TheoremStatement: Literature-sourcedProof: Literature-sourcedPipeline-generatedjudge pass (gpt-5.6-terra)audited 2026-09-06‡ rests on unproved material (inherited)

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.

‡ Rests on 3 statements not proved in this library, by way of the results it cites. This item cites no such statement directly; it depends on results that do. The unproved premises it inherits are ‡ Cohen's first model: an infinite Dedekind-finite set of reals, ‡ Sierpiński 1947: the generalised continuum hypothesis implies the Axiom of Choice and ‡ The continuum hypothesis and its generalisation are independent of ZFC. Each is recorded with a citation to the literature and is not established here, because the track that would prove it has not yet been developed in this library. Everything else in this proof is proved here.

Duality induces an anti-isomorphism on the Grothendieck ring

Statement

For a tensor category, $[X] \mapsto [X^{\lor}]$ induces an additive anti-isomorphism of $K_{0} (C)$ with its opposite ring. It need not be an involution: its inverse is induced by right duals.

Facts & Assumptions

Given: A tensor category $C$ and chosen left duals.

[F1]

Dualization is exact (Dualization in a multitensor category is exact).

[F2]

$(X \otimes Y)^{\lor} ≅ Y^{\lor} \otimes X^{\lor}$ (Left duality is a contravariant antimonoidal functor).

[F3]

The Grothendieck group is defined by exact-sequence relations (The Grothendieck ring of a tensor category).

Proof

technique · direct

1.1

By [F1], dualization takes each short exact relation in [F3] to a short exact relation, so $[X] \mapsto [X^{\lor}]$ is a well-defined additive map.

F1F3given

2.1

By [F2], it reverses products: $([X] [Y])^{*} = [Y]^{*} [X]^{*}$ . Chosen right duals give the inverse map on isomorphism classes and hence on $K_{0}$ . Therefore this is an anti-isomorphism; no identification of $X$ with $X^{\lor \lor}$ is asserted.

step 1.1F2given∎

Depends on

Used by

The Grothendieck ring of a tensor category is always commutative False statement

Dependency tree · two levels

9 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

Etingof, Gelaki, Nikshych, Ostrik, Tensor Categories, Section 4.5 (standard reference, not scraped)