805: For C∞ Manifold with Boundary and Chart, Restriction of Chart on Open Subset Domain Is Chart

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description/proof of that for $C^{\mathrm{\infty }}$ manifold with boundary and chart, restriction of chart on open subset domain is chart

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Topics

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

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The table of contents of this article

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

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Starting Context

• The reader knows a definition of $C^{\mathrm{\infty }}$ manifold with boundary.
• The reader admits the proposition that for any topological space and any topological subspace that is open on the base space, any subset of the subspace is open on the subspace if and only if it is open on the base space.
• The reader admits the proposition that any restriction of any continuous map on the domain and the codomain is continuous.

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Target Context

• The reader will have a description and a proof of the proposition that for any $C^{\mathrm{\infty }}$ manifold with boundary and its any chart, the restriction of the chart on any open subset domain is a chart.

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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.

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Main Body

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1: Structured Description

Here is the rules of Structured Description.

Entities:
$M$: $\in \{\text{ the }d\text{ -dimensional }{C}^{\mathrm{\infty }}\text{ manifolds with boundary }\}$
$(U\subseteq M,\varphi )$: $\in \{\text{ the charts of }M\}$
$U^{\prime }$: $\in \{\text{ the open subsets of }U\}$
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Statements:
$(U^{\prime }\subseteq M,\varphi {|}_{U^{\prime }})\in \{\text{ the charts of }M\}$
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$U^{\prime }\subseteq U$ is an open subset of $U$ if and only if it is an open subset of $M$, because $U$ is an open subspace of $M$, by the proposition that for any topological space and any topological subspace that is open on the base space, any subset of the subspace is open on the subspace if and only if it is open on the base space.

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2: Proof

Whole Strategy: Step 1: see that $U^{\prime }\subseteq M$ and $\varphi {|}_{U^{\prime }}(U^{\prime })\subseteq {\mathbb{R}}^d\text{ or }{\mathbb{H}}^d$ are open and $\varphi {|}_{U^{\prime }}:U^{\prime }\to \varphi (U^{\prime })$ is homeomorphic; Step 2: see that $(U^{\prime }\subseteq M,\varphi {|}_{U^{\prime }})$ and $(U\subseteq M,\varphi )$ are $C^{\mathrm{\infty }}$ compatible.

Step 1:

$U^{\prime }$ is open on $M$, by the proposition that for any topological space and any topological subspace that is open on the base space, any subset of the subspace is open on the subspace if and only if it is open on the base space.

Let us see that $\varphi {|}_{U^{\prime }}(U^{\prime })\subseteq {\mathbb{R}}^d\text{ or }{\mathbb{H}}^d$ is open on ${\mathbb{R}}^d$ or ${\mathbb{H}}^d$.

As $\varphi$ is homeomorphic, $\varphi {|}_{U^{\prime }}(U^{\prime })=\varphi (U^{\prime })\subseteq \varphi (U)$ is open on $\varphi (U)$. As $\varphi (U)$ is an open subspace of ${\mathbb{R}}^d$ or ${\mathbb{H}}^d$, $\varphi {|}_{U^{\prime }}(U^{\prime })$ is open on ${\mathbb{R}}^d$ or ${\mathbb{H}}^d$, by the proposition that for any topological space and any topological subspace that is open on the base space, any subset of the subspace is open on the subspace if and only if it is open on the base space.

Let us see that $\varphi {|}_{U^{\prime }}:U^{\prime }\to \varphi (U^{\prime })$ is homeomorphic.

As $\varphi$ is homeomorphic, $\varphi {|}_{U^{\prime }}$ is homeomorphic, by the proposition that any restriction of any continuous map on the domain and the codomain is continuous.

Step 2:

Let us see that $(U^{\prime }\subseteq M,\varphi {|}_{U^{\prime }})$ and $(U\subseteq M,\varphi )$ are $C^{\mathrm{\infty }}$ compatible.

The transition map, $\varphi \circ {\varphi {|}_{U^{\prime }}}^{-1}:\varphi {|}_{U^{\prime }}(U^{\prime })\to \varphi (U^{\prime })$, is $C^{\mathrm{\infty }}$, because it is $id$.

The transition map, $\varphi {|}_{U^{\prime }}\circ {\varphi {|}_{U^{\prime }}}^{-1}:\varphi {|}_{U^{\prime }}(U^{\prime })\to \varphi {|}_{U^{\prime }}(U^{\prime })$, is $C^{\mathrm{\infty }}$, because it is $id$.

So, $(U^{\prime }\subseteq M,\varphi {|}_{U^{\prime }})$ and $(U\subseteq M,\varphi )$ are indeed $C^{\mathrm{\infty }}$ compatible.

So, $(U^{\prime }\subseteq M,\varphi {|}_{U^{\prime }})$ is in the maximal atlas.

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References

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