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LemmaStatement: Literature-sourcedProof: AI-generatedPipeline-generatedjudge pass (gpt-6-luna)verified 2026-10-22 (gpt-6.1-sol)
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The knot exterior has a peripheral torus basis

Statement

Assume countable choice. For a smooth oriented knot with a chosen product tubular neighbourhood N(K)≅S1×D2, let μ be the positive fibre meridian and let σ be the product section s↦(s,1) of TK=∂N(K)≅S1×S1. Then TK=∂EK is a torus and [μ],[σ] form a basis of H1(TK;Z). The curve σ is a product section; this statement does not assert that it is the preferred zero-homology longitude.

Facts & Assumptions

Given: Countable choice, a smooth oriented knot K, its chosen tubular exterior, and the product trivialization of the normal disk bundle.

[A1]

Countable choice, written ACω, is The Axiom of Countable Choice (ACω).

[F1]

Knot exterior retains TK=∂EK=∂N(K) and the tube coordinates as part of the exterior data.

[F2]

Meridian and preferred longitude orients the fibre meridian as the boundary of its oriented normal disk.

[F3]

Tubular exterior and boundary torus supplies N(K)≅S1×D2 and TK≅S1×S1 under [A1].

[F4]

Cellular homology computes singular homology identifies cellular and singular homology for the finite CW structure on the torus.

[F5]

Cellular homology defines cellular homology as ker⁡dn/im⁡dn+1.

[F6]

Cellular boundary from three consecutive skeleta defines the cellular differential by the connecting map of successive skeleta followed by the quotient map.

[F7]

Cellular boundary is the incidence degree matrix computes each cellular boundary coefficient as the degree of the corresponding attaching map after collapsing the other cells.

Proof

technique · direct
1.1A1F1F2F3construct

In product coordinates on TK=S1×S1, the meridian μ is the second factor loop and the section σ is the first factor loop. Give the torus its standard product CW structure: take the quotient of a square by identifying opposite edges, with one vertex, the two oriented factor loops as its 1-cells, and the open square as its 2-cell attached by the commutator μσμ−1σ−1.

2.1F5F6F7step 1.1algebra

The cellular chain groups in dimensions 2,1,0 are Z,Z2,Z. The differential d1 is zero because each 1-cell starts and ends at the only vertex. By [F7], each coefficient of d2 is the degree of the attaching map after collapsing the other 1-cell. The commutator traverses each factor once in each orientation, so both degrees are zero. Thus by [F5] and [F6], H1cell(TK;Z)=Z2 with the factor loops as basis.

3.1F4step 2.1∎

By [F4], the same two classes form a basis of singular H1(TK;Z).

Peripheral torus diagram

Use coordinates (u,v)∈R2/Z2 on the product torus, with u in the oriented knot direction and v in the oriented meridian direction. The square's opposite edges are identified in the same direction. The horizontal and vertical loops are the product section σ and meridian μ; the diagonal is a representative of [σ]+[μ]. The diagram records the basis convention and is not used in the proof.

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