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CounterexampleConstruction: Literature-sourcedVerification: AI-generatedPipeline-generatedjudge pass (gpt-6-luna)verified 2026-10-22 (gpt-6.1-sol)
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An arbitrary parallel push-off need not be preferred

Statement refuted

Assume countable choice. For every smooth oriented knot K with a chosen product tubular neighbourhood N(K)≅S1×D2 using its product Euclidean fibre norm, define its exterior by EK=S3∖int⁡N(K) and its boundary torus by TK=∂EK. Let i:TK↪EK be inclusion and let p:TK→S1 be the product projection to the oriented core. Every primitive class c∈H1(TK;Z) with p∗c=1∈H1(S1;Z) is in ker⁡i∗. Here primitive means that c is not an integer multiple md of another class with ∣m∣>1.

Facts & Assumptions

Given: Countable choice, a smooth oriented knot with a chosen product tubular neighbourhood, and its oriented boundary torus.

[A1]

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

[F1]

Knot exterior defines EK=S3∖int⁡N(K) and retains its boundary torus TK.

[F2]

Meridian and preferred longitude orients the fibre meridian and defines the preferred longitude by degree one over the oriented core and zero image in H1(EK;Z).

[F3]

The knot exterior has a peripheral torus basis identifies H1(TK;Z) with the product meridian and section basis ([μ],[σ]).

[F4]

Homology and preferred longitude of a knot exterior gives H1(EK;Z)≅Z, generated by [μ], with [λ] a degree-one kernel class, and says ([μ],[λ]) is a basis of H1(TK;Z).

Counterexample

technique · direct
1.1A1F1F2F3F4givenconstruct

By [F1] and [F2], the exterior inclusion and the meridian and longitude classes have the stated meanings. In the basis from [F3], the degree-one property in [F4] writes [λ]=[σ]+q[μ] for a unique integer q. The curve γ(t)=(e2πit,e2πi(q+1)t) is embedded because its first coordinate is injective on S1, and it represents [σ]+(q+1)[μ]=[λ]+[μ]. Its projection to the oriented core has degree +1.

2.1F4step 1.1algebra

In the basis ([μ],[λ]) from [F4], the class [λ]+[μ] has coordinates (1,1). If it were md for an integer ∣m∣>1, both coordinates would be divisible by m, which is impossible; hence it is primitive.

3.1F4step 1.1step 2.1algebra∎

The inclusion sends [λ]+[μ] to [μ] because [λ]∈ker⁡i∗. The latter is a generator of H1(EK;Z)≅Z, so this image is nonzero. The primitive degree-one class constructed above therefore contradicts the refuted assertion.

Depends on

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Sources