CounterexampleConstruction: AI-adaptedVerification: AI-adaptedPipeline-generatedjudge pass (gpt-5.6-terra)audited 2026-09-13

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Proper endpoint maps joined by a nonproper combined homotopy

Statement refuted

Properness of the endpoint maps does not imply properness of the combined homotopy. There are proper smooth maps $F_{0}, F_{1} : R \to R$ and a smooth homotopy between them whose combined map $R \times [0, 1] \to R$ is not proper.

Facts & Assumptions

Given: Define $H : R \times [0, 1] \to R$ by $H (x, t) = (2 t - 1)^{2} x$ , and write $F_{t} (x) = H (x, t)$ .

[F1]

Degree is invariant under proper smooth homotopy requires the combined map $H$ to be proper and explicitly warns that proper endpoint maps alone do not suffice.

[F2]

Heine-Borel in $R^{n}$ : with the Euclidean metric a subset of $R^{n}$ is compact if and only if it is closed and bounded, and the proof by bisection uses no choice principle; the same holds on the real line makes every singleton in $R$ compact and the unbounded real line noncompact.

Counterexample

1.1

The displayed polynomial formula is smooth. At both parameter endpoints, $F_{0} (x) = H (x, 0) = x = H (x, 1) = F_{1} (x) .$ Thus both endpoint maps are the identity, and each is proper because its inverse image of any compact set is that same compact set.

given

1.2

The compact singleton ${0}$ has inverse image $H^{- 1} ({0}) = ({0} \times [0, 1]) \cup (R \times {1 / 2}) .$ This inverse image is not compact: for $n \geq 1$ , let $U_{n} = H^{- 1} ({0}) \cap ((- n, n) \times (1 / 4, 3 / 4))$ , and let $V = H^{- 1} ({0}) \cap (R \times ([0, 1] ∖ {1 / 2}))$ . These sets are open in the inverse-image subspace and ${V, U_{1}, U_{2}, \dots}$ covers it, but any finite subfamily misses $(x, 1 / 2)$ once $∣ x ∣$ exceeds every selected index. Hence $H$ is not proper.

F2given

2.1

The two proper endpoint maps are therefore joined by a nonproper combined homotopy, so the endpoint-only inference fails and [F1]'s hypothesis is indispensable. At $t = 1 / 2$ the slice is the constant zero map, which pinpoints the degeneracy; at $t = 0, 1$ it is the identity. The source and compact test set are nonempty, all endpoints and the zero fibre are explicit, and the countable cover is specified by a formula rather than selected, so no choice principle is used.

F1step 1.1step 1.2∎

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Sources

Robbin–Salamon, Introduction to Differential Topology, proper-homotopy clause following Theorem 5.4.1 (standard reference, not scraped)