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

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Constant-rank kernels and images of bundle maps over one base are subbundles

Statement

Let $Φ : E \to F$ be a smooth vector bundle map over ${id}_{M}$ , and assume that the fibre rank of $Φ_{p} : E_{p} \to F_{p}$ is the same integer $k$ for every $p \in M$ . Then $\ker Φ$ is a smooth vector subbundle of $E$ and $im Φ$ is a smooth vector subbundle of $F$ .

Facts & Assumptions

Given: A smooth bundle map $Φ : E \to F$ over ${id}_{M}$ with constant fibre rank $k$ .

[L1]

In local frames, $Φ$ is represented by a smooth matrix-valued function (Smoothness of a bundle map is equivalent to smooth local matrices).

[L2]

A smooth matrix-valued map has smooth inverse matrix entries wherever its determinant never vanishes (Matrix inversion preserves $C^{k}$ regularity where the determinant is nonzero).

Proof

technique · direct

1.1L1given

If $k = 0$ , then every fibre map is zero, so $\ker Φ = E$ and $im Φ$ is the zero subbundle of $F$ . Assume now that $k \geq 1$ , fix $p \in M$ , and write $Φ$ in local frames near $p$ as a smooth matrix $A (x)$ . Because $rank A (p) = k$ , after reordering coordinates some $k \times k$ minor $B (x)$ is nonzero at $p$ , hence nonzero on a smaller neighborhood.

2.1L2step 1.1algebra

Writing source coordinates as $(u, w)$ for that split, the kernel equation becomes $B (x) u + C (x) w = 0$ , so [L2] makes $B (x)^{- 1}$ smooth and gives $u = - B (x)^{- 1} C (x) w$ . Therefore the kernel fibres are spanned by smooth local sections depending on the free variables $w$ .

3.1L1step 1.1step 2.1algebra∎

The same chosen $k$ columns of $A (x)$ remain linearly independent nearby, so they form a smooth local frame of the image bundle. Thus in the positive-rank case both the kernel and the image are locally spanned by part of a frame, and step 1.1 already handled $k = 0$ . Therefore $\ker Φ$ and $im Φ$ are smooth vector subbundles.

Depends on

Used by

Every short exact sequence of smooth vector bundles splits Corollary
A rank-jumping kernel is not a vector subbundle Counterexample
The graph of a bundle map as a subbundle of a Whitney sum Example
Normal bundle of the zero locus of a transverse section Lemma
The normal bundle of the diagonal is canonically the tangent bundle Lemma
The Euler class of an oriented even-rank normal bundle controls self-intersection Proposition
The pullback foliation under a transverse map Proposition
Symmetric and alternating images are smooth subbundles Theorem

Dependency tree · two levels

32 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)