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LemmaStatement: Literature-sourcedProof: AI-generatedPipeline-generatedverified 2026-10-22 (gpt-6.1-sol)
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The exterior of a smooth proper disk has the homology of a circle

Statement

Assume AC. Let D⊂B4 be a smooth proper disk and let XD be its compact exterior. Then XD is connected, H1(XD;Z)≅Z generated by the positive meridian, and Hi(XD;Z)=0 for every i≥2.

Facts & Assumptions

Given: Full AC, a smooth proper disk D⊂B4, its chosen tubular neighborhood N(D), and its exterior XD.

[A1]

Full AC is assumed wherever the cover and orientation conventions use the choice axiom. The Axiom of Choice

[F1]

The tubular neighborhood is N(D)≅D2×D2 and N(D)∩XD≃D2×S1≃S1, whose generator is the positive meridian. Exterior of a smooth proper disk

[F2]

Mayer–Vietoris is exact for the collar-thickened decomposition B4=N(D)∪XD. Mayer–Vietoris sequence in singular homology

Proof

1.1F1F2givenA1

Both B4 and N(D) are contractible, and N(D)∩XD is connected and homotopy equivalent to S1 by [F1]. The reduced degree-zero part of Mayer–Vietoris [F2] therefore gives H~0(XD;Z)=0, so XD is connected.

2.1F1F2step 1.1algebra

In degree one, the sequence has H2(B4)=0, then H1(N(D)∩XD)≅Z, then H1(N(D))⊕H1(XD), then H1(B4)=0. Since H1(N(D))=0, the inclusion of the overlap induces an isomorphism H1(N(D)∩XD)→H1(XD). Its generator is the positive meridian by [F1].

3.1F1F2step 2.1algebra∎

For every i≥2, the adjacent groups Hi(N(D)∩XD), Hi(N(D)), and Hi(B4) vanish. Exactness of [F2] then forces Hi(XD)=0. This proves the asserted homology in all degrees.

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Dependency tree · two levels

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Sources