TheoremStatement: AI-adaptedProof: AI-generatedprecheck passjudge pass (gpt-5.6-terra)audited 2026-09-01

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.

Every braided monoidal category is monoidally equivalent to a strict braided one

Statement

Let $(C, c)$ be a braided monoidal category. Then there exists a strict monoidal category $D$ , a braiding $c^{s t r}$ on $D$ , and a monoidal equivalence $F : C \to D$ such that $F$ is a braided monoidal functor.

Facts & Assumptions

Given: A braided monoidal category $(C, c)$ .

[L1]

Mac Lane strictification gives a monoidal equivalence from $C$ to a strict monoidal category (Mac Lane strictification).

[L2]

A monoidal equivalence consists of a strong monoidal functor with strong monoidal quasi-inverse data (Monoidal equivalence and monoidal quasi-inverse data).

[L3]

A braided monoidal functor is a strong monoidal functor whose tensor constraint intertwines the two braidings (Braided monoidal functor).

[F1]

The braided strictification theorem in the cited monoidal-category source states that Mac Lane's strictification carries a unique transported braiding for which the strictification equivalence is braided monoidal.

Proof

technique · direct

1.1

Apply [L1] and [L2] to obtain a monoidal equivalence $F : C \to D$ with $D$ strict. By [F1], the braiding $c$ transports across the full strong monoidal equivalence to a braiding $c^{s t r}$ on $D$ .

L1L2F1givenchoose

2.1

The transported braiding is characterized by the compatibility square $J_{Y, X} \circ c_{F (X), F (Y)}^{s t r} = F (c_{X, Y}) \circ J_{X, Y} .$ Thus [L3] makes $F$ braided monoidal, and $C$ is braided-monoidally equivalent to the strict braided monoidal category $(D, c^{s t r})$ .

L3F1step 1.1∎

Depends on

Used by

FALSE: every braided monoidal category is equivalent to a strict commutative one False statement
Braided coherence is controlled by underlying braids Theorem
Symmetric coherence Theorem

Dependency tree · two levels

13 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

Michael Muger, Tensor Categories: A Selective Guided Tour, Section 4 (standard reference, not scraped)
P. Etingof, S. Gelaki, D. Nikshych, and V. Ostrik, Tensor Categories, Chapter 8.1 (standard reference, not scraped)