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Noncompact intersections can escape during a homotopy

Statement refuted

The transverse intersection count of a smooth homotopy is preserved whenever the endpoint maps are proper, even when the source is noncompact and the combined homotopy is not proper.

Facts & Assumptions

Given: The standard oriented R as source and target, Z={0} with positive point orientation, and F(x,t)=x−(1−t)x2 on R×[0,1].

[F1]

Euclidean spaces have their standard smooth structure and polynomial evaluation formulas give smooth families (Euclidean spaces and Euclidean open subsets as smooth manifolds, Smooth families of maps and their evaluation maps).

[F2]

At a zero, a real-valued slice is transverse to {0} if its derivative there is nonzero; its finite signed count is the sum of those derivative signs and its finite parity count is the number of zeros modulo two (The local oriented intersection sign). These are finite transverse counts, without asserting the compact-source invariants of The oriented intersection number and The mod 2 intersection number are defined for this noncompact source.

[F3]

A compact trace is what permits the boundary-count arguments for invariance; proper endpoints alone do not supply it (Properness can replace compactness only when the intersection trace is compact, The mod 2 intersection number is homotopy invariant, The oriented intersection number is homotopy invariant).

Counterexample

1.1F1F2givenalgebra

For t<1, Ft−1(0)={0,1/(1−t)} and ∂xFt is +1 at 0 and −1 at 1/(1−t). At t=1, F1(x)=x has the single zero 0 with derivative +1. Every slice is transverse at its zeros. The cardinalities change from 2 to 1, the signed counts from 0 to 1, and the parity counts from 0 to 1 as t reaches 1.

2.1F1F3step 1.1algebra∎

Every slice is proper. For t<1, ∣x−(1−t)x2∣→∞ as ∣x∣→∞; for t=1, the map is the identity. Thus each preimage of a compact real set is closed and bounded, hence compact. The trace contains {(1/(1−t),t):0≤t<1}, an unbounded branch whose intersection point escapes to +∞ as t↑1. Consequently F−1({0}) is noncompact and the combined map is not proper, although both endpoint maps, indeed all slices, are proper. This satisfies the refuted claim's hypotheses and disproves its conclusion.

The arctangent family G(x,t)=arctan⁡x−t on R×[0,2] gives another escape: its zero is tan⁡t for t<π/2 and absent for t≥π/2. It has compact individual regular fibres but improper endpoint maps, so it does not by itself refute the proper-endpoint assertion.

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