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 $- \times B ⊣ (-)^{B}$ are mutually inverse. Equivalently, for every $f : A \times B \to C$ and $g : A \to C^{B}$ ,

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

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 $- \times B$ 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

1.1

By [L1] and [L2], currying is the transpose map $C (A \times B, C) \to C (A, C^{B})$ , and uncurrying is its inverse transpose. For any adjunction, transpose followed by inverse transpose and inverse transpose followed by transpose are the identity.

givenL1L2

2.1

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

step 1.1algebra

2.2

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

step 1.1L3algebra

3.1

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

step 2.1step 2.2∎

Depends on

Used by

Currying for sets of three variables Example

Dependency tree · two levels

12 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

S. Mac Lane, Categories for the Working Mathematician, 2nd ed., IV.6 (standard reference, not scraped)
Emily Riehl, Category Theory in Context, 2nd ed., Section 4.4 (standard reference, not scraped)