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TheoremStatement: Literature-sourcedProof: AI-adaptedprecheck passaudited 2026-09-01
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Currying and uncurrying are mutually inverse

Statement

In a cartesian closed category, currying and uncurrying for the adjunction −×B⊣(−)B are mutually inverse. Equivalently, for every f:A×B→C and g:A→CB,

uncurry⁡(curry⁡(f))=f,curry⁡(uncurry⁡(g))=g.

Repeated currying is associative after the canonical reassociation of products.

Facts & Assumptions

Given: A cartesian closed category and objects A,B,C,D.

[L1]

Cartesian closed means that the cartesian product is a closed monoidal tensor, so −×B has right adjoint (−)B (Cartesian closed category, A category with finite products is monoidal).

[L2]

The internal-hom adjunction comes with evaluation and inverse transposition operations (The internal hom and its evaluation morphism).

[L3]

Internal hom composition is obtained by transposing iterated evaluation and is compatible with reassociation (The internal-hom composition morphism).

Proof

technique · direct
1.1givenL1L2

By [L1] and [L2], currying is the transpose map C(A×B,C)→C(A,CB), and uncurrying is its inverse transpose. For any adjunction, transpose followed by inverse transpose and inverse transpose followed by transpose are the identity.

2.1step 1.1algebra

Therefore uncurry⁡(curry⁡(f))=f and curry⁡(uncurry⁡(g))=g.

2.2step 1.1L3algebra

For a morphism h:A×B×C→D, first curry in the C-variable and then in the B-variable. The resulting map A→(DC)B is the transpose of the same iterated evaluation map that produces A→DB×C after reassociating products. By [L3], these coincide under the canonical internal-hom composition isomorphism.

3.1step 2.1step 2.2∎

So currying and uncurrying are mutually inverse, and repeated currying is associative up to the canonical reassociation.

Depends on

Used by

Dependency tree · two levels

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Sources