CorollaryStatement: AI-adaptedProof: AI-adaptedPipeline-generatedjudge pass (gpt-5.6-terra)audited 2026-09-07

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.

Index n handles cap boundary spheres

Statement

An $n$ -handle attaches along its whole $S^{n - 1}$ boundary. For $n \geq 2$ it fills a boundary component diffeomorphic to $S^{n - 1}$ . For $n = 1$ its attaching $S^{0}$ is a pair of boundary points, possibly in different components. For $n = 0$ it is the same disjoint point attachment as a $0$ -handle.

Facts & Assumptions

[F1]

Attaching a smooth handle with corner rounding: Assume ${A C}_{ω}$ . Let $X$ be a smooth $n$ -manifold with boundary. Attach the handle of def-k-handle-core-cocore-attaching-region-and-belt-sphere by a smooth embedding $h : S^{k - 1} \times D^{n - k} \to \partial X$ that extends to a neighborhood of the disk factor. Form the quotient of $X ⊔ (D^{k} \times D^{n - k})$ identifying $z$ with $h (z)$ in the attaching region. The disk coordinates trivialize the normal bundle of the attaching sphere; this framing is part of the data. Use collars from thm-collar-neighborhood-theorem to give the seam its product smooth charts, then round the compact codimension-two corner. A compatible rounding is a smooth monotone planar profile, transverse to a common diagonal direction, agreeing with the two faces away from a small corner neighborhood. In coordinates along that diagonal it is a graph. This convention fixes the gluing and collar data; changing the attaching embedding is a different question. There is no corner to round when $k = 0$ or $k = n$ .

Proof

Given: The objects and hypotheses in the statement.

1.1

For $k = n$ the disk factor $D^{n - k} = D^{0}$ is a point, so the attaching region is all of $S^{n - 1} = \partial D^{n}$ . For $n \geq 2$ , its smooth embedding into $\partial X$ is locally a diffeomorphism (the dimensions agree and its derivative is injective). Its image is open and is also closed by compactness. Since $S^{n - 1}$ is connected, that image is one boundary component. Gluing the disk fills it.

F1algebra

2.1

For $n = 1$ , $S^{0}$ has two points, whose images are two distinct boundary points; nothing forces them to lie in the same component of $X$ . For $n = 0$ , $S^{- 1}$ is empty and the attached $D^{0}$ is a new isolated point. Thus the connected-sphere formulation is restricted exactly as stated.

F1step 1.1algebra∎

Depends on

Attaching a smooth handle with corner rounding

Used by

Sublevels of height on the sphere Example

Dependency tree · two levels

3 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

Benedetti, Lectures on Differential Topology (standard reference, not scraped)