TheoremStatement: Literature-sourcedProof: AI-adaptedprecheck passjudge pass (deepseek-v4-pro + claude-sonnet-5)audited 2026-08-17

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The cofree–forgetful co-Eilenberg–Moore adjunction induces the given comonad

Statement

For a comonad $(G, ε, δ)$ on $C$ , the forgetful functor $U^{G} : C_{G} \to C$ has a right adjoint $F^{G}$ , where $F^{G} (A) = (G A, δ_{A})$ . The comonad induced by $U^{G} ⊣ F^{G}$ is $(G, ε, δ)$ on the nose.

Facts & Assumptions

Given: A comonad $(G, ε, δ)$ on $C$ .

[L1]

A comonad on $C$ is a monad on $C^{o p}$ (Comonad on a category).

[L2]

The Eilenberg–Moore free–forgetful adjunction of a monad induces that monad on the nose (The free–forgetful Eilenberg–Moore adjunction induces the given monad).

Proof

technique · direct

1.1L1L2

Regard $G$ as a monad on $C^{o p}$ and apply [L2] there.

2.1L1L2step 1.1

Taking opposites translates algebras into coalgebras, the free algebra into the cofree coalgebra $F^{G} (A) = (G A, δ_{A})$ , and the free–forgetful adjunction into $U^{G} ⊣ F^{G}$ . Its unit at $(A, c)$ is $c : (A, c) \to (G A, δ_{A})$ and its counit at $A$ is $ε_{A} : G A \to A$ .

3.1step 2.1∎

The triangle equations translate to $G ε \circ δ = 1_{G}$ and $ε G \circ δ = 1_{G}$ . Therefore the induced endofunctor is $U^{G} F^{G} = G$ , its counit is $ε$ , and its comultiplication $U^{G} η F^{G}$ is $δ$ .

Depends on

Used by

Co-Kleisli and co-Eilenberg–Moore adjunctions have the dual extremal universal properties Corollary

Dependency tree · two levels

9 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 5.2.iii (standard reference, not scraped)