Alphabeta Math
PropositionStatement: Literature-sourcedProof: AI-adaptedprecheck passaudited 2026-08-31
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.

Smoothness into an embedded submanifold is an initial property

Statement

Let SM be an embedded submanifold with inclusion i:SM, and let G:NS be a map from a smooth manifold N. Then G is smooth if and only if iG:NM is smooth.

Facts & Assumptions

Given: An embedded submanifold SM, its inclusion i, and a map G:NS.

[F1]

Embedded submanifolds are locally cut out by slice charts (Embedded submanifolds and slice charts).

[L1]

The restricted slice charts define the smooth structure on S (Slice-chart restrictions form a smooth atlas).

[L2]

Ambient charts are diffeomorphisms onto open Euclidean sets (Chart maps are diffeomorphisms onto Euclidean open sets).

Proof

technique · direct
1.1

Assume first that G is smooth. Choose a chart θ on N and a slice chart φ on M around G(x)S. In the corresponding restricted chart on S from [L1], the representative of G has values in Rk, and the representative of iG is obtained by appending mk zero coordinates. Hence iG is smooth.

F1L1L2given
1.2

Conversely, assume iG is smooth. In a slice chart on M, the image of S is Rk×{0} by [F1]. Therefore the representative of iG has last mk coordinates identically zero, and its first k coordinates are exactly the representative of G in the restricted slice chart from [L1]. Those first k coordinates are smooth, so G is smooth.

F1L1L2given
2.1

Steps 1.1 and 1.2 prove both directions.

step 1.1step 1.2

Depends on

Used by

Dependency tree · two levels

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Sources