CounterexampleConstruction: Literature-sourcedVerification: Literature-sourcedPipeline-generatedprecheck passjudge 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.

Algebraic intersection one with three geometric points

Statement refuted

Assume ${A C}_{ω}$ . In the standard $4$ -dimensional model $W = D^{4} \cup h^{2}$ whose middle boundary is $N ≅ S^{2} \times S^{1}$ with belt sphere $B = {p} \times S^{1}$ of the $2$ -handle, let $A \subset N$ be the $2$ -sphere obtained from $S^{2} \times {u_{0}}$ by a finger move across $B$ . Then $A$ meets $B$ transversely in exactly three points with local signs $+ 1, + 1, - 1$ ; the oriented intersection number is $I (A, B) = 1$ and the mod-2 number is $1$ , while the attaching sphere $A$ and the belt sphere $B$ do not meet in one point. Hence the attached $3$ -handle and the $2$ -handle form a pair whose matrix entry is a unit but which is not geometrically cancelling: the cancellation theorem does not apply, although reversing this specific finger isotopy removes the extra pair. A general algebraic-to-geometric conversion is a separate Whitney-trick issue with additional hypotheses.

Facts & Assumptions

Given: The standard $4$ -dimensional model $W = D^{4} \cup h^{2}$ with middle boundary $N ≅ S^{2} \times S^{1}$ , belt sphere $B = {p} \times S^{1}$ , and a $2$ -sphere $A \subseteq N$ obtained from $S^{2} \times {u_{0}}$ by a finger move across $B$ .

[F1]

Attaching-belt intersection matrix of adjacent-index handles and K handle core cocore attaching region and belt sphere: the belt sphere of the $2$ -handle in the middle boundary is a circle, and the attaching sphere of a $3$ -handle attached to $N$ is a $2$ -sphere; the matrix entry is the oriented, respectively mod-2, intersection number of the attaching sphere with the belt sphere.

[F2]

Algebraic cancellation does not yet give geometric cancellation: on $N = S^{2} \times S^{1}$ there are an embedded $2$ -sphere and an embedded circle meeting transversely in exactly three points with local signs $+ 1, + 1, - 1$ , so that the oriented intersection number is $1$ although the geometric intersection has three points; the configuration is realized with the sphere as the attaching sphere of a $3$ -handle and the circle as the belt sphere of the $2$ -handle in the standard model.

[F3]

The oriented intersection number, The local oriented intersection sign and The mod 2 intersection number: the oriented number is the sum of local signs over the transverse intersection, and the mod-2 number is its cardinality modulo two.

[F4]

Geometric cancellation is a unit entry in the handle matrix and Handle cancellation: a single transverse point gives a unit entry, and only then does the cancellation theorem apply; a unit entry does not by itself supply a single geometric intersection point.

[F5]

The Axiom of Countable Choice ( ${A C}_{ω}$ ): ${A C}_{ω}$ is assumed; it is used through the intersection-number and cancellation suppliers.

Counterexample

Given: The configuration of the statement.

1.1F2given

The model $W = D^{4} \cup h^{2}$ has middle boundary $N ≅ S^{2} \times S^{1}$ with belt sphere $B = {p} \times S^{1}$ of the $2$ -handle; the sphere $A$ obtained by the finger move meets $B$ transversely in exactly three points with local signs $+ 1, + 1, - 1$ .

2.1F3step 1.1

By [F3] the oriented intersection number is the sum $1 + 1 - 1 = 1$ , a unit in $Z$ , and the mod-2 number is the cardinality $3 \equiv 1$ modulo $2$ , a unit in $Z_{2}$ ; the geometric intersection set has three points and the two spheres do not meet in one point.

3.1F1F4F5step 2.1∎

Reading the sphere as the attaching sphere of a $3$ -handle attached to $N$ and the circle as the belt sphere of the $2$ -handle, [F1] gives matrix entry $1$ : a unit entry, but not a geometrically cancelling configuration. Hence the attaching spheres do not satisfy the single-point hypothesis of [F4], the cancellation theorem does not apply, and no single-point conclusion follows from the matrix alone. This inserted finger pair can be removed by the inverse finger isotopy of [F2].

Depends on

Used by

Nothing in the library uses this result yet.

Dependency tree · two levels

41 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

John Milnor, Lectures on the h-Cobordism Theorem (notes by L. Siebenmann and J. Sondow; scanned edition with text layer) (standard reference, not scraped)
C. T. C. Wall, Differential Topology (Cambridge Studies in Advanced Mathematics 156; complete PDF) (standard reference, not scraped)