PropositionStatement: Literature-sourcedProof: Literature-sourcedPipeline-generatedprecheck pass

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.

Connected cobordisms admit presentations without superfluous zero handles

Statement

Assume ${A C}_{ω}$ . Let $(W; M_{0}, M_{1})$ be a nonempty compact connected triad with $M_{0} \neq \emptyset$ connected. Then $W$ admits a handle decomposition relative to $M_{0}$ with no $0$ -handles. If $M_{0}$ has $k$ components the same argument leaves no $0$ -handles and begins with $k - 1$ connecting $1$ -handles; if $M_{0} = \emptyset$ , exactly the $0$ -handles needed to create the components of $W$ remain, one per component.

Facts & Assumptions

[F1]

Adapted excellent Morse functions exist on compact cobordisms supplies an initial adapted excellent pair. Self-indexing Morse functions exist and Morse functions and handle decompositions correspond: Assume ${A C}_{ω}$ . An adapted Morse function on the compact triad may be taken self-indexed, so that all critical points of index $k$ lie at one level and the levels increase with $k$ ; the associated decomposition has all $0$ -handles attached at the first level, then all $1$ -handles at the next, and so on, one handle per critical point.

[F2]

Index zero handles create components: a $0$ -handle attaches along the empty set and adds one disjoint $n$ -disk component.

[F3]

Handles of equal index can be attached on one level: if a common critical level contains no critical point of another index, handles of equal index attached at one level may be regarded as attached simultaneously or successively in any order, with the same result up to diffeomorphism relative to the lower stage; in particular the $1$ -handles may be reordered freely.

[F4]

A one-handle between distinct manifold components is a boundary connected sum: a $1$ -handle whose attaching disks lie in two distinct manifold components of the lower stage performs their boundary connected sum.

[F5]

Boundary connected sum with a disk does not change the diffeomorphism type: for a connected manifold $N$ with nonempty boundary and an embedded disk $D \subseteq \partial N$ , the boundary connected sum $N ♮ D^{n}$ is diffeomorphic to $N$ by a diffeomorphism equal to the identity outside a collar neighbourhood of $D$ .

[F6]

Handle decomposition relative to the incoming boundary: a decomposition relative to $M_{0}$ is an ordered list of handles attached to the successive stages, starting from the collar $M_{0} \times [0, ε]$ .

[F7]

Collar neighborhood theorem: Assume ${A C}_{ω}$ . Every smooth manifold with boundary has a smooth collar.

[F8]

The Axiom of Countable Choice ( ${A C}_{ω}$ ): ${A C}_{ω}$ : every at most countable family of nonempty sets has a choice function.

[A1]

Connectivity of the first stages. After the $0$ -handles and $1$ -handles have been attached, the stage $W_{1}$ is connected whenever $W$ is connected: a handle of index $k \geq 2$ attaches along $S^{k - 1} \times D^{n - k}$ , whose first factor is a sphere of dimension $k - 1 \geq 1$ and hence connected, so such a handle cannot join two components; the handle bodies themselves are connected. Consequently the graph whose vertices are the components of the collar $⊔$ $0$ -handles and whose edges are the $1$ -handles is connected, and it has a spanning tree.

Proof

Given: The nonempty compact connected triad $(W; M_{0}, M_{1})$ , a self-indexed presentation of [F1], and the resulting graph of [A1].

1.1F1F2F6givenalgebra

Present $W$ with all $0$ -handles first and all $1$ -handles next, by [F1]. Let $m_{0}$ be the number of $0$ -handles and let $k$ be the number of components of $M_{0}$ ( $k = 0$ if $M_{0} = \emptyset$ ). After the $0$ -handle stage the first stage is the disjoint union of the $k$ collar components $M_{0}^{(1)} \times [0, ε], \dots, M_{0}^{(k)} \times [0, ε]$ and $m_{0}$ disjoint $n$ -disks, hence it has $k + m_{0}$ components; by [F2] each disk is created by exactly one $0$ -handle.

2.1F3A1step 1.1choose

By [A1] the graph $G$ of components and $1$ -handles is connected; choose a spanning tree $T$ of $G$ . Reorder the $1$ -handles by [F3] so that the tree edges come first, in an order in which each new edge joins the component accumulated so far to one further component of the first stage; this is possible by rooting the tree at one vertex and listing the edges in the order in which a breadth-first search reaches new vertices.

3.1F4F5F8step 2.1construct

Process the tree edges in that order. Each processed $1$ -handle joins two distinct components and realizes a boundary connected sum of them by [F4]. If one of the two components is a new bare $0$ -handle, that is the $n$ -disk $D^{n}$ created by that handle, then the summand $D^{n}$ is absorbed by [F5]: the stage is diffeomorphic to the other component, so the $0$ -handle and this $1$ -handle may be deleted from the presentation without changing the diffeomorphism type relative to $M_{0}$ . If neither component is a bare $0$ -handle, the sum is a genuine boundary connected sum of two components, each either a collar component or a component already obtained by a previous sum; such a step is retained. Applying the two cases successively to all edges of the tree makes the first stage connected, and the retained edges are exactly the $1$ -handles that join two components neither of which is a bare $0$ -handle.

4.1F4F5F6step 2.1step 3.1algebra

If $k > 0$ , root the tree at a collar vertex. Every bare disk is then a new vertex reached from the accumulated component and is absorbed with its incident tree edge. Thus precisely $m_{0}$ zero-handles and $m_{0}$ tree edges disappear, leaving $(k + m_{0} - 1) - m_{0} = k - 1$ connecting edges between collar components. If $k = 0$ , root at one disk and retain it; precisely the other $m_{0} - 1$ disks and tree edges are absorbed, leaving one zero-handle and no tree edge. Each absorption fixes the incoming face; transport every remaining attaching map through the stage diffeomorphism rather than assuming its support is disjoint from later handles.

5.1F2F3F4step 4.1algebra

The three cases. If $M_{0} \neq \emptyset$ is connected, then $k = 1$ , all $m_{0}$ disks are deleted, no edge is retained, and the resulting decomposition of $W$ relative to $M_{0}$ has no $0$ -handles. If $M_{0}$ has $k \geq 2$ components, all $m_{0}$ disks are deleted, no $0$ -handle remains, and after reordering the retained $1$ -handles first (legal by [F3]) the decomposition begins with $k - 1$ $1$ -handles that connect the $k$ collar components into one boundary sum. If $M_{0} = \emptyset$ , then $k = 0$ and every component of the first stage is a bare disk: every processed edge deletes one disk, so $m_{0} - 1$ disks disappear and exactly one $0$ -handle remains, which is the number of components of the connected manifold $W$ ; the claim that exactly the needed $0$ -handles remain is the statement $m_{0} - (m_{0} - 1) = 1$ .

6.1F1F2F6F7step 5.1algebra∎

Conclusion. In every case the presentation obtained by deleting the absorbed $0$ -handles and $1$ -handles and reordering the retained $1$ -handles is a handle decomposition of $W$ relative to $M_{0}$ (still with one handle per remaining handle body, all indices preserved) with the asserted number of $0$ -handles, and with $k - 1$ connecting $1$ -handles coming first when $M_{0}$ has $k$ components. This is Wall's normalization of a presentation relative to a nonempty incoming boundary; the argument uses the collar, handle and absorption suppliers, and through them ${A C}_{ω}$ .

Depends on

Used by

Dependency tree · two levels

55 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

C. T. C. Wall, Differential Topology (Cambridge Studies in Advanced Mathematics 156), Sections 5.1-5.4, printed pp. 129-148 (standard reference, not scraped)
John Milnor, Lectures on the h-Cobordism Theorem (notes by L. Siebenmann and J. Sondow), Sections 2-4, printed pp. 10-48 (standard reference, not scraped)