CounterexampleConstruction: Literature-sourcedVerification: AI-adaptedPipeline-generatedjudge pass (gpt-6.1-sol)

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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.

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Equal total degree does not classify maps from a disconnected domain

Statement refuted

For every closed oriented smooth $m$ -manifold $M$ , possibly disconnected, two maps $M \to S^{m}$ are homotopic if and only if their total degree, the sum of the signed degrees on the connected components, agrees. Counterexample: for $m \geq 1$ take $M = S^{m} ⊔ S^{m}$ , let $c$ be a constant map of $S^{m}$ into $S^{m}$ , and put $f = id ⊔ c$ and $g = c ⊔ id$ . Both maps have total degree $1$ , but they are not homotopic.

Facts & Assumptions

Given: An integer $m \geq 1$ , the closed oriented smooth manifold $M = S^{m} ⊔ S^{m}$ with the orientation of each copy, the identity $id$ of $S^{m}$ and a constant map $c : S^{m} \to S^{m}$ (Euclidean spheres and closed balls as subspaces of $R^{n}$ , Smooth manifolds and their smooth charts).

[F1]

The identity map of an oriented closed manifold has degree $1$ , while a constant map has degree $0$ : the identity has degree $1$ by the cited composition proposition, and a constant map factors through a point and has empty regular fibre over any value other than its constant, so the regular-value formula gives degree $0$ (Degree is multiplicative under composition, Regular-value formula for degree, Degree of a proper smooth map by compact-support cohomology).

[F2]

The total degree of a map on a disjoint union of closed oriented components is the sum of the degrees of its restrictions, and a homotopy of maps of $M$ restricts on each component to a homotopy of the restrictions; degrees of proper smooth homotopic maps between closed oriented manifolds agree, and for continuous self-maps of a sphere homotopic maps have equal degree (Degree of a proper smooth map by compact-support cohomology, Homotopies of continuous maps, homotopies relative to a subspace, and path homotopies relative to the endpoints, Degree is invariant under proper smooth homotopy, Degree is homotopy invariant and multiplicative under composition).

[F3]

The Hopf classification requires a connected domain, so it does not apply to $M$ ; the connectedness hypothesis is recorded as load-bearing (The Hopf degree theorem for oriented domains, Connectedness is needed for a single degree invariant).

Counterexample

technique · constructive

1.1F1F2givenconstruct

(Equal total degrees.) Let $f = id ⊔ c$ and $g = c ⊔ id$ on $M = S^{m} ⊔ S^{m}$ . The restrictions to the two components have degrees $1$ and $0$ in the first case and $0$ and $1$ in the second, so by [F2] both total degrees equal $1 + 0 = 1 = 0 + 1$ .

2.1F1F2step 1.1

(No homotopy exists.) Suppose $H : M \times I \to S^{m}$ were a homotopy from $f$ to $g$ . Its restriction to the first copy $S^{m} \times I$ is a homotopy from $id$ to $c$ between continuous self-maps of the sphere; by the sphere homotopy invariance recorded in [F2], $\deg (id) = \deg (c)$ , contradicting $1 \neq 0$ from [F1].

3.1F3step 2.1discharge-construct∎

There is therefore a closed oriented smooth $m$ -manifold, namely $S^{m} ⊔ S^{m}$ , and two maps on it whose total degrees agree but which are not homotopic; the total degree is not a complete invariant for disconnected domains, exactly as recorded in the connectedness remark.

Depends on

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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)