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.
Parametric transversality
Statement
Assume the Axiom of Countable Choice (The Axiom of Countable Choice ()). Let be a smooth family of maps and let be an embedded submanifold. If , then the set of parameters for which the slice fails to be transverse to is a null subset of .
Facts & Assumptions
Given: and a smooth family transverse to an embedded submanifold .
The slice maps come from the evaluation map (Smooth families of maps and their evaluation maps).
Transversality to an embedded submanifold means a tangent-space spanning condition at each point of the preimage (A smooth map transverse to an embedded submanifold).
The preimage is an embedded submanifold, and under the critical values of the projection form a null set (The transverse preimage theorem, Morse-Sard for smooth manifolds).
Proof
Because , [L1] makes an embedded submanifold. Let be the restriction of the second projection.
Fix and write . The fibre of over is which identifies with by [F1]. A pair lies in exactly when , so is surjective exactly when every admits some with .
If at , then [F2] gives For any , choose so that . Then , so step 2.1 makes a submersion at .
Conversely, assume is a submersion at . Given , the transversality of in [F2] gives and with . By step 2.1 choose with , so . Then which is exactly the transversality condition for at .
Therefore is a regular value of if and only if the slice is transverse to at every point of its fibre. Applying the Sard statement in [L1] to shows that the bad parameters form a null subset of .