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PropositionStatement: AI-adaptedProof: AI-generatedprecheck passaudited 2026-08-31
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A fibrewise bijective smooth bundle map over a diffeomorphism is a bundle isomorphism

Statement

Let Φ:EF be a smooth vector bundle map over a diffeomorphism f:MN. If each fibre map Φp:EpFf(p) is bijective, then Φ is a vector bundle isomorphism.

Facts & Assumptions

Given: A smooth bundle map Φ:EF over a diffeomorphism f:MN, with each Φp bijective.

[L1]

In local frames, smooth bundle maps are given by smooth matrix-valued functions (Smoothness of a bundle map is equivalent to smooth local matrices).

[L2]

A real square matrix is invertible exactly when its determinant is nonzero (A finite square real matrix is invertible if and only if its determinant is nonzero).

[L3]

A smooth matrix-valued map has smooth inverse matrix entries wherever its determinant never vanishes (Matrix inversion preserves Ck regularity where the determinant is nonzero).

Proof

technique · direct
1.1

If the common fibre rank is 0, then every fibre is the zero vector space, so Φ is already the unique smooth bundle map between zero bundles over f and hence a bundle isomorphism. Otherwise choose local frames so that on one trivializing neighborhood, Φ(p,v)=(f(p),A(p)v). Fibrewise bijectivity means that each matrix A(p) is invertible, so [L2] gives detA(p)0 for every p.

L1L2given
2.1

In the positive-rank case, [L3] makes the entries of A1 smooth on the same neighborhood. Thus the local inverse is (q,w)(f1(q),A(f1(q))1w), which is smooth because f1 is smooth. These local inverses agree on overlaps, so Φ is a smooth bundle isomorphism. Together with the rank-0 branch of step 1.1, this proves the proposition.

L3step 1.1algebra

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