784: For C∞ Manifold, Regular Domain, C∞ Manifold with Boundary, and C∞ Map from Regular Domain into C∞ Manifold with Boundary, Corresponding Map with Domain Regarded as Subset of Manifold Is C∞

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description/proof of that for $C^{\mathrm{\infty }}$ manifold, regular domain, $C^{\mathrm{\infty }}$ manifold with boundary, and $C^{\mathrm{\infty }}$ map from regular domain into $C^{\mathrm{\infty }}$ manifold with boundary, corresponding map with domain regarded as subset of manifold 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: Natural Language Description
• 3: Note
• 4: 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 $C^k$ map between arbitrary subsets of $C^{\mathrm{\infty }}$ manifolds with boundary, where $k$ includes $\mathrm{\infty }$.
• The reader admits the proposition that for any $C^{\mathrm{\infty }}$ manifold, any subset, and any point on the subset, if a chart satisfies the local slice condition for embedded submanifold or the local slice condition for embedded submanifold with boundary, its any sub-open-neighborhood does so.

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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, its any regular domain, any $C^{\mathrm{\infty }}$ manifold with boundary, and any $C^{\mathrm{\infty }}$ map from the regular domain into the $C^{\mathrm{\infty }}$ manifold with boundary, the corresponding map with the domain regarded as the subset of the manifold 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:
$M_1^{\prime }$: $\in \{\text{ the }{d}^{\prime }\text{ -dimensional }{C}^{\mathrm{\infty }}\text{ manifolds }\}$
$M_1$: $\in \{\text{ the regular domains of }{M}_1^{\prime }\}$
$M_2$: $\in \{\text{ the }{C}^{\mathrm{\infty }}\text{ manifolds with boundary }\}$
$f$: $:M_1\to M_2$, $\in \{\text{ the }{C}^{\mathrm{\infty }}\text{ maps }\}$
$S$: $=M_1$, as the subset of $M_1^{\prime }$
$f^{\prime }$: $:S\to M_2,s\mapsto f(s)$
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Statements:
$f^{\prime }\in \{\text{ the }{C}^{\mathrm{\infty }}\text{ maps }\}$
//

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2: Natural Language Description

For any $d^{\prime }$-dimensional $C^{\mathrm{\infty }}$ manifold, $M_1^{\prime }$, any regular domain of $M_1^{\prime }$, $M_1$, any $C^{\mathrm{\infty }}$ manifold with boundary, $M_2$, any $C^{\mathrm{\infty }}$ map, $f:M_1\to M_2$, the subset of $M_1^{\prime }$, $S=M_1$, and the map, $f^{\prime }:S\to M_2,s\mapsto f(s)$, $f^{\prime }$ is $C^{\mathrm{\infty }}$.

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3: Note

Although $f$ and $f^{\prime }$ map each point to the same point, they are not the same map: whether $f$ is $C^{\mathrm{\infty }}$ or not depends on the atlas of $M_1$, while whether $f^{\prime }$ is $C^{\mathrm{\infty }}$ or not depends on the atlas of $M_1^{\prime }$, so, the proposition is not trivial.

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4: Proof

Whole Strategy: Step 1: for each $m\in M_1$, take an adopted chart, $(U_m^{\prime }\subseteq M_1^{\prime },{\varphi }_m^{\prime })$, for $M_1$, take any chart, $(U_{f(m)}\subseteq M_2,{\varphi }_{f(m)})$, take an open neighborhood of $m$, $V_m=V_m^{\prime }\cap M_1\subseteq M_1$, such that $f(V_m)\subseteq U_{f(m)}$, and take the adopted chart, $(U_m^{\prime }\cap V_m^{\prime }\subseteq M_1^{\prime },{\varphi }_m^{\prime }{|}_{U_m^{\prime }\cap V_m^{\prime }})$, and the corresponding adopting chart, $(U_m^{\prime }\cap V_m^{\prime }\cap M_1\subseteq M_1,{\pi }_J\circ {\varphi }_m^{\prime }{|}_{U_m^{\prime }\cap V_m^{\prime }\cap M_1})$; Step 2: see what $f$'s being $C^{\mathrm{\infty }}$ at $m$ means with respect to $(U_m:=U_m^{\prime }\cap V_m^{\prime }\cap M\subseteq M_1,{\varphi }_m:={\pi }_J\circ {\varphi }_m^{\prime }{|}_{U_m})$; Step 3: see what $f^{\prime }$' being $C^{\mathrm{\infty }}$ at $m$ means with respect to $(U_m^{\prime }\cap V_m^{\prime }\subseteq M_1^{\prime },{\varphi }_m^{\prime }{|}_{U_m^{\prime }\cap V_m^{\prime }})$; Step 4: conclude the proposition.

Step 1:

Let $m\in M_1$ be any.

As $M_1$ is an embedded submanifold with boundary of $M_1^{\prime }$, there is an adopted chart, $(U_m^{\prime }\subseteq M_1^{\prime },{\varphi }_m^{\prime })$, for $M_1$.

Let us take any chart, $(U_{f(m)}\subseteq M_2,{\varphi }_{f(m)})$.

As $f$ is continuous, there is an open neighborhood of $m$, $V_m\subseteq M_1$, such that $f(V_m)\subseteq U_{f(m)}$. As $M_1$ is a topological subspace of $M_1^{\prime }$, $V_m=V_m^{\prime }\cap M_1$, where $V_m^{\prime }\subseteq M_1^{\prime }$ is an open neighborhood of $m$ on $M_1^{\prime }$.

$(U_m^{\prime }\cap V_m^{\prime }\subseteq M_1^{\prime },{\varphi }_m^{\prime }{|}_{U_m^{\prime }\cap V_m^{\prime }})$ is an adopted chart, by the proposition that for any $C^{\mathrm{\infty }}$ manifold, any subset, and any point on the subset, if a chart satisfies the local slice condition for embedded submanifold or the local slice condition for embedded submanifold with boundary, its any sub-open-neighborhood does so.

The corresponding adopting chart is $(U_m:=U_m^{\prime }\cap V_m^{\prime }\cap M_1\subseteq M_1,{\varphi }_m:={\pi }_J\circ {\varphi }_m^{\prime }{|}_{U_m})$.

Step 2:

$f(U_m^{\prime }\cap V_m^{\prime }\cap M_1)\subseteq f(V_m^{\prime }\cap M_1)=f(V_m)\subseteq U_{f(m)}$.

So, by the definition of $C^k$ map between arbitrary subsets of $C^{\mathrm{\infty }}$ manifolds with boundary, where $k$ includes $\mathrm{\infty }$, ${\varphi }_{f(m)}\circ f\circ {{\varphi }_m}^{-1}:{\varphi }_m(U_m)\to {\varphi }_{f(m)}(U_{f(m)})$ is $C^{\mathrm{\infty }}$ at ${\varphi }_m(m)$.

Step 3:

$f^{\prime }(U_m^{\prime }\cap V_m^{\prime }\cap S)=f(U_m^{\prime }\cap V_m^{\prime }\cap M_1)\subseteq U_{f(m)}$.

By the definition of $C^k$ map between arbitrary subsets of $C^{\mathrm{\infty }}$ manifolds with boundary, where $k$ includes $\mathrm{\infty }$, $f^{\prime }$ is $C^{\mathrm{\infty }}$, if ${\varphi }_{f(m)}\circ f^{\prime }\circ {{\varphi }_m^{\prime }{|}_{U_m^{\prime }\cap V_m^{\prime }}}^{-1}{|}_{{\varphi }_m^{\prime }{|}_{U_m^{\prime }\cap V_m^{\prime }}(U_m^{\prime }\cap V_m^{\prime }\cap S)}:{\varphi }_m^{\prime }{|}_{U_m^{\prime }\cap V_m^{\prime }}(U_m^{\prime }\cap V_m^{\prime }\cap S)\to {\varphi }_{f(m)}(U_{f(m)})$ is $C^{\mathrm{\infty }}$ at ${\varphi }_m^{\prime }{|}_{U_m^{\prime }\cap V_m^{\prime }}(m)={\varphi }_m^{\prime }(m)$.

Step 4:

${\varphi }_m(U_m)={\pi }_J\circ {\varphi }_m^{\prime }{|}_{U_m^{\prime }\cap V_m^{\prime }\cap M_1}(U_m^{\prime }\cap V_m^{\prime }\cap M_1)={\varphi }_m^{\prime }{|}_{U_m^{\prime }\cap V_m^{\prime }\cap M_1}(U_m^{\prime }\cap V_m^{\prime }\cap M_1)$, because as $M_1$ is a regular domain, which means that the codimension of $M_1$ is $0$, ${\pi }_J$ does nothing.

On the other hand, ${\varphi }_m^{\prime }{|}_{U_m^{\prime }\cap V_m^{\prime }}(U_m^{\prime }\cap V_m^{\prime }\cap S)={\varphi }_m^{\prime }{|}_{U_m^{\prime }\cap V_m^{\prime }\cap M_1}(U_m^{\prime }\cap V_m^{\prime }\cap M_1)$.

So, in fact, the domain of ${\varphi }_{f(m)}\circ f\circ {{\varphi }_m}^{-1}$ and the domain of ${\varphi }_{f(m)}\circ f^{\prime }\circ {{\varphi }_m^{\prime }{|}_{U_m^{\prime }\cap V_m^{\prime }}}^{-1}{|}_{{\varphi }_m^{\prime }{|}_{U_m^{\prime }\cap V_m^{\prime }}(U_m^{\prime }\cap V_m^{\prime }\cap S)}$ are the same, and the 2 maps are the same.

So, ${\varphi }_{f(m)}\circ f^{\prime }\circ {{\varphi }_m^{\prime }{|}_{U_m^{\prime }\cap V_m^{\prime }}}^{-1}{|}_{{\varphi }_m^{\prime }{|}_{U_m^{\prime }\cap V_m^{\prime }}(U_m^{\prime }\cap V_m^{\prime }\cap S)}$ is $C^{\mathrm{\infty }}$ at ${\varphi }_m^{\prime }(m)$, as ${\varphi }_{f(m)}\circ f\circ {{\varphi }_m}^{-1}$ is $C^{\mathrm{\infty }}$ at ${\varphi }_m(m)={\varphi }_m^{\prime }(m)$.

So, $f^{\prime }$ is $C^{\mathrm{\infty }}$.

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References

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