860: Identity Map from Subset of Euclidean C∞ Manifold or Closed Upper Half Euclidean C∞ Manifold with Boundary into Subset of Euclidean C∞ Manifold or Closed Upper Half Euclidean C∞ Manifold with Boundary Is C∞

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description/proof of that identity map from subset of Euclidean $C^{\mathrm{\infty }}$ manifold or closed upper half Euclidean $C^{\mathrm{\infty }}$ manifold with Boundary into subset of Euclidean $C^{\mathrm{\infty }}$ manifold or closed upper half Euclidean $C^{\mathrm{\infty }}$ manifold with boundary is $C^{\mathrm{\infty }}$

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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: Note
• 3: Proof

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Starting Context

• The reader knows a definition of Euclidean $C^{\mathrm{\infty }}$ manifold.
• The reader knows a definition of closed upper half Euclidean $C^{\mathrm{\infty }}$ manifold with boundary.
• The reader knows a definition of $C^k$ map between arbitrary subsets of $C^{\mathrm{\infty }}$ manifolds with boundary, where $k$ includes $\mathrm{\infty }$.

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Target Context

• The reader will have a description and a proof of the proposition that the identity map from any subset of any Euclidean $C^{\mathrm{\infty }}$ manifold or any closed upper half Euclidean $C^{\mathrm{\infty }}$ manifold with Boundary into any subset of the same-dimensional Euclidean $C^{\mathrm{\infty }}$ manifold or the same-dimensional closed upper half Euclidean $C^{\mathrm{\infty }}$ manifold with boundary is $C^{\mathrm{\infty }}$.

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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:
${\mathbb{R}}^d$: $=\text{ the Euclidean }{C}^{\mathrm{\infty }}\text{ manifold }$
${\mathbb{H}}^d$: $=\text{ the closed upper half Euclidean }{C}^{\mathrm{\infty }}\text{ manifold with boundary }$
$S_1$: $\subseteq {\mathbb{R}}^d\text{ or }{\mathbb{H}}^d$
$S_2$: $\subseteq {\mathbb{R}}^d\text{ or }{\mathbb{H}}^d$
$f$: $:S_1\to S_2,s\mapsto s$
//

Statements:
$f\in \{\text{ the }{C}^{\mathrm{\infty }}\text{ maps }\}$
//

$S_1$ and $S_2$ are such that $f$ is possible.

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2: Note

This proposition may seem obvious, but let us be ease at conscience by proving it once and for all, based on the definition of '$C^k$ map between arbitrary subsets of $C^{\mathrm{\infty }}$ manifolds with boundary, where $k$ includes $\mathrm{\infty }$'.

The point is that a subset may be of ${\mathbb{H}}^d$ instead of of ${\mathbb{R}}^d$ and $S_1$ or $S_2$ may be an arbitrary subset instead of an open subset.

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3: Proof

Whole Strategy: Step 1: suppose that $S_1\subseteq {\mathbb{R}}^d$ and $S_2\subseteq {\mathbb{R}}^d$, take the charts, $({\mathbb{R}}^d\subseteq {\mathbb{R}}^d,id)$ and $({\mathbb{R}}^d\subseteq {\mathbb{R}}^d,id)$, and see that $id\circ f\circ {id}^{-1}{|}_{id(S_1)}:id(S_1)\to id({\mathbb{R}}^d)$ is $C^{\mathrm{\infty }}$; Step 2: suppose that $S_1\subseteq {\mathbb{R}}^d$ and $S_2\subseteq {\mathbb{H}}^d$, take the charts, $({\mathbb{R}}^d\subseteq {\mathbb{R}}^d,id)$ and $({\mathbb{H}}^d\subseteq {\mathbb{H}}^d,id)$, and see that $id\circ f\circ {id}^{-1}{|}_{id(S_1)}:id(S_1)\to id({\mathbb{H}}^d)$ is $C^{\mathrm{\infty }}$; Step 3: suppose that $S_1\subseteq {\mathbb{H}}^d$ and $S_2\subseteq {\mathbb{R}}^d$, take the charts, $({\mathbb{H}}^d\subseteq {\mathbb{H}}^d,id)$ and $({\mathbb{R}}^d\subseteq {\mathbb{R}}^d,id)$, and see that $id\circ f\circ {id}^{-1}{|}_{id(S_1)}:id(S_1)\to id({\mathbb{R}}^d)$ is $C^{\mathrm{\infty }}$; Step 4: suppose that $S_1\subseteq {\mathbb{H}}^d$ and $S_2\subseteq {\mathbb{H}}^d$, take the charts, $({\mathbb{H}}^d\subseteq {\mathbb{H}}^d,id)$ and $({\mathbb{H}}^d\subseteq {\mathbb{H}}^d,id)$, and see that $id\circ f\circ {id}^{-1}{|}_{id(S_1)}:id(S_1)\to id({\mathbb{H}}^d)$ is $C^{\mathrm{\infty }}$.

While all the Steps are almost the same, let us do all of them diligently.

Step 1:

Let us suppose that $S_1\subseteq {\mathbb{R}}^d$ and $S_2\subseteq {\mathbb{R}}^d$.

Let $s\in S_1$ be any.

Let us take the chart around $s$, $({\mathbb{R}}^d\subseteq {\mathbb{R}}^d,id)$, and the chart around $f(s)$, $({\mathbb{R}}^d\subseteq {\mathbb{R}}^d,id)$.

Let us see that $id\circ f\circ i{d}^{-1}{|}_{id(S_1)}:id(S_1)\to id({\mathbb{R}}^d)$ is $C^{\mathrm{\infty }}$ at $id(s)$, which is the proposition is about by the definition of $C^k$ map between arbitrary subsets of $C^{\mathrm{\infty }}$ manifolds with boundary, where $k$ includes $\mathrm{\infty }$.

Let us take the open neighborhood of $id(s)$, $U_{id(s)}:={\mathbb{R}}^d\subseteq {\mathbb{R}}^d$ and the $C^{\mathrm{\infty }}$ extension of $id\circ f\circ i{d}^{-1}{|}_{id(S_1)}$, $f^{\prime }:{\mathbb{R}}^d\to {\mathbb{R}}^d,r\mapsto r$.

$f^{\prime }$ is indeed $C^{\mathrm{\infty }}$, because it is the identity map from ${\mathbb{R}}^d$ onto ${\mathbb{R}}^d$.

$f^{\prime }{|}_{id(S_1)}=id\circ f\circ i{d}^{-1}{|}_{id(S_1)}$, obviously.

So, $f$ is $C^{\mathrm{\infty }}$ at each $s\in S_1$, and so, $f$ is $C^{\mathrm{\infty }}$ all over.

Step 2:

Let us suppose that $S_1\subseteq {\mathbb{R}}^d$ and $S_2\subseteq {\mathbb{H}}^d$.

Let $s\in S_1$ be any.

Let us take the chart around $s$, $({\mathbb{R}}^d\subseteq {\mathbb{R}}^d,id)$, and the chart around $f(s)$, $({\mathbb{H}}^d\subseteq {\mathbb{H}}^d,id)$.

Let us see that $id\circ f\circ i{d}^{-1}{|}_{id(S_1)}:id(S_1)\to id({\mathbb{H}}^d)$ is $C^{\mathrm{\infty }}$ at $id(s)$, which is the proposition is about by the definition of $C^k$ map between arbitrary subsets of $C^{\mathrm{\infty }}$ manifolds with boundary, where $k$ includes $\mathrm{\infty }$.

Let us take the open neighborhood of $id(s)$, $U_{id(s)}:={\mathbb{R}}^d\subseteq {\mathbb{R}}^d$ and the $C^{\mathrm{\infty }}$ extension of $id\circ f\circ i{d}^{-1}{|}_{id(S_1)}$, $f^{\prime }:{\mathbb{R}}^d\to {\mathbb{R}}^d,r\mapsto r$.

$f^{\prime }$ is indeed $C^{\mathrm{\infty }}$, because it is the identity map from ${\mathbb{R}}^d$ onto ${\mathbb{R}}^d$.

$f^{\prime }{|}_{id(S_1)}=id\circ f\circ i{d}^{-1}{|}_{id(S_1)}$, obviously.

So, $f$ is $C^{\mathrm{\infty }}$ at each $s\in S_1$, and so, $f$ is $C^{\mathrm{\infty }}$ all over.

Step 3:

Let us suppose that $S_1\subseteq {\mathbb{H}}^d$ and $S_2\subseteq {\mathbb{R}}^d$.

Let $s\in S_1$ be any.

Let us take the chart around $s$, $({\mathbb{H}}^d\subseteq {\mathbb{H}}^d,id)$, and the chart around $f(s)$, $({\mathbb{R}}^d\subseteq {\mathbb{R}}^d,id)$.

Let us see that $id\circ f\circ i{d}^{-1}{|}_{id(S_1)}:id(S_1)\to id({\mathbb{R}}^d)$ is $C^{\mathrm{\infty }}$ at $id(s)$, which is the proposition is about by the definition of $C^k$ map between arbitrary subsets of $C^{\mathrm{\infty }}$ manifolds with boundary, where $k$ includes $\mathrm{\infty }$.

Let us take the open neighborhood of $id(s)$, $U_{id(s)}:={\mathbb{R}}^d\subseteq {\mathbb{R}}^d$ and the $C^{\mathrm{\infty }}$ extension of $id\circ f\circ i{d}^{-1}{|}_{id(S_1)}$, $f^{\prime }:{\mathbb{R}}^d\to {\mathbb{R}}^d,r\mapsto r$.

$f^{\prime }$ is indeed $C^{\mathrm{\infty }}$, because it is the identity map from ${\mathbb{R}}^d$ onto ${\mathbb{R}}^d$.

$f^{\prime }{|}_{id(S_1)}=id\circ f\circ i{d}^{-1}{|}_{id(S_1)}$, obviously.

So, $f$ is $C^{\mathrm{\infty }}$ at each $s\in S_1$, and so, $f$ is $C^{\mathrm{\infty }}$ all over.

Step 4:

Let us suppose that $S_1\subseteq {\mathbb{H}}^d$ and $S_2\subseteq {\mathbb{H}}^d$.

Let $s\in S_1$ be any.

Let us take the chart around $s$, $({\mathbb{H}}^d\subseteq {\mathbb{H}}^d,id)$, and the chart around $f(s)$, $({\mathbb{H}}^d\subseteq {\mathbb{H}}^d,id)$.

Let us see that $id\circ f\circ i{d}^{-1}{|}_{id(S_1)}:id(S_1)\to id({\mathbb{H}}^d)$ is $C^{\mathrm{\infty }}$ at $id(s)$, which is the proposition is about by the definition of $C^k$ map between arbitrary subsets of $C^{\mathrm{\infty }}$ manifolds with boundary, where $k$ includes $\mathrm{\infty }$.

Let us take the open neighborhood of $id(s)$, $U_{id(s)}:={\mathbb{R}}^d\subseteq {\mathbb{R}}^d$ and the $C^{\mathrm{\infty }}$ extension of $id\circ f\circ i{d}^{-1}{|}_{id(S_1)}$, $f^{\prime }:{\mathbb{R}}^d\to {\mathbb{R}}^d,r\mapsto r$.

$f^{\prime }$ is indeed $C^{\mathrm{\infty }}$, because it is the identity map from ${\mathbb{R}}^d$ onto ${\mathbb{R}}^d$.

$f^{\prime }{|}_{id(S_1)}=id\circ f\circ i{d}^{-1}{|}_{id(S_1)}$, obviously.

So, $f$ is $C^{\mathrm{\infty }}$ at each $s\in S_1$, and so, $f$ is $C^{\mathrm{\infty }}$ all over.

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References

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