405: Map from Open Subset of C^\infty Manifold onto Open Subset of Euclidean C^\infty Manifold Is Chart Map iff It Is Diffeomorphism

<The previous article in this series | The table of contents of this series | The next article in this series>

A description/proof of that map from open subset of $C^{\mathrm{\infty }}$ manifold onto open subset of Euclidean $C^{\mathrm{\infty }}$ manifold is chart map iff it is diffeomorphism

---

Topics

About: $C^{\mathrm{\infty }}$ manifold

---

The table of contents of this article

• Starting Context
• Target Context
• Orientation
• Main Body
• 1: Description
• 2: Proof
• 3: Note

---

Starting Context

• The reader knows a definition of $C^{\mathrm{\infty }}$ manifold.
• The reader knows a definition of diffeomorphism.

---

Target Context

• The reader will have a description and a proof of the proposition that for any $C^{\mathrm{\infty }}$ manifold, any map from any open subset of the manifold onto any open subset of the corresponding-dimensional Euclidean $C^{\mathrm{\infty }}$ manifold is a chart map if and only if it is a diffeomorphism.

---

Orientation

There is a list of definitions discussed so far in this site.

There is a list of propositions discussed so far in this site.

---

Main Body

---

1: Description

For any $C^{\mathrm{\infty }}$ manifold, $M$, any open subset, $U\subseteq M$, any open subset of the corresponding-dimensional Euclidean $C^{\mathrm{\infty }}$ manifold, $U^{\prime }\subseteq {\mathbb{R}}^d$, and any map, $f:U\to U^{\prime }$, $(U,f)$ is a chart if and only if $f$ is a diffeomorphism.

---

2: Proof

Let us suppose that $(U,f)$ is a chart. $f$ is a homeomorphism, by the definition of chart. Around any point, $p\in U$, there is the chart, $(U,f)$, and around $f(p)\in U^{\prime }$, there is the chart on the Euclidean $C^{\mathrm{\infty }}$ manifold, $(U^{\prime },id)$ where $id$ is the identity map, $id:U^{\prime }\to U^{\prime }$. $id\circ f\circ f^{-1}$ is $C^{\mathrm{\infty }}$, because it is $id$, so, $f$ is $C^{\mathrm{\infty }}$. $f\circ f^{-1}\circ i{d}^{-1}$ is $C^{\mathrm{\infty }}$, because it is $id$, so, $f^{-1}$ is $C^{\mathrm{\infty }}$.

Let us suppose that $f$ is a diffeomorphism. $f$ is a homeomorphism. For any chart, $(U^{″},\varphi )$, such that $U^{″}\cap U\ne \mathrm{\varnothing }$, $f\circ {\varphi }^{-1}{|}_{\varphi (U^{″}\cap U)}$ is $C^{\mathrm{\infty }}$ as a composition of $C^{\mathrm{\infty }}$ maps, and $\varphi \circ f^{-1}{|}_{f(U^{″}\cap U)}$ is $C^{\mathrm{\infty }}$ as a composition of $C^{\mathrm{\infty }}$ maps. So, $(U,f)$ is compatible with $(U^{″},\varphi )$, and $(U,f)$ is included in the maximal atlas.

---

3: Note

Talking of "corresponding-dimensional", in fact, there cannot be any diffeomorphism onto any open subset of any different-dimensional Euclidean $C^{\mathrm{\infty }}$ manifold, by the $C^{\mathrm{\infty }}$ invariance of dimension theorem, so, if there is a diffeomorphism onto an open subset of the Euclidean $C^{\mathrm{\infty }}$ manifold of a dimension, that dimension is nothing but the corresponding dimension.

---

References

<The previous article in this series | The table of contents of this series | The next article in this series>