PropositionStatement: AI-adaptedProof: AI-generatedprecheck passjudge pass (gpt-5.6-terra)audited 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.

The canonical map to a quotient bundle is a smooth bundle map

Statement

If $S \subseteq E$ is a smooth vector subbundle, then the fibrewise quotient map $q : E \to E / S$ is a smooth vector bundle map over ${id}_{M}$ , and its kernel is $S$ .

Facts & Assumptions

Given: A smooth vector bundle $E \to M$ and a smooth subbundle $S \subseteq E$ .

[L1]

The quotient $E / S \to M$ is a smooth vector bundle (A vector bundle quotient by a subbundle is a smooth vector bundle).

Proof

technique · direct

1.1L1given

In an adapted local frame $(e_{1}, \dots, e_{r})$ with $S$ spanned by the first $k$ vectors, the quotient bundle chart from [L1] identifies $q$ with the map $(p, u, w) \mapsto (p, w)$ , where $u \in R^{k}$ and $w \in R^{r - k}$ . Thus $q$ is smooth and fibrewise linear.

2.1step 1.1algebra∎

In the same coordinates, $q (p, u, w) = 0$ exactly when $w = 0$ , which means that the vector lies in the span of $e_{1}, \dots, e_{k}$ , namely in $S_{p}$ . Therefore $\ker q = S$ .

The canonical map to a quotient bundle is a smooth bundle map

Statement

Facts & Assumptions

Proof

Depends on

Dependency tree · 2 levels

Sources