PropositionStatement: AI-adaptedProof: AI-generatedprecheck 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 of a bundle map is equivalent to smooth local matrices

Statement

Let $Φ : E \to F$ be a fibrewise linear map over a smooth base map $f : M \to N$ . Choose local frames $(e_{1}, \dots, e_{r})$ for $E$ on $U \subseteq M$ and $(u_{1}, \dots, u_{s})$ for $F$ on $V \subseteq N$ with $f (U) \subseteq V$ . Then $Φ$ is smooth on $E ∣_{U}$ if and only if there are smooth scalar functions $a_{j i} : U \to R$ such that

$Φ (e_{i} (p)) = \sum_{j = 1}^{s} a_{j i} (p) u_{j} (f (p))$

for every $p \in U$ .

Facts & Assumptions

Given: A fibrewise linear map $Φ : E \to F$ over a smooth map $f : M \to N$ and local frames on $U$ and $V$ as above.

[L1]

Local frames are equivalent to local trivializations (Local frames and local trivializations are equivalent data).

[L2]

A section is smooth exactly when its local components are smooth (Smoothness of a section is equivalent to smooth local components).

Proof

technique · direct

1.1

By [L1], the chosen frames identify $E ∣_{U}$ with $U \times R^{r}$ and $F ∣_{V}$ with $V \times R^{s}$ . In these trivializations, fibrewise linearity forces $Φ$ to have the form $Φ (p, v) = (f (p), A (p) v)$ for a unique matrix $A (p) = (a_{j i} (p))$ .

L1given

2.1

The local representative $(p, v) \mapsto (f (p), A (p) v)$ is smooth exactly when its matrix entries $a_{j i}$ are smooth on $U$ . Equivalently, the images $Φ (e_{i}) = \sum_{j} a_{j i} u_{j} \circ f$ have smooth local components, which is the criterion in [L2].

L2step 1.1algebra∎

Depends on

Used by

A continuous fibrewise linear map over a smooth base map is automatically smooth False statement
A fibrewise bijective smooth bundle map over a diffeomorphism is a bundle isomorphism Proposition
Constant-rank kernels and images of bundle maps over one base are subbundles Proposition
Sections of Hom are the same as smooth fibrewise linear maps Proposition

Dependency tree · two levels

13 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

John M. Lee, Introduction to Smooth Manifolds (standard reference, not scraped)
Will J. Merry, Differential Geometry (standard reference, not scraped)