Alphabeta Math
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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.

Knot cobordism maps and functoriality — Examples and Counterexamples

1 · Prerequisites

2 · Logical flowchart

3 · Definitions, theorems and proofs

None yet.

4 · Examples, counterexamples and false statements

ExampleConstruction: Literature-sourcedVerification: AI-adaptedjudge pass (gpt-6-luna)verified 2026-10-22 (gpt-6.1-sol)Open item page →

A birth followed by a merging saddle

Example

Let A=Z[X]/(X2) be the Frobenius algebra with unit η(1)=1 and multiplication m. In the algebraic TQFT, consider the movie that first creates a disjoint circle and then attaches a saddle band merging that circle with one existing circle. The induced local map is

A→id⁡A⊗ηA⊗A→mA,

and this composite is the identity on A.

The diagram records the two successive surface events; it is only a description of the movie, while the verification below is algebraic.

diagram failed to render:
\begin{tikzcd}
\mathcal A
\arrow[r, "{\operatorname{id}\otimes\eta}"]
& \mathcal A\otimes\mathcal A
\arrow[r, "m"]
& \mathcal A
\end{tikzcd}

Facts & Assumptions

Given: The rank-two algebra A=Z[X]/(X2), its unit η(1)=1, and its merge map m.

[F1]

Khovanov Frobenius algebra and cube of resolutions gives the basis {1,X}, the unit 1↦1, and multiplication m(a⊗1)=a for a∈A.

Proof

1.1F1constructalgebra

On the basis element 1, (id⁡⊗η)(1)=1⊗1, so m(id⁡⊗η)(1)=m(1⊗1)=1. On the basis element X, (id⁡⊗η)(X)=X⊗1, so m(id⁡⊗η)(X)=m(X⊗1)=X.

2.1step 1.1algebra∎

The two basis elements generate A as an abelian group, and both are fixed by the composite in step 1.1. By linearity the composite fixes every element of A, hence equals id⁡A.

CounterexampleConstruction: Literature-sourcedVerification: AI-adaptedverified 2026-10-22 (gpt-6.1-sol)Open item page →

A birth and a death do not cancel

Statement refuted

The assertion that a birth of a disjoint circle followed by a death of that same circle induces the identity up to an overall sign on an unaffected circle is false. Let A=Z[X]/(X2) with unit η(1)=1 and counit ϵ(1)=0,ϵ(X)=1. On a movie with one unaffected circle, birth a disjoint circle and then cap that same circle by a death. The composite operator is

A→id⁡A⊗ηA⊗A→id⁡A⊗ϵA,

which is zero and therefore is not the identity, even after allowing multiplication by an overall sign.

diagram failed to render:
\begin{tikzcd}
\mathcal A
\arrow[r, "{\operatorname{id}\otimes\eta}"]
& \mathcal A\otimes\mathcal A
\arrow[r, "{\operatorname{id}\otimes\epsilon}"]
& \mathcal A
\end{tikzcd}

The movie is a cylinder on the original circle disjoint union a born-and-capped sphere; the diagram specifies the local birth and death sequence and is not used as a proof.

Facts & Assumptions

Given: The free abelian group A=Z[X]/(X2), with η(1)=1, ϵ(1)=0, and ϵ(X)=1.

[F1]

Khovanov Frobenius algebra and cube of resolutions gives the unit and counit formulas and the nonzero basis element 1∈A.

Counterexample

1.1F1algebra

The birth map sends 1∈Z to 1∈A, and the following death sends this element to ϵ(1)=0. Thus ϵη=0:Z→Z.

2.1F1step 1.1algebra∎

Tensoring the zero map ϵη with id⁡A gives the zero endomorphism of A, which is the composite in the statement. The identity of A is nonzero because it sends 1 to 1≠0; its negative also sends 1 to −1≠0. Hence the composite is neither the identity nor its negative.

Sources