ \circ T η = 1_{T} = μ \circ η T$ and $μ \circ T μ = μ \circ μ T$ (Monad on a category).

Proof

technique · direct

1.1

The codensity unit $η$ is defined by the identity on $G$ : by [F1], $1_{G} = ε \circ (η G)$ .

2.1

The codensity multiplication $μ$ is defined by pasting the counit with itself: by [F1], $ε \circ (μ G) = ε \circ (T ε)$ .

F1step 1.1

3.1

To prove the unit laws, compare natural transformations $T \Rightarrow T$ after whiskering with $G$ and composing with $ε$ , which [F2] makes a uniqueness test: $ε \circ ((μ \circ T η) G) = ε \circ (μ G) \circ T (η G) = ε \circ T ε \circ T (η G) = ε$ by [F1], so $μ \circ T η = 1_{T}$ ; likewise $ε \circ ((μ \circ η T) G) = ε \circ (μ G) \circ η T G = ε \circ T ε \circ η T G = ε \circ (η G) \circ ε = ε$ , where the third equality is naturality of $η$ at $ε$ , so $μ \circ η T = 1_{T}$ .

F1F2step 2.1

4.1

For associativity, both composites $μ \circ T μ$ and $μ \circ μ T$ are natural transformations $T^{3} \Rightarrow T$ . After whiskering with $G$ and composing with $ε$ , the left composite gives $ε \circ ((μ \circ T μ) G) = ε \circ (μ G) \circ T (μ G) = ε \circ T ε \circ T^{2} ε$ by [F1], while the right composite gives $ε \circ ((μ \circ μ T) G) = ε \circ (μ G) \circ μ T G = ε \circ T ε \circ μ T G = ε \circ (μ G) \circ T^{2} ε = ε \circ T ε \circ T^{2} ε$ , where the third equality is naturality of $μ$ at $ε$ and the last uses [F1] again. By the uniqueness clause [F2], the two composites are equal. Thus $μ \circ T μ = μ \circ μ T$ . Together with step 3.1, this is exactly the monad law package [F3].

F1F2F3step 3.1∎

Depends on

Used by

The codensity monad of the small skeleton of finite sets is the ultrafilter monad Theorem

Cited to discharge well-definedness by Codensity monad.

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

E. Riehl, Category Theory in Context, 2nd ed., Exercise 6.5(viii) (standard reference, not scraped)
T. Leinster, Codensity and the ultrafilter monad, §2 (standard reference, not scraped)