PropositionStatement: AI-adaptedProof: AI-generatedprecheck passaudited 2026-08-31

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Literature-sourced: the exact statement appears in a cited source; only wording and notation differ.
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A vector bundle projection is a surjective submersion

Statement

If $π : E \to M$ is a smooth vector bundle, then $π$ is a surjective submersion.

Facts & Assumptions

Given: A smooth vector bundle $π : E \to M$ .

[L1]

A smooth vector bundle is, in particular, a smooth fibre bundle with local trivializations over the identity on the base (Smooth vector bundles, rank, fibres, and trivial bundles).

[L2]

A smooth map is a submersion exactly when its differential is surjective at every point (Immersions, submersions, and constant-rank maps).

Proof

technique · direct

1.1L1given

Surjectivity is part of the definition of a smooth fibre bundle, so $π$ is surjective. Fix $e \in E$ with $p = π (e)$ , and choose a local trivialization $Φ : π^{- 1} (U) \to U \times R^{r}$ around $e$ . In this chart, $π$ becomes the product projection ${pr}_{1} : U \times R^{r} \to U$ .

2.1L2step 1.1algebra∎

In product coordinates the differential of ${pr}_{1}$ is the coordinate projection $T_{p} U \oplus T_{v} R^{r} \to T_{p} U$ , which is surjective. Therefore $d π_{e}$ is surjective, and since $e$ was arbitrary, $π$ is a submersion by [L2].

Depends on

Used by

Nothing in the library uses this result yet.

Dependency tree · two levels

19 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)