PropositionStatement: Literature-sourcedProof: AI-generatedprecheck passaudited 2026-09-05

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Local diffeomorphisms carry distributions and integral manifolds

Statement

Let $F : M \to N$ be a local diffeomorphism, let $D$ be a smooth distribution on $M$ , and let $U \subseteq M$ be open such that $F ∣_{U} : U \to V : = F (U)$ is a diffeomorphism. Then:

the family $(F_{*} D)_{F (p)} : = d F_{p} (D_{p}) (p \in U)$ is a smooth distribution on $V$ , and
if $i : P \to U$ is an integral manifold of $D ∣_{U}$ , then $F \circ i : P \to V$ is an integral manifold of $F_{*} D$ .

Facts & Assumptions

Given: A local diffeomorphism $F : M \to N$ , a smooth distribution $D$ on $M$ , and an open set $U$ on which $F$ is a diffeomorphism onto $V$ .

[A1]

Let $i : P \to U$ be an integral manifold of $D ∣_{U}$ .

Proof

technique · direct

1.1

Because $F ∣_{U}$ is a diffeomorphism, its differential identifies [given] $T U \to T V$ fibrewise by linear isomorphisms. Transporting the rank- $k$ subbundle $D ∣_{U}$ through those isomorphisms yields a rank- $k$ smooth subbundle of $T V$ , namely $F_{*} D$ .

given

1.2

The composite $F \circ i$ is an injective immersion, because both factors [given] are immersions and $F ∣_{U}$ is injective. For each $q \in P$ , $d (F \circ i)_{q} (T_{q} P) = d F_{i (q)} (d i_{q} (T_{q} P)) = d F_{i (q)} (D_{i (q)}) = (F_{*} D)_{F (i (q))} .$ Hence $F \circ i$ is an integral manifold of the transported distribution.

givenalgebra

2.1

Therefore local diffeomorphisms preserve the regular-distribution and [given] integral-manifold structure on any neighborhood where they are genuine diffeomorphisms.

given∎

Depends on

Used by

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Dependency tree · two levels

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Sources

Will J. Merry, Differential Geometry (standard reference, not scraped)