TheoremStatement: Literature-sourcedProof: AI-adaptedprecheck passjudge pass (gpt-5.6-terra)audited 2026-08-26

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Adjunctions as absolute Kan extensions, with the preserved converse

Statement

Let $F : C \to D$ and $G : D \to C$ be functors.

If $F ⊣ G$ with unit $η : 1_{C} \Rightarrow G F$ and counit $ε : F G \Rightarrow 1_{D}$ , then:

$(G, η)$ is a left Kan extension of $1_{C}$ along $F$ , and it is absolute;
$(F, ε)$ is a right Kan extension of $1_{D}$ along $G$ , and it is absolute.

Conversely, if $(G, η)$ is a left Kan extension of $1_{C}$ along $F$ and is preserved by $F$ , then $F ⊣ G$ . Dually, if $(F, ε)$ is a right Kan extension of $1_{D}$ along $G$ and is preserved by $G$ , then $F ⊣ G$ .

The preservation clause in the converse is load-bearing.

Facts & Assumptions

Given: Functors $F : C \to D$ and $G : D \to C$ .

[F1]

An adjunction $F ⊣ G$ consists of a unit $η : 1_{C} \Rightarrow G F$ and a counit $ε : F G \Rightarrow 1_{D}$ satisfying $ε F \circ F η = 1_{F}$ and $G ε \circ η G = 1_{G}$ (Adjunction by unit, counit, and the triangle identities).

[F2]

A left Kan extension of $1_{C}$ along $F$ is initial among natural transformations $1_{C} \Rightarrow H F$ , and a right Kan extension of $1_{D}$ along $G$ is terminal among natural transformations $H G \Rightarrow 1_{D}$ (Left and right Kan extensions).

[F3]

An absolute Kan extension is one preserved by every functor out of its codomain (Absolute Kan extension, Covariant functor, identity functor, composite functor, and contravariant functor).

Proof

technique · direct

1.1F1F2

Assume $F ⊣ G$ with unit $η$ and counit $ε$ . Let $H : D \to C$ and $α : 1_{C} \Rightarrow H F$ be given. Define ${\overline{α}}_{d} : = H (ε_{d}) \circ α_{G d} : G d \to H d$ . Naturality of $α$ at $η_{c} : c \to G F c$ and the triangle identity $ε_{F c} \circ F (η_{c}) = 1_{F c}$ give $(\overline{α} F) \circ η = α$ . If $γ : G \Rightarrow H$ also satisfies $(γ F) \circ η = α$ , then for each $d$ naturality of $γ$ at $ε_{d} : F G d \to d$ and the triangle identity $G (ε_{d}) \circ η_{G d} = 1_{G d}$ force $γ_{d} = H (ε_{d}) \circ α_{G d} = {\overline{α}}_{d}$ . So $(G, η)$ is a left Kan extension of $1_{C}$ along $F$ . Dually, for $H : C \to D$ and $β : H G \Rightarrow 1_{D}$ , the formula ${\overline{β}}_{c} : = β_{F c} \circ H (η_{c}) : H c \to F c$ gives the unique factorization $ε \circ (\overline{β} G) = β$ , so $(F, ε)$ is a right Kan extension of $1_{D}$ along $G$ .

2.1F1F3step 1.1

The same formulas prove absoluteness. Let $S : C \to Z$ and $α : S \Rightarrow H F$ with $H : D \to Z$ . Then ${\overline{α}}_{d} : = H (ε_{d}) \circ α_{G d} : S G d \to H d$ gives the unique factorization $(\overline{α} F) \circ S η = α$ , so $(S G, S η)$ is a left Kan extension of $S$ along $F$ . Thus $(G, η)$ is absolute by [F3]. The right-handed argument is dual.

3.1F1F2step 2.1∎

Conversely, suppose $(G, η)$ is a left Kan extension of $1_{C}$ along $F$ and is preserved by $F$ . Applying the preserved Kan-extension property to the identity transformation $1_{F} : F \Rightarrow F$ yields a unique natural transformation $ε : F G \Rightarrow 1_{D}$ with $ε F \circ F η = 1_{F}$ , the first triangle identity. To obtain the second, note that both $1_{G}$ and $G ε \circ η G$ are natural transformations $G \Rightarrow G$ whose composites with $η$ agree: componentwise, naturality of $η$ at $η_{c}$ and the first triangle identity give $G (ε_{F c}) \circ η_{G F c} \circ η_{c} = G (ε_{F c} \circ F η_{c}) \circ η_{c} = η_{c} .$ By the uniqueness clause in the left Kan universal property, $G ε \circ η G = 1_{G}$ . Hence $F ⊣ G$ by [F1]. The right-handed converse is dual.

Depends on

Used by

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Dependency tree · two levels

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Sources

E. Riehl, Category Theory in Context, 2nd ed., Proposition 6.5.2 (standard reference, not scraped)
B. Richter, From Categories to Homotopy Theory, Proposition 4.7.3 (standard reference, not scraped)