PropositionStatement: Literature-sourcedProof: AI-generatedprecheck passaudited 2026-08-29

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Chart maps are diffeomorphisms onto Euclidean open sets

Statement

Let $(M, S)$ be a smooth manifold and let $(U, φ)$ be a smooth chart. Give $U$ the restricted smooth structure of the open submanifold and $φ (U)$ its standard smooth structure. Then the corestriction $φ : U \to φ (U)$ is a diffeomorphism.

Facts & Assumptions

Given: A smooth manifold $(M, S)$ and a smooth chart $(U, φ)$ .

[F1]

A smooth chart is a chart of the maximal atlas, hence a homeomorphism of the open set $U$ onto the open subset $φ (U) \subseteq R^{n}$ (Smooth manifolds and their smooth charts, Manifold charts, coordinate domains, and coordinate functions).

[F2]

The open subset $φ (U)$ carries the standard smooth structure generated by the identity chart (Open subsets of Euclidean space have the standard smooth structure), and the open subset $U$ carries the restricted structure of $M$ (An open subset of a smooth manifold has a canonical restricted smooth structure).

[F3]

A map between smooth manifolds is smooth when its representative with respect to suitable smooth charts is smooth ( $C^{r}$ and smooth maps between smooth manifolds).

[F4]

A diffeomorphism is a bijective smooth map with smooth inverse (Diffeomorphisms and local diffeomorphisms of manifolds).

[A1]

The identity map of an open subset of $R^{n}$ is smooth, each iterated coordinate derivative being a constant function.

Proof

technique · direct

1.1givenF1

By [F1] the corestriction $φ : U \to φ (U)$ is a bijective [given, F1] homeomorphism.

1.2givenF2F3A1

$φ$ is smooth: with the smooth chart $(U, φ)$ of $U$ from [F2] and the identity chart of $φ (U)$ from [F2], the representative is $i d \circ φ \circ φ^{- 1} = {i d}_{φ (U)}$ , smooth by [A1], so [F3] applies.

1.3givenF2F3A1

$φ^{- 1}$ is smooth: with the identity chart of $φ (U)$ and the [given, F2, F3, A1] chart $(U, φ)$ of $U$ , the representative is $φ \circ φ^{- 1} \circ i d = {i d}_{φ (U)}$ , again smooth by [A1], so [F3] applies.

2.1F4step 1.1step 1.2step 1.3∎

Steps 1.1-1.3 give a bijective smooth map with smooth inverse, which is [F4, step 1.1, step 1.2, step 1.3] exactly a diffeomorphism by [F4].

Chart maps are diffeomorphisms onto Euclidean open sets

Statement

Facts & Assumptions

Proof

Depends on

Dependency tree · 2 levels

Sources