Alphabeta Math
PropositionStatement: AI-adaptedProof: AI-generatedprecheck passaudited 2026-08-31
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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.

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A vector bundle projection is a surjective submersion

Statement

If π:EM is a smooth vector bundle, then π is a surjective submersion.

Facts & Assumptions

Given: A smooth vector bundle π:EM.

[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
1.1

Surjectivity is part of the definition of a smooth fibre bundle, so π is surjective. Fix eE with p=π(e), and choose a local trivialization Φ:π1(U)U×Rr around e. In this chart, π becomes the product projection pr1:U×RrU.

L1given
2.1

In product coordinates the differential of pr1 is the coordinate projection TpUTvRrTpU, which is surjective. Therefore dπe is surjective, and since e was arbitrary, π is a submersion by [L2].

L2step 1.1algebra

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