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 zero handles create components

Statement

Assume ${A C}_{ω}$ . A $0$ -handle on a smooth $n$ -manifold with boundary attaches along the empty set and adds one disjoint $n$ -disk component. This includes an empty starting manifold and $n = 0$ .

Facts & Assumptions

[F1]

Attaching a smooth handle with corner rounding: Assume ${A C}_{ω}$ . Let $X$ be a smooth $n$ -manifold with boundary, and let $k$ be an integer with $0 \leq k \leq n$ . 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.

1.1F1algebra

At $k = 0$ the handle is $D^{0} \times D^{n} ≅ D^{n}$ and its attaching region is $S^{- 1} \times D^{n} = \emptyset$ . Thus the defining quotient makes no identifications between the old manifold and the disk.

2.1step 1.1algebra∎

The quotient is their disjoint union. The disk is connected and nonempty, so it supplies exactly one new component, including when the old manifold is empty. At $n = 0$ the new disk is a single point. There is no corner seam to round.

Depends on

Attaching a smooth handle with corner rounding

Used by

An empty incoming boundary requires zero handles Example
Dual handle presentations of a genus-g surface Example
Reordering independent one-handles Example
Sublevels of height on the sphere Example
Connected cobordisms admit presentations without superfluous zero handles Proposition
Handle duality from negating a Morse function Theorem

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)