781: For C∞ Manifold with Boundary and Regular Domain, Differential of Inclusion at Point on Regular Domain Is 'Vectors Spaces - Linear Morphisms' Isomorphism

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description/proof of that for $C^{\mathrm{\infty }}$ manifold with boundary and regular domain, differential of inclusion at point on regular domain 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: Natural Language Description
• 3: Proof

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Starting Context

• The reader knows a definition of regular domain 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 $C^{\mathrm{\infty }}$ manifold with boundary, any embedded submanifold with boundary of the manifold with boundary is properly embedded if and only if it is closed.
• The reader admits the proposition that for any $C^{\mathrm{\infty }}$ manifold with boundary, any closed subset, and any open neighborhood of the subset, any $C^{\mathrm{\infty }}$ map from the subset into ${\mathbb{R}}^n$ can be extended to the whole space supported in the neighborhood.
• The reader admits the proposition that any tangent vector at any point on any $C^{\mathrm{\infty }}$ manifold with boundary is the velocity of a $C^{\mathrm{\infty }}$ curve, especially from a half closed interval, especially as linear in coordinates.
• The reader admits the proposition that any bijective linear map 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 regular domain, the differential of the inclusion at each point on the regular domain 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:
$M^{\prime }$: $\in \{\text{ the }d\text{ -dimensional }{C}^{\mathrm{\infty }}\text{ manifolds with boundary }\}$
$M$: $\in \{\text{ the regular domains of }{M}^{\prime }\}$
$\iota :$: $:M\to M^{\prime }$, $=\text{ the inclusion }$
$m$: $\in M$
$d{\iota }_m$: $:T_{m}M\to T_m{M}^{\prime }$, $=\text{ the differential }$
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Statements:
$d{\iota }_m\in \{\text{ the 'vectors spaces - linear morphisms' isomorphisms }\}$
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2: Natural Language Description

For any $d$-dimensional $C^{\mathrm{\infty }}$ manifold with boundary, $M^{\prime }$, any regular domain, $M$, the inclusion, $\iota :M\to M^{\prime }$, any $m\in M$, and the differential, $d{\iota }_m:T_{m}M\to T_m{M}^{\prime }$, $d{\iota }_m$ is a 'vectors spaces - linear morphisms' isomorphism.

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3: 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 $vf=v^{\prime }\tilde{f}$ is mapped to $v^{\prime }$, where $\tilde{f}$ 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. As $M$ was closed on $M^{\prime }$ by the proposition that for any $C^{\mathrm{\infty }}$ manifold with boundary, any embedded submanifold of the manifold is properly embedded if and only if it is closed, there would be a $C^{\mathrm{\infty }}$ extension of $f$ on $M^{\prime }$, $\tilde{f}\in C^{\mathrm{\infty }}(M^{\prime })$, such that $f=\tilde{f}{|}_M$, by the proposition that for any $C^{\mathrm{\infty }}$ manifold with boundary, any closed subset, and any open neighborhood of the subset, any $C^{\mathrm{\infty }}$ map from the subset into ${\mathbb{R}}^n$ can be extended to the whole space supported in the neighborhood. $(v_1-v_2)f=(v_1-v_2)\tilde{f}{|}_M=(v_1-v_2)\tilde{f}\circ \iota =d{\iota }_m(v_1-v_2)\tilde{f}=0$, which would imply that $v_1-v_2=0$, so, $v_1=v_2$, 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 $f\in C^{\mathrm{\infty }}(M)$, $vf=v^{\prime }\tilde{f}$, where $\tilde{f}\in C^{\mathrm{\infty }}(M^{\prime })$ is any extension of $f$.

Let us see that $v$ is well-defined. An $\tilde{f}$ exists as before. As $M$ is an embedded submanifold with boundary of $M^{\prime }$, let us take an adopted chart, $(U_m^{\prime }\subseteq M^{\prime },{\varphi }_m^{\prime })$. $v^{\prime }=v^{\prime j}\partial /\partial x^{\prime j}$ is the velocity of a $C^{\mathrm{\infty }}$ curve, $\gamma :J\to M^{\prime },t\mapsto {{\varphi }_m^{\prime }}^{-1}(({\varphi }_m^{\prime }(m{)}^1+v^{\prime 1}(t-t_0),...,{\varphi }_m^{\prime }(m{)}^d+v^{\prime d}(t-t_0)))$, where $J=[t_0,t_2)\text{ or }(t_1,t_0]$ according to $0\le v^{\prime d}$ or $v^{\prime d}<0$, by the proposition that any tangent vector at any point on any $C^{\mathrm{\infty }}$ manifold with boundary is the velocity of a $C^{\mathrm{\infty }}$ curve, especially from a half closed interval, especially as linear in coordinates. As $0\le v^{\prime d}(t-t_0)$, which means that $\gamma$ is into $M$, $v^{\prime }\tilde{f}=d\gamma (d/dt{|}_{t_0})\tilde{f}=d/dt{|}_{t_0}(\tilde{f}\circ \gamma )$, which is the one-sided derivative, $=d/dt{|}_{t_0}(\tilde{f}{|}_M\circ \gamma )=d/dt{|}_{t_0}(f\circ \gamma )$, which does not depend on the choice of $\tilde{f}$.

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

$v(rf)=v^{\prime }\widetilde{rf}=v^{\prime }(r\tilde{f})=r{v}^{\prime }\tilde{f}=rvf$. $v(f_1{f}_2)=v^{\prime }(\widetilde{f_1{f}_2})=v^{\prime }(\widetilde{f_1}\widetilde{f_2})=(v^{\prime }\widetilde{f_1})\widetilde{f_2}(m)+\widetilde{f_1}(m)v^{\prime }\widetilde{f_2}=(v{f}_1)f_2(m)+f_1(m)v{f}_2$.

So, $v$ is indeed a derivation.

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

For each $f^{\prime }\in C^{\mathrm{\infty }}(M^{\prime })$, $d{\iota }_{m}v(f^{\prime })=v(f^{\prime }\circ \iota )=v^{\prime }\widetilde{f^{\prime }\circ \iota }$, 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 map is a 'vectors spaces - linear morphisms' isomorphism.

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References

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