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PropositionStatement: AI-adaptedProof: AI-generatedprecheck passaudited 2026-08-31
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Bundle maps over f are sections of the pulled-back Hom bundle

Statement

Let Φ:E→F be a smooth vector bundle map over a smooth map f:M→N. Then Φ is naturally equivalent to a smooth section of the pulled-back Hom bundle Hom⁡(E,f∗F)→M.

Facts & Assumptions

Given: Smooth vector bundles E→M, F→N, a smooth map f:M→N, and a bundle map Φ:E→F over f.

[L1]

The pullback fibre product f∗F is a smooth vector bundle over M (The pullback fibre product is a smooth vector bundle).

[L2]

Sections of a Hom bundle are the same as fibrewise linear bundle maps over the identity (Sections of Hom are the same as smooth fibrewise linear maps).

Proof

technique · direct
1.1L1givenconstruct

For each p∈M, the fibre map Φp:Ep→Ff(p) may be viewed as a linear map Ep→(f∗F)p because (f∗F)p is canonically Ff(p). Hence p↦Φp defines a section of Hom⁡(E,f∗F)→M.

2.1L1L2step 1.1algebra∎

In local trivializations the matrix of this section is exactly the local matrix of Φ, so the section is smooth exactly when Φ is smooth. Therefore [L2] applied to the bundles E and f∗F yields the required bijection.

Depends on

Used by

Dependency tree · two levels

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Sources