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The comparison from the Kleisli category is fully faithful with image the free algebras

Statement

For a monad $T$ on $C$ , the canonical comparison functor

$M : C_{T} \to C^{T}$

sends $A$ to the free algebra $(T A, μ_{A})$ and is fully faithful. Its strict image is exactly the full subcategory of free $T$ -algebras.

Facts & Assumptions

Given: The Kleisli and Eilenberg–Moore resolutions, their comparison functors (The Kleisli factorisation functor for an adjunction inducing a monad exists and is unique, The comparison functor to the Eilenberg–Moore category exists and is unique), and the definition of a free algebra (Free algebra for a monad).

Proof

technique · direct

1.1given

The comparison sends $A$ to $(T A, μ_{A})$ and a Kleisli arrow $f : A \to T B$ to the algebra homomorphism $μ_{B} \circ T (f) : T A \to T B$ .

2.1step 1.1given

For every $A, B$ , the map $f \mapsto μ_{B} \circ T (f)$ is a bijection from $C (A, T B) = C_{T} (A, B)$ to the algebra homomorphisms $(T A, μ_{A}) \to (T B, μ_{B})$ ; its inverse sends $h$ to $h \circ η_{A}$ . The monad unit laws show the two composites are identities, using the algebra-homomorphism equation for $h$ in one direction.

3.1step 2.1∎

Thus every induced hom-set map is bijective, so $M$ is fully faithful (Faithful, full, fully faithful, essentially surjective, and split essentially surjective functors); step 1.1 also shows that its objects are precisely the free algebras, giving the claimed strict image.

Depends on

Used by

Dependency tree · two levels

13 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., Lemma 5.2.14 (standard reference, not scraped)
B. Richter, From Categories to Homotopy Theory, Proposition 6.3.5 (standard reference, not scraped)