873: Restricted C∞ Vectors Bundle w.r.t. Embedded Submanifold with Boundary Is Embedded Submanifold with Boundary

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description/proof of that restricted $c^{\mathrm{\infty }}$ vectors bundle w.r.t. embedded submanifold with boundary is embedded submanifold with boundary

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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 restricted (C^\infty\) vectors bundle.
• The reader knows a definition of embedded submanifold with boundary of $C^{\mathrm{\infty }}$ manifold with boundary.
• The reader knows the definition of open submanifold with boundary of $C^{\mathrm{\infty }}$ manifold with boundary.
• The reader admits the proposition that for any topological space contained in any ambient topological space, if the space is ambient-space-wise locally topological subspace of the ambient space, the space is the topological subspace of the space.
• The reader admits the proposition that for any $C^{\mathrm{\infty }}$ manifold with boundary and its any embedded submanifold with boundary, around each point on the submanifold with boundary, there is a trivializing open subset for the manifold with boundary whose intersection with the submanifold with boundary is a chart domain for the submanifold with boundary.
• The reader admits the proposition that for any $C^{\mathrm{\infty }}$ manifold with boundary, any embedded submanifold with boundary of any embedded submanifold with boundary is an embedded submanifold with boundary of the manifold with boundary.
• The reader admits the proposition that for any map, the map preimage of any intersection of sets is the intersection of the map preimages of the sets.
• The reader admits the proposition that any restriction of any continuous map on the domain and the codomain is continuous.

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Target Context

• The reader will have a description and a proof of the proposition that for any $C^{\mathrm{\infty }}$ vectors bundle, the restricted $C^{\mathrm{\infty }}$ vectors bundle w.r.t. any embedded submanifold with boundary is an embedded submanifold with boundary.

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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:
$(E^{\prime },M^{\prime },{\pi }^{\prime })$: $\in \{\text{ the }{C}^{\mathrm{\infty }}\text{ vectors bundles }\}$
$M$: $\in \{\text{ the }d\text{ -dimensional embedded submanifolds with boundary of }{M}^{\prime }\}$
$(E,M,\pi )$: $=\text{ the restricted }{C}^{\mathrm{\infty }}\text{ vectors bundle }$
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Statements:
$E\in \{\text{ the embedded submanifolds with boundary of }{E}^{\prime }\}$
//

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2: Note

We already know that $E$ is an immersed submanifold with boundary of $E^{\prime }$, by Note for the definition of restricted (C^\infty\) vectors bundle.

Definition-wise, $E$ does not need to be an embedded submanifold with boundary of $E^{\prime }$ for an embedded submanifold with boundary $M$, but in fact, $E$ is an embedded submanifold with boundary of $E^{\prime }$, which we are going to prove.

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3: Proof

Whole Strategy: Step 1: see that $E$ has the subspace topology of $E^{\prime }$ by the proposition that for any topological space contained in any ambient topological space, if the space is ambient-space-wise locally topological subspace of the ambient space, the space is the topological subspace of the space; Step 2: see that the inclusion $\iota :E\to E^{\prime }$ is a $C^{\mathrm{\infty }}$ embedding.

Step 1:

Let us see that $E$ has the subspace topology of $E^{\prime }$.

We are going to apply the proposition that for any topological space contained in any ambient topological space, if the space is ambient-space-wise locally topological subspace of the ambient space, the space is the topological subspace of the space.

Let $m\in M$ be any.

There is a trivializing open subset around $m$ for $M^{\prime }$, $U_m^{\prime }\subseteq M^{\prime }$, such that $U_m:=U_m^{\prime }\cap M$ is a chart domain for $M$, by the proposition that for any $C^{\mathrm{\infty }}$ manifold with boundary and its any embedded submanifold with boundary, around each point on the submanifold with boundary, there is a trivializing open subset for the manifold with boundary whose intersection with the submanifold with boundary is a chart domain for the submanifold with boundary. Let the chart be $(U_m\subseteq M,{\varphi }_m)$.

The corresponding $U_{\beta }^{\prime }$ s and $(U_{\beta }\subseteq M,{\varphi }_{\beta })$ s are legitimate as the ones used for constructing the topology and the atlas of $E$ in the definition of restricted $C^{\mathrm{\infty }}$ vectors bundle: $U_{\beta }$ is an embedded submanifold with boundary of $M$, by Note for the definition of open submanifold with boundary of $C^{\mathrm{\infty }}$ manifold with boundary and $U_{\beta }$ is an embedded submanifold with boundary of $M^{\prime }$, by the proposition that for any $C^{\mathrm{\infty }}$ manifold with boundary, any embedded submanifold with boundary of any embedded submanifold with boundary is an embedded submanifold with boundary of the manifold with boundary. We take $\{(U_{\beta }\subseteq M,{\varphi }_{\beta })|\beta \in B\}$ and $\{U_{\beta }^{\prime }|\beta \in B\}$ and $\{{\mathrm{\Phi }}_{\beta }^{\prime }:{\pi }^{\prime -1}(U_{\beta }^{\prime })\to U_{\beta }^{\prime }\times {\mathbb{R}}^k|\beta \in B\}$, accordingly.

For each point on $E$, let us take the open neighborhood of the point on $E^{\prime }$ for the proposition that for any topological space contained in any ambient topological space, if the space is ambient-space-wise locally topological subspace of the ambient space, the space is the topological subspace of the space as ${\pi }^{\prime -1}(U_{\beta }^{\prime })$.

Let us see that ${\pi }^{\prime -1}(U_{\beta }^{\prime })$ is indeed one the theorem requires.

${\pi }^{\prime -1}(U_{\beta }^{\prime })$ is indeed an open neighborhood of the point on $E^{\prime }$, because the point is contained in it and $U_{\beta }^{\prime }$ is open on $M^{\prime }$ and ${\pi }^{\prime }$ is continuous.

Is ${\pi }^{\prime -1}(U_{\beta }^{\prime })\cap E$ an open subset of $E$?

Yes, because ${\pi }^{\prime -1}(U_{\beta }^{\prime })\cap E={\pi }^{\prime -1}(U_{\beta }^{\prime })\cap {\pi }^{\prime -1}(M)={\pi }^{\prime -1}(U_{\beta }^{\prime }\cap M)$, by the proposition that for any map, the map preimage of any intersection of sets is the intersection of the map preimages of the sets, $={\pi }^{-1}(U_{\beta })$, which is open on $E$, because $U_{\beta }$ is open on $M$ and $\pi$ is continuous.

Does ${\pi }^{\prime -1}(U_{\beta }^{\prime })\cap E\subseteq E\text{ as the subspace of }E$ equal ${\pi }^{\prime -1}(U_{\beta }^{\prime })\cap E\subseteq {\pi }^{\prime -1}(U_{\beta }^{\prime })\text{ as the subspace of }{\pi }^{\prime -1}(U_{\beta }^{\prime })$?

${\pi }^{\prime -1}(U_{\beta }^{\prime })\cap E\subseteq E$ is nothing but ${\pi }^{-1}(U_{\beta })$, and ${\mathrm{\Phi }}_{\beta }:{\pi }^{-1}(U_{\beta })\to U_{\beta }\times {\mathbb{R}}^k$ is homeomorphic.

For each subset, $S\subseteq {\pi }^{\prime -1}(U_{\beta }^{\prime })\cap E$, $S\cap E=S\cap {\pi }^{-1}(U_{\beta })$, because while $S\cap {\pi }^{-1}(U_{\beta })\subseteq S\cap E$ is obvious, for each $p\in S\cap E$, $p\in S\subseteq {\pi }^{\prime -1}(U_{\beta }^{\prime })$, which implies that ${\pi }^{\prime }(p)=\pi (p)\in U_{\beta }^{\prime }$, but also $\pi (p)\in M$, so, $\pi (p)\in U_{\beta }^{\prime }\cap M=U_{\beta }$, so, $p\in {\pi }^{-1}(U_{\beta })$, and $p\in S\cap {\pi }^{-1}(U_{\beta })$, and so, $S\cap E\subseteq S\cap {\pi }^{-1}(U_{\beta })$.

Especially, ${\pi }^{\prime -1}(U_{\beta }^{\prime })\cap E={\pi }^{\prime -1}(U_{\beta }^{\prime })\cap {\pi }^{-1}(U_{\beta })$.

${\mathrm{\Phi }}_{\beta }^{\prime }:{\pi }^{\prime -1}(U_{\beta }^{\prime })\to U_{\beta }^{\prime }\times {\mathbb{R}}^k$ is homeomorphic, and ${\mathrm{\Phi }}_{\beta }^{\prime }{|}_{{\pi }^{\prime -1}(U_{\beta }^{\prime })\cap E}:{\pi }^{\prime -1}(U_{\beta }^{\prime })\cap E={\pi }^{\prime -1}(U_{\beta }^{\prime })\cap {\pi }^{-1}(U_{\beta })\to {\mathrm{\Phi }}_{\beta }^{\prime }({\pi }^{\prime -1}(U_{\beta }^{\prime })\cap {\pi }^{-1}(U_{\beta }))={\mathrm{\Phi }}_{\beta }^{\prime }({\pi }^{-1}(U_{\beta }))={\mathrm{\Phi }}_{\beta }({\pi }^{-1}(U_{\beta }))=U_{\beta }\times {\mathbb{R}}^k\subseteq U_{\beta }^{\prime }\times {\mathbb{R}}^k$ is homeomorphic, by the proposition that any restriction of any continuous map on the domain and the codomain is continuous. Note that it is homeomorphic with the domain, ${\pi }^{\prime -1}(U_{\beta }^{\prime })\cap E$, regarded as the subspace of ${\pi }^{\prime -1}(U_{\beta }^{\prime })$ and the codomain, $U_{\beta }\times {\mathbb{R}}^k$, regarded as the subspace of $U_{\beta }^{\prime }\times {\mathbb{R}}^k$.

Let us think of the identity map, $id:U_{\beta }\times {\mathbb{R}}^k\to U_{\beta }\times {\mathbb{R}}^k\subseteq U_{\beta }^{\prime }\times {\mathbb{R}}^k$. In fact, $U_{\beta }\times {\mathbb{R}}^k\subseteq U_{\beta }^{\prime }\times {\mathbb{R}}^k$ regarded as the subspace of $U_{\beta }^{\prime }\times {\mathbb{R}}^k$ is nothing but $U_{\beta }\times {\mathbb{R}}^k$, because $M$ is a topological subspace of $M^{\prime }$.

Now, for each open subset of ${\pi }^{\prime -1}(U_{\beta }^{\prime })\cap E\subseteq E={\pi }^{-1}(U_{\beta })$, $U$, ${\mathrm{\Phi }}_{\beta }(U)\subseteq U_{\beta }\times {\mathbb{R}}^k$ is open; $id({\mathrm{\Phi }}_{\beta }(U))\subseteq U_{\beta }\times {\mathbb{R}}^k\subseteq U_{\beta }^{\prime }\times {\mathbb{R}}^k$ is open; ${{\mathrm{\Phi }}_{\beta }^{\prime }{|}_{{\pi }^{\prime -1}(U_{\beta }^{\prime })\cap E}}^{-1}(id({\mathrm{\Phi }}_{\beta }(U)))\subseteq {\pi }^{\prime -1}(U_{\beta }^{\prime })\cap E\subseteq {\pi }^{\prime -1}(U_{\beta }^{\prime })$ is open. In fact, that ${{\mathrm{\Phi }}_{\beta }^{\prime }{|}_{{\pi }^{\prime -1}(U_{\beta }^{\prime })\cap E}}^{-1}\circ id\circ {\mathrm{\Phi }}_{\beta }$ is the identity map, because ${\mathrm{\Phi }}_{\beta }$ is the restriction of ${\mathrm{\Phi }}_{\beta }^{\prime }$. That shows that each open subset of ${\pi }^{\prime -1}(U_{\beta }^{\prime })\cap E\subseteq E$ is open on ${\pi }^{\prime -1}(U_{\beta }^{\prime })\cap E\subseteq {\pi }^{\prime -1}(U_{\beta }^{\prime })$.

Likewise, for each open subset of ${\pi }^{\prime -1}(U_{\beta }^{\prime })\cap E\subseteq {\pi }^{\prime -1}(U_{\beta }^{\prime })$, $U$, ${\mathrm{\Phi }}_{\beta }^{\prime }{|}_{{\pi }^{\prime -1}(U_{\beta }^{\prime })\cap E}(U)\subseteq U_{\beta }\times {\mathbb{R}}^k\subseteq U_{\beta }^{\prime }\times {\mathbb{R}}^k$ is open; ${id}^{-1}({\mathrm{\Phi }}_{\beta }^{\prime }{|}_{{\pi }^{\prime -1}(U_{\beta }^{\prime })\cap E}(U))\subseteq U_{\beta }\times {\mathbb{R}}^k$ is open; ${{\mathrm{\Phi }}_{\beta }}^{-1}({id}^{-1}({\mathrm{\Phi }}_{\beta }^{\prime }{|}_{{\pi }^{\prime -1}(U_{\beta }^{\prime })\cap E}(U))\subseteq {\pi }^{\prime -1}(U_{\beta }^{\prime })\cap E\subseteq E$ is open. In fact, that ${{\mathrm{\Phi }}_{\beta }}^{-1}\circ {id}^{-1}\circ {\mathrm{\Phi }}_{\beta }^{\prime }{|}_{{\pi }^{\prime -1}(U_{\beta }^{\prime })\cap E}$ is the identity map, because ${\mathrm{\Phi }}_{\beta }$ is the restriction of ${\mathrm{\Phi }}_{\beta }^{\prime }$. That shows that each open subset of ${\pi }^{\prime -1}(U_{\beta }^{\prime })\cap E\subseteq {\pi }^{\prime -1}(U_{\beta }^{\prime })$ is open on ${\pi }^{\prime -1}(U_{\beta }^{\prime })\cap E\subseteq E$.

So, yes, ${\pi }^{\prime -1}(U_{\beta }^{\prime })\cap E\subseteq E\text{ as the subspace of }E$ equals ${\pi }^{\prime -1}(U_{\beta }^{\prime })\cap E\subseteq {\pi }^{\prime -1}(U_{\beta }^{\prime })\text{ as the subspace of }{\pi }^{\prime -1}(U_{\beta }^{\prime })$.

So, $E$ is the topological subspace of $E^{\prime }$.

Step 2:

Let us see that the inclusion $\iota :E\to E^{\prime }$ is a $C^{\mathrm{\infty }}$ embedding.

We already know that $\iota$ is a $C^{\mathrm{\infty }}$ immersion, because $E$ is an immersed submanifold with boundary of $E^{\prime }$.

The codomain restriction, ${\iota }^{\prime }:E\to \iota (E)\subseteq E^{\prime }$, is homeomorphic, because $E$ has the subspace topology of $E^{\prime }$ by Step 1, and ${\iota }^{\prime }$ is the identity map.

So, $\iota$ is a $C^{\mathrm{\infty }}$ embedding.

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References

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