LemmaStatement: Literature-sourcedProof: AI-generatedprecheck passjudge pass (gpt-5.6-terra)audited 2026-09-01

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.

A tubular target produces a submersive finite-dimensional perturbation family

Statement

Let $f : M \to N$ be smooth. Then there exist an open ball

B \subseteq R^{m}

containing $0$ and a smooth family of maps

F : M \times B \to N

such that $F_{0} = f$ and, for every $p \in M$ , the parameter map

F_{p} : B \to N, a ⟼ F (p, a),

is a submersion. In particular, the evaluation map $F$ is a submersion, so it is transverse to every closed embedded submanifold $Z \subseteq N$ .

Facts & Assumptions

Given: A smooth map $f : M \to N$ .

[F1]

The standard transversality-family construction provides an open ball $B \subseteq R^{m}$ and a smooth map $F : M \times B \to N$ with $F (p, 0) = f (p)$ and such that, for each fixed $p \in M$ , the map $a \mapsto F (p, a)$ is a submersion.

[F2]

A smooth family of maps is its evaluation map on a product manifold (Smooth families of maps and their evaluation maps).

[L2]

Every submersion is transverse to every embedded submanifold (A submersion is transverse to every embedded submanifold).

Proof

technique · direct

1.1

By [F2], the data of [F1] is exactly a smooth family of maps with $F_{0} = f$ .

F1F2given

2.1

Because each parameter map $F_{p} : B \to N$ is a submersion, the full evaluation map $F$ is a submersion as well. Therefore [L2] shows that it is transverse to every closed embedded submanifold of $N$ .

L2step 1.1∎

Depends on

Used by

Relative transversality preserves a map on a closed good region Proposition
The transversality homotopy theorem Theorem

Dependency tree · two levels

6 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

Marco Gualtieri, Topology I: Smooth Manifolds, Part 10, Corollary 3.27 (standard reference, not scraped)