705: For Euclidean C∞ Manifold, Open Ball Is Diffeomorphic to Whole Space

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description/proof of that for Euclidean $C^{\mathrm{\infty }}$ manifold, open ball is diffeomorphic to whole space

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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: Natural Language Description
• 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 diffeomorphism between arbitrary subsets of $C^{\mathrm{\infty }}$ manifolds with boundary.
• The reader admits the proposition that for any open subset of the $d_1$-dimensional Euclidean $C^{\mathrm{\infty }}$ manifold, any $C^{\mathrm{\infty }}$ map into the $d_2$-dimensional Euclidean $C^{\mathrm{\infty }}$ manifold divided by any never-zero $C^{\mathrm{\infty }}$ map into the 1-dimensional Euclidean $C^{\mathrm{\infty }}$ manifold is $C^{\mathrm{\infty }}$.

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Target Context

• The reader will have a description and a proof of the proposition that for any Euclidean $C^{\mathrm{\infty }}$ manifold, any open ball is diffeomorphic to the whole space.

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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:
${\mathbb{R}}^d$: $=\text{ the Euclidean }{C}^{\mathrm{\infty }}\text{ manifold }$
$B_{p,ϵ}$: $=\text{ the open ball around }p\text{ on }{\mathbb{R}}^d$
$f$: $:B_{p,ϵ}\to {\mathbb{R}}^d,p^{\prime }\mapsto (p^{\prime }-p)/\sqrt{{ϵ}^2-\Vert p^{\prime }-p{\Vert }^2}$
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Statements:
$B_{p,ϵ}{\cong }_f{\mathbb{R}}^d$, where ${\cong }_f$ denotes being diffeomorphic by $f$
//

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2: Natural Language Description

For the Euclidean $C^{\mathrm{\infty }}$ manifold, ${\mathbb{R}}^d$, any open ball around any $p\in {\mathbb{R}}^d$, $B_{p,ϵ}\subseteq {\mathbb{R}}^d$, is diffeomorphic to ${\mathbb{R}}^d$ by $f:B_{p,ϵ}\to {\mathbb{R}}^d,p^{\prime }\mapsto (p^{\prime }-p)/\sqrt{{ϵ}^2-\Vert p^{\prime }-p{\Vert }^2}$.

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3: Proof

Whole Strategy: Step 1: see that $f$ is bijective; Step 2: get the inverse of $f$; Step 3: see that $f$ is $C^{\mathrm{\infty }}$; Step 4: see that $f^{-1}$ is $C^{\mathrm{\infty }}$.

$f$ is well-defined, because $0<{ϵ}^2-\Vert p^{\prime }-p{\Vert }^2$ on $B_{p,ϵ}$.

Step 1:

Let us see that $f$ is injective.

Let $p^{\prime },p^{″}\in B_{p,ϵ}$ be any such that $p^{\prime }\ne p^{″}$.

There are 2 cases: Case 1: $\Vert p^{\prime }-p\Vert \ne \Vert p^{″}-p\Vert$; Case 2: $\Vert p^{\prime }-p\Vert =\Vert p^{″}-p\Vert$.

Let us think of Case 1.

$\Vert (p^{\prime }-p)/\sqrt{{ϵ}^2-\Vert p^{\prime }-p{\Vert }^2}{\Vert }^2=\Vert (p^{\prime }-p){\Vert }^2/({ϵ}^2-\Vert p^{\prime }-p{\Vert }^2)=1/({ϵ}^2/\Vert p^{\prime }-p{\Vert }^2-1)$. Likewise, $\Vert (p^{″}-p)/\sqrt{{ϵ}^2-\Vert p^{″}-p{\Vert }^2}{\Vert }^2=1/({ϵ}^2/\Vert p^{″}-p{\Vert }^2-1)$. Obviously, $1/({ϵ}^2/\Vert p^{\prime }-p{\Vert }^2-1)\ne 1/({ϵ}^2/\Vert p^{″}-p{\Vert }^2-1)$, which means that $\Vert f(p^{\prime }){\Vert }^2\ne \Vert f(p^{″}){\Vert }^2$, which implies that $f(p^{\prime })\ne f(p^{″})$.

Let us think of Case 2.

The directions of $p^{\prime }-p$ and $p^{″}-p$ are different. So, the directions of $f(p^{\prime })$ and $f(p^{″})$ are different. So, $f(p^{\prime })\ne f(p^{″})$.

So, $f$ is injective.

Let us see that $f$ is surjective.

Let $p^{″}\in {\mathbb{R}}^d$ be any. Let us see that there is a $p^{\prime }\in B_{p,ϵ}$ such that $f(p^{\prime })=p^{″}$.

Let $p^{\prime }:=(p^{″}/\sqrt{1+\Vert p^{″}{\Vert }^2})ϵ+p$. $f(p^{\prime })=(p^{″}/\sqrt{1+\Vert p^{″}{\Vert }^2})ϵ/\sqrt{{ϵ}^2-\Vert (p^{″}/\sqrt{1+\Vert p^{″}{\Vert }^2})ϵ{\Vert }^2}=p^{″}/\sqrt{1+\Vert p^{″}{\Vert }^2}ϵ/\sqrt{{ϵ}^2-(\Vert p^{″}{\Vert }^2/(1+\Vert p^{″}{\Vert }^2)){ϵ}^2}=p^{″}/\sqrt{1+\Vert p^{″}{\Vert }^2}ϵ/\sqrt{({ϵ}^2+{ϵ}^2\Vert p^{″}{\Vert }^2-{ϵ}^2\Vert p^{″}{\Vert }^2)/(1+\Vert p^{″}{\Vert }^2)}=p^{″}/\sqrt{1+\Vert p^{″}{\Vert }^2}ϵ/\sqrt{{ϵ}^2/(1+\Vert p^{″}{\Vert }^2)}=p^{″}$.

Step 2:

By Step 1, $f^{-1}:p^{″}\mapsto (p^{″}/\sqrt{1+\Vert p^{″}{\Vert }^2})ϵ+p$.

Step 3:

Let us see that $f$ is $C^{\mathrm{\infty }}$.

$p^{\prime }-p$ is obviously $C^{\mathrm{\infty }}$.

${ϵ}^2-\Vert p^{\prime }-p{\Vert }^2$ is obviously $C^{\mathrm{\infty }}$.

$g:x\mapsto \sqrt{x}$ is $C^{\mathrm{\infty }}$ on $\mathbb{R}/\{0\}$, because $dg/dx=1/2x^{-1/2}$, $d^{2}g/d{x}^2=1/2(-1/2)x^{-3/2}$, ..., while $x\ne 0$.

So, the composition, $\sqrt{{ϵ}^2-\Vert p^{\prime }-p{\Vert }^2}$, is $C^{\mathrm{\infty }}$.

So, $f$ is $C^{\mathrm{\infty }}$, by the proposition that for any open subset of the $d_1$-dimensional Euclidean $C^{\mathrm{\infty }}$ manifold, any $C^{\mathrm{\infty }}$ map into the $d_2$-dimensional Euclidean $C^{\mathrm{\infty }}$ manifold divided by any never-zero $C^{\mathrm{\infty }}$ map into the 1-dimensional Euclidean $C^{\mathrm{\infty }}$ manifold is $C^{\mathrm{\infty }}$.

Step 4:

Let us see that $f^{-1}$ is $C^{\mathrm{\infty }}$.

$p^{″}ϵ$ is obviously $C^{\mathrm{\infty }}$.

$1+\Vert p^{″}{\Vert }^2$ is obviously $C^{\mathrm{\infty }}$.

$g:x\mapsto \sqrt{x}$ is $C^{\mathrm{\infty }}$ on $\mathbb{R}/\{0\}$ as before.

So, the composition, $\sqrt{1+\Vert p^{″}{\Vert }^2}$, is $C^{\mathrm{\infty }}$.

$p^{″}/\sqrt{1+\Vert p^{″}{\Vert }^2}ϵ$ is $C^{\mathrm{\infty }}$, by the proposition that for any open subset of the $d_1$-dimensional Euclidean $C^{\mathrm{\infty }}$ manifold, any $C^{\mathrm{\infty }}$ map into the $d_2$-dimensional Euclidean $C^{\mathrm{\infty }}$ manifold divided by any never-zero $C^{\mathrm{\infty }}$ map into the 1-dimensional Euclidean $C^{\mathrm{\infty }}$ manifold is $C^{\mathrm{\infty }}$.

$f^{-1}$ is obviously $C^{\mathrm{\infty }}$.

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References

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