830: For C∞ Manifold with Boundary and Open Submanifold with Boundary, Differential of Inclusion at Point on Open Submanifold with Boundary Is 'Vectors Spaces- Linear Morphisms' Isomorphism

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description/proof of that for $C^{\mathrm{\infty }}$ manifold with boundary and open submanifold with boundary, differential of inclusion at point on open submanifold with boundary is 'vectors spaces- linear morphisms' isomorphism

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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 open submanifold with boundary of $C^{\mathrm{\infty }}$ manifold with boundary.
• The reader knows a definition of differential of $C^{\mathrm{\infty }}$ map between $C^{\mathrm{\infty }}$ manifolds with boundary at point.
• The reader knows a definition of %category name% isomorphism.
• The reader admits the proposition that for any $C^{\mathrm{\infty }}$ function on any point open neighborhood of any $C^{\mathrm{\infty }}$ manifold with boundary, there exists a $C^{\mathrm{\infty }}$ function on the whole manifold with boundary that equals the original function on a possibly smaller neighborhood of the point.
• The reader admits the proposition that any bijective linear morphism is a 'vectors spaces - linear morphisms' isomorphism.

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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 any open submanifold with boundary, the differential of the inclusion at each point on the open submanifold with boundary is a 'vectors spaces - linear morphisms' isomorphism.

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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:
$M^{\prime }$: $\in \{\text{ the }d\text{ -dimensional }{C}^{\mathrm{\infty }}\text{ manifolds with boundary }\}$
$M$: $\in \{\text{ the open submanifolds with boundary of }{M}^{\prime }\}$
$\iota$: $:M\to M^{\prime }$, $=\text{ the inclusion }$
$m$: $\in M$
$d{\iota }_m$: $:T_{m}M\to T_m{M}^{\prime }$, $=\text{ the differential }$
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Statements:
$d{\iota }_m\in \{\text{ the 'vectors spaces - linear morphisms' isomorphisms }\}$
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2: Proof

Whole Strategy: Step 1: see that $d{\iota }_m$ is injective, by seeing that $d{\iota }_m(v_1)=d{\iota }_m(v_2)$ implies that $v_1=v_2$; Step 2: see that $d{\iota }_m$ is surjective, by seeing that for each $v^{\prime }\in T_m{M}^{\prime }$, $v\in T_{m}M$ as that for each $f\in C^{\mathrm{\infty }}(M)$, $vf=v^{\prime }\bar{f}$ is mapped to $v^{\prime }$, where $\bar{f}$ is any "extension" of $f$ on $M^{\prime }$; Step 3: conclude the proposition.

Step 1:

Let us see that $d{\iota }_m$ is injective.

Let $v_1,v_2\in T_{m}M$ be any such that $v_1\ne v_2$.

Let us suppose that $d{\iota }_m(v_1)=d{\iota }_m(v_2)$.

$d{\iota }_m(v_1-v_2)=0$.

Let $f\in C^{\mathrm{\infty }}(M)$ be any.

$f$ would be $C^{\mathrm{\infty }}$ also as a function with $M$ regarded as the subset of $M^{\prime }$ (instead of as the submanifold): each chart of $M$ is also a chart of $M^{\prime }$.

By the proposition that for any $C^{\mathrm{\infty }}$ function on any point open neighborhood of any $C^{\mathrm{\infty }}$ manifold with boundary, there exists a $C^{\mathrm{\infty }}$ function on the whole manifold with boundary that equals the original function on a possibly smaller neighborhood of the point, there would be an open neighborhood of $m$, $U_m\subseteq M^{\prime }$, such that $U_m\subseteq \iota (M)$, and a $C^{\mathrm{\infty }}$ function, $\bar{f}:M^{\prime }\to \mathbb{R}$, such that $f{|}_{U_m}=\bar{f}{|}_{U_m}$.

$v_1(\bar{f}\circ \iota )=d{\iota }_m(v_1)\bar{f}=d{\iota }_m(v_2)\bar{f}=v_2(\bar{f}\circ \iota )$.

But $\bar{f}\circ \iota {|}_{U_m}=f{|}_{U_m}$.

So, $v_j(\bar{f}\circ \iota )=v_{j}f$, because any tangent vector is a local operator.

So, $v_{1}f=v_1(\bar{f}\circ \iota )=v_2(\bar{f}\circ \iota )=v_{2}f$, a contradiction.

So, $d{\iota }_m(v_1)\ne d{\iota }_m(v_2)$.

Step 2:

Let us see that $d{\iota }_m$ is surjective.

Let $v^{\prime }\in T_m{M}^{\prime }$ be any.

Let us define $v\in T_{m}M$ as that for each $f\in C^{\mathrm{\infty }}(M)$, $vf=v^{\prime }\bar{f}$, where $\bar{f}\in C^{\mathrm{\infty }}(M^{\prime })$ is gotten from $f$ as above.

Let us see that $v$ is well-defined. An $\bar{f}$ exists as before. $v^{\prime }\bar{f}$ does not depend on the choice of $\bar{f}$, because $v^{\prime }$ is a local operator.

Let us see that $v$ is indeed a derivation.

$v(rf)=v^{\prime }\overline{rf}=v^{\prime }(r\bar{f})=r{v}^{\prime }\bar{f}=rvf$. $v(f_1{f}_2)=v^{\prime }(\overline{f_1{f}_2})=v^{\prime }(\overline{f_1}\overline{f_2})=(v^{\prime }\overline{f_1})\overline{f_2}(m)+\overline{f_1}(m)v^{\prime }\overline{f_2}=(v{f}_1)f_2(m)+f_1(m)v{f}_2$: although $\overline{f_1}$ and $\overline{f_2}$ may have some different open neighborhoods of $m$, the intersection of the open neighborhoods is an open neighborhood of $m$ on which $\overline{f_1}\overline{f_2}$ equals $f_1{f}_2$.

So, $v$ is indeed a derivation.

Let us see that $d{\iota }_{m}v=v^{\prime }$.

For each $f^{\prime }\in C^{\mathrm{\infty }}(M^{\prime })$, $d{\iota }_{m}v(f^{\prime })=v(f^{\prime }\circ \iota )=v^{\prime }\overline{f^{\prime }\circ \iota }$, but $f^{\prime }$ is a $C^{\mathrm{\infty }}$ extension of $f^{\prime }\circ \iota$, so, $=v^{\prime }{f}^{\prime }$.

So, $d{\iota }_{m}v=v^{\prime }$.

Step 3:

$d{\iota }_m$ is a 'vectors spaces - linear morphisms' isomorphism, by the proposition that any bijective linear morphism is a 'vectors spaces - linear morphisms' isomorphism.

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References

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