LemmaStatement: Literature-sourcedProof: AI-generatedPipeline-generatedprecheck passverified 2026-11-05 (gpt-6.1-sol)

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.

Uniqueness of a representing dual fusion product

Statement

Let $C$ be a category and let $F : C^{o p} \to {V e c t}_{C}$ be a contravariant functor. A representing pair for $F$ is an object $D$ with an element $γ \in F (D)$ such that, for every object $M$ , the map ${H o m}_{C} (M, D) \to F (M)$ sending $T$ to $F (T) (γ)$ is a bijection. If $(D, γ)$ and $(D^{'}, γ^{'})$ both represent $F$ , then there is a unique isomorphism $f : D \to D^{'}$ satisfying $γ = F (f) (γ^{'})$ .

Statement references

The universal property used here is the abstract representing property that also defines a dual fusion product on this page.

Facts & Assumptions

Given: A contravariant functor $F$ and two representing pairs $(D, γ)$ and $(D^{'}, γ^{'})$ .

[A1]

Both pairs satisfy the representing bijection in the Statement for every object of $C$ .

Proof

1.1givenA1construct

Apply the representing bijection for $(D^{'}, γ^{'})$ with $M = D$ to $γ \in F (D)$ . It gives a unique morphism $f : D \to D^{'}$ with $γ = F (f) (γ^{'})$ .

1.2givenA1construct

Apply the representing bijection for $(D, γ)$ with $M = D^{'}$ to $γ^{'} \in F (D^{'})$ . It gives a unique morphism $g : D^{'} \to D$ with $γ^{'} = F (g) (γ)$ .

2.1givenA1step 1.1step 1.2algebra∎

By contravariance, $F (g \circ f) = F (f) \circ F (g)$ . Steps 1.1–1.2 give $F (g \circ f) (γ) = F (f) (γ^{'}) = γ$ , and the identity morphism has the same image; uniqueness for $(D, γ)$ gives $g \circ f = {i d}_{D}$ . The same argument gives $f \circ g = {i d}_{D^{'}}$ . Any other compatible morphism is equal to $f$ by uniqueness in the representing bijection for $(D^{'}, γ^{'})$ . Thus $f$ is the unique compatible isomorphism.

Dependency tree · 0 levels

Nothing. This result depends on no other item in the library.

Sources

Bin Gui and Hao Zhang, Analytic Conformal Blocks of C2-cofinite Vertex Operator Algebras I: Propagation and Dual Fusion Products