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The free–forgetful Eilenberg–Moore adjunction induces the given monad

Statement

For a monad $(T, η, μ)$ on $C$ , the assignment $A \mapsto (T A, μ_{A})$ defines a functor $F^{T} : C \to C^{T}$ left adjoint to the forgetful functor $U^{T} : C^{T} \to C$ . The monad induced by $F^{T} ⊣ U^{T}$ is $(T, η, μ)$ on the nose.

Facts & Assumptions

Given: A monad $(T, η, μ)$ , its Eilenberg–Moore category and forgetful functor (Eilenberg–Moore category of a monad), and its free algebras (Free algebra for a monad).

Proof

technique · direct

1.1

Define $F^{T} (A) = (T A, μ_{A})$ and $F^{T} (f) = T (f)$ . Naturality of $μ$ gives $T (f) \circ μ_{A} = μ_{B} \circ T^{2} (f)$ , so $T (f)$ is an algebra homomorphism; the functor laws follow from those of $T$ .

given

2.1

At an algebra $(A, a)$ define the counit component $ϵ_{(A, a)} : = a : (T A, μ_{A}) \to (A, a)$ ; the algebra associativity law makes it an algebra homomorphism and the homomorphism equation makes these components natural. The equations $a \circ η_{A} = 1_{A}$ and $μ_{A} \circ T (η_{A}) = 1_{T A}$ are the two triangle identities, so $F^{T} ⊣ U^{T}$ .

step 1.1given

3.1

The composite $U^{T} F^{T}$ equals $T$ , the adjunction unit is $η$ , and $U^{T} ϵ F^{T}$ has component $ϵ_{(T A, μ_{A})} = μ_{A}$ ; hence the induced multiplication is $μ$ and the induced monad is the given one on the nose.

step 1.1step 2.1∎

Depends on

Used by

Dependency tree · next 3 levels

Direct dependencies and their dependencies through the next three levels: 11 results over 7 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., Lemma 5.2.9 (standard reference, not scraped)
B. Richter, From Categories to Homotopy Theory, Theorem 6.2.5 (standard reference, not scraped)