TheoremStatement: Literature-sourcedProof: AI-adaptedprecheck passaudited 2026-08-16

How statement and proof provenance work

The first chip identifies the source of the statement or construction; the second identifies the source of its local proof or verification.

Literature-sourced: the exact statement appears in a cited source; only wording and notation differ.
AI-adapted: a semantically identical restatement of literature-sourced material, modulo indexing, notation, and boundary cases adopted by the library.
AI-generated: a genuinely novel statement formulated by AI, with no source for the claim itself.

These labels describe origin, not correctness: citations and verification chips remain separate evidence.

Adjunctions compose with the composite unit and counit formulas

Statement

Let $F : C \to D$ be left adjoint to $G : D \to C$ , with unit $η$ and counit $ε$ , and let $F^{'} : D \to E$ be left adjoint to $G^{'} : E \to D$ , with unit $η^{'}$ and counit $ε^{'}$ . Then

$F^{'} F ⊣ G G^{'}$

with unit and counit

$\overset{ˉ}{η} = (G η^{'} F) \circ η : 1_{C} \Rightarrow G G^{'} F^{'} F,$

$\overset{ˉ}{ε} = ε^{'} \circ (F^{'} ε G^{'}) : F^{'} F G G^{'} \Rightarrow 1_{E} .$

Facts & Assumptions

Given: The two adjunctions and their units and counits as in the Statement.

[L1]

An adjunction is determined by a unit, a counit, and the two triangle identities (Adjunction by unit, counit, and the triangle identities).

[F1]

Whenever the expressions are defined, the interchange identity is $(β^{'} \circ β) * (α^{'} \circ α) = (β^{'} * α^{'}) \circ (β * α)$ (Horizontal and vertical composition of natural transformations satisfy the interchange law).

Proof

technique · direct

1.1F1

Whiskering gives $G η^{'} F : G F \Rightarrow G G^{'} F^{'} F$ and $F^{'} ε G^{'} : F^{'} F G G^{'} \Rightarrow F^{'} G^{'}$ , so the displayed composites have the required types; they are natural because whiskering and vertical composition preserve naturality.

2.1F1step 1.1

Expanding the first triangle composite for $F^{'} F$ gives $(\overset{ˉ}{ε} F^{'} F) \circ (F^{'} F \overset{ˉ}{η}) = (ε^{'} F^{'} F) \circ (F^{'} ε G^{'} F^{'} F) \circ (F^{'} F G η^{'} F) \circ (F^{'} F η)$ . Interchange rewrites the two middle factors as $F^{'} [(ε G^{'} F^{'} F) \circ (F G η^{'} F)]$ , and naturality of $ε$ at the component $η_{F c}^{'}$ turns that bracket into $(η^{'} F) \circ (ε F)$ .

2.2F1L1step 1.1

Expanding the second triangle composite for $G G^{'}$ gives $(G G^{'} \overset{ˉ}{ε}) \circ (\overset{ˉ}{η} G G^{'}) = (G G^{'} ε^{'}) \circ (G G^{'} F^{'} ε G^{'}) \circ (G η^{'} F G G^{'}) \circ (η G G^{'})$ . Interchange rewrites the two middle factors as $G [(G^{'} F^{'} ε G^{'}) \circ (η^{'} F G G^{'})]$ , and naturality of $η^{'}$ at the component $ε_{G^{'} d}$ turns that bracket into $(η^{'} G^{'}) \circ (ε G^{'})$ . The composite becomes $G [(G^{'} ε^{'}) \circ (η^{'} G^{'})] \circ [(G ε) \circ (η G)] G^{'}$ , which by the two second triangle identities $G ε \circ η G = 1_{G}$ and $G^{'} ε^{'} \circ η^{'} G^{'} = 1_{G^{'}}$ is $1_{G G^{'}}$ .

3.1L1step 2.1

By step 2.1 the first composite becomes $[(ε^{'} F^{'}) \circ (F^{'} η^{'})] F \circ F^{'} [(ε F) \circ (F η)]$ , which by the two first triangle identities $ε F \circ F η = 1_{F}$ and $ε^{'} F^{'} \circ F^{'} η^{'} = 1_{F^{'}}$ is $1_{F^{'} F}$ .

4.1step 3.1step 2.2L1∎

Thus $\overset{ˉ}{η}$ and $\overset{ˉ}{ε}$ satisfy both triangle identities, and [L1] gives $F^{'} F ⊣ G G^{'}$ .

Depends on

Used by

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

Emily Riehl, Category Theory in Context, 2nd ed., Proposition 4.3.4 (standard reference, not scraped)
Saunders Mac Lane, Categories for the Working Mathematician, 2nd ed., Chapter IV.8 (standard reference, not scraped)