TheoremStatement: Literature-sourcedProof: AI-generatedprecheck passjudge pass (gpt-5.6-terra)audited 2026-09-04

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The double dual is a monoidal functor

Statement

With chosen left duals, the double-dual assignment $X \mapsto X^{\lor \lor}$ is a monoidal endofunctor.

Facts & Assumptions

Given: Chosen left duals on a left rigid monoidal category.

[L1]

Left duality is a contravariant antimonoidal functor (Left duality is a contravariant antimonoidal functor).

Proof

technique · direct

1.1

By [L1], $(-)^{\lor} : C^{o p} \to C$ is contravariant, so applying it twice yields a covariant endofunctor $(-)^{\lor \lor} : C \to C$ .

givenL1

1.2

Again by [L1], there are compatible isomorphisms $(X \otimes Y)^{\lor} ≅ Y^{\lor} \otimes X^{\lor}$ and $1^{\lor} ≅ 1$ . Dualizing once more reverses the order a second time, so the composite comparison gives $X^{\lor \lor} \otimes Y^{\lor \lor} ≅ (Y^{\lor} \otimes X^{\lor})^{\lor} ≅ (X \otimes Y)^{\lor \lor}$ together with a unit isomorphism $1 ≅ 1^{\lor \lor}$ .

L1algebra

2.1

Since the monoidal comparison maps are obtained by composing those of the antimonoidal functor with itself, their coherence is inherited from the coherence in [L1]. Therefore $(-)^{\lor \lor}$ is a monoidal endofunctor.

step 1.1step 1.2∎

Depends on

Left duality is a contravariant antimonoidal functor

Used by

Left and right duals, and double duals, need not collapse Counterexample
Pivotal structure Definition
The categorical trace of a morphism into the double dual Definition
A braided rigid category has a Drinfeld morphism Theorem

Dependency tree · two levels

4 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

P. Etingof, S. Gelaki, D. Nikshych, and V. Ostrik, Tensor Categories, Definition 4.7.7 (standard reference, not scraped)