TheoremStatement: AI-adaptedProof: AI-adaptedPipeline-generatedjudge pass (gpt-6.1-sol)

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.

The Hopf degree theorem for oriented domains

Statement

Assume ${A C}_{ω}$ . Let $M$ be a nonempty closed connected oriented smooth $m$ -manifold, $m \geq 1$ . (i) Two smooth maps $f, g : M \to S^{m}$ are smoothly homotopic if and only if $\deg (f) = \deg (g)$ ; equivalently degree induces a bijection from smooth homotopy classes to $Z$ . (ii) Every integer occurs as $\deg (f)$ for some smooth $f : M \to S^{m}$ . (iii) Consequently degree induces a bijection from the set $[M, S^{m}]$ of free homotopy classes of continuous maps to $Z$ : two continuous maps $M \to S^{m}$ are homotopic if and only if they have the same degree. Here the degree of a continuous map is the degree of any homotopic smooth representative; part (i) and the approximation theorems make this independent of the representative.

Facts & Assumptions

Given: A nonempty closed connected oriented smooth $m$ -manifold $M$ with $m \geq 1$ and the compact-support degree of Degree of a proper smooth map by compact-support cohomology (Oriented smooth manifolds and oriented charts, Open cover, subcover, and compact topological space; a compact subset is a subspace that is compact in its own right, The Axiom of Countable Choice ( ${A C}_{ω}$ )).

[F1]

The compact-support degree is invariant under proper smooth homotopy, and any homotopy $M \times I \to S^{m}$ is proper because $M$ is compact (Degree is invariant under proper smooth homotopy, Homotopies of continuous maps, homotopies relative to a subspace, and path homotopies relative to the endpoints).

[F2]

For a smooth map $f : M \to S^{m}$ , a regular value $y$ with a positive basis $b$ and the framed regular preimage $(f^{- 1} (y), f_{*} b)$ , the signed count of the framed preimage equals $\deg (f)$ (The signed preimage count equals the degree, Orientation of a finite-dimensional real vector space, The Axiom of Countable Choice ( ${A C}_{ω}$ )).

[F3]

The signed count is a complete invariant of framed cobordism classes of closed framed $0$ -manifolds in $M$ : it is a bijection onto $Z$ and framed null-cobordism is exactly vanishing signed count (Framed zero-dimensional bordism in an oriented manifold is the integers).

[F4]

If two smooth maps have framed cobordant regular preimages at some regular values and positive bases, then they are smoothly homotopic (A framed cobordism of regular preimages produces a homotopy).

[F5]

Every integer is realized as the degree of a smooth map $M \to S^{m}$ ; regular values exist by Sard's theorem; every continuous map is homotopic to a smooth map and continuously homotopic smooth maps are smoothly homotopic (Every integer is realized by a map to the sphere, Morse-Sard for smooth manifolds, Every continuous map between smooth manifolds is homotopic to a smooth map, Continuously homotopic smooth maps are smoothly homotopic, Homotopies of continuous maps, homotopies relative to a subspace, and path homotopies relative to the endpoints).

Proof

technique · direct

1.1F1givenalgebra

(Forward direction.) If $f$ and $g$ are smoothly homotopic then their degrees agree by [F1], since the homotopy is proper.

1.2F2F3F5

(Converse.) Suppose $\deg (f) = \deg (g)$ . By [F5] choose regular values $y$ of $f$ and $y^{'}$ of $g$ and positive bases there; by [F2] the signed counts of the framed preimages $(f^{- 1} (y), f_{*} b)$ and $(g^{- 1} (y^{'}), g_{*} b^{'})$ equal $\deg (f)$ and $\deg (g)$ , hence are equal, and by [F3] the two framed preimages are framed cobordant.

2.1F4F5step 1.1step 1.2∎

Applying [F4] to the framed cobordism of step 1.2 gives a smooth homotopy $f ≃ g$ , which proves (i) for smooth maps; (ii) is [F5], and (iii) follows because [F5] lets every continuous map be replaced by a homotopic smooth map and every continuous homotopy by a smooth one, after which (i) applies. No homotopy invariance is used in the converse: that direction is the framed-cobordism classification together with the inverse Pontryagin-Thom construction.

Depends on

Used by

Dependency tree · two levels

84 results within two dependency steps of this one, each drawn at its shortest distance from it. An arrow runs from a result to what uses it, so the chart reads left to right and ends at this result, which carries a heavier outline. Every node is a link to that result. Click elsewhere on the chart to enlarge it.

Sources

Daniel S. Freed, Bordism: Old and New (lecture notes, UT Austin, Fall 2012) (standard reference, not scraped)
John Milnor, Topology from the Differentiable Viewpoint (standard reference, not scraped)
Victor Guillemin and Alan Pollack, Differential Topology (standard reference, not scraped)