PropositionStatement: Literature-sourcedProof: AI-adaptedPipeline-generatedprecheck passjudge pass (gpt-5.6-terra)audited 2026-09-14

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.

Thom spaces of zero and trivial bundles

Statement

Naturally in $B$ , $Th (0_{B}) = B_{+}; Th (B \times R^{n}) = B_{+} \land S^{n} = Σ^{n} B_{+},$ and the trivial disk/sphere pair is $(B \times D^{n}, B \times S^{n - 1})$ .

Facts & Assumptions

Given: A space $B$ , the zero bundle, and the product bundle with its standard Euclidean metric.

[F1]

Disk, sphere, and Thom spaces of a metric vector bundle defines the disk, sphere, and based quotient, including the empty-sphere convention in rank zero.

Proof

technique · calculate the defining quotients

1.1F1

In rank zero each fiber is the singleton zero vector, so [F1] gives $D (0_{B}) = B$ and $S (0_{B}) = \emptyset$ . By the based-quotient convention, $Th (0_{B}) = B_{+}$ .

1.2F1

For $B \times R^{n}$ with the product metric, the norm depends only on the second coordinate, hence $(D, S) = (B \times D^{n}, B \times S^{n - 1})$ . Collapsing the second subspace gives $(B \times D^{n}) / (B \times S^{n - 1}) ≅ B_{+} \land (D^{n} / S^{n - 1}) = B_{+} \land S^{n}$ .

2.1F1step 1.1step 1.2∎

By definition $B_{+} \land S^{n} = Σ^{n} B_{+}$ . Every displayed map sends $(b, v)$ by the identity formula and therefore commutes with pullback along a map $B^{'} \to B$ . For $B = \emptyset$ both sides are the one-point based space; for $n = 0$ , $S^{- 1} = \emptyset$ and step 1.2 reduces to step 1.1; for $n = 1$ the boundary consists of the two endpoints. Zero and unit radii and the quotient basepoint are preserved, and no choice is used.

Thom spaces of zero and trivial bundles

Statement

Facts & Assumptions

Proof

Depends on

Dependency tree · 2 levels

Sources