833: For C∞ Vectors Bundle and Trivializing Open Subsets Cover, Preimages Under Trivializations of Products of Basis of Open Subset and Basis of Rk Constitute Basis of Total Space

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description/proof of that for $C^{\mathrm{\infty }}$ vectors bundle and trivializing open subsets cover, preimages under trivializations of products of basis of open subset and basis of $R^k$ constitute basis of total space

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Topics

About: topological space

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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 vectors bundle of rank $k$.
• The reader knows a definition of basis of topological space.
• The reader knows a definition of product topology.

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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 and any trivializing open subsets cover, the preimages under trivializations of the products of any basis of open subset and any basis of $R^k$ constitute a basis of the total 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:
$(E,T,\pi )$: $\in \{\text{ the }{C}^{\mathrm{\infty }}\text{ vectors bundles of rank }k\}$
$B$: $\in \{\text{ the possibly uncountable index sets }\}$
$\{U_{\beta }|\beta \in B\}$: $\in \{\text{ the trivializing open subsets covers of }T\}$
$\{{\mathrm{\Phi }}_{\beta }|\beta \in B\}$: ${\mathrm{\Phi }}_{\beta }:{\pi }^{-1}(U_{\beta })\to U_{\beta }\times {\mathbb{R}}^k\in \{\text{ the trivializations over }{U}_{\beta }\}$
$\{B_{\beta }|\beta \in B\}$: $B_{\beta }\in \{\text{ the bases of }{U}_{\beta }\}$
$\{B_{\beta }^{\prime }|\beta \in B\}$: $B_{\beta }^{\prime }\in \{\text{ the bases of }{\mathbb{R}}^k\}$
$\{B_{\beta }^{″}|\beta \in B\}$: $=\{V\times V^{\prime }|V\in B_{\beta },V^{\prime }\in B_{\beta }^{\prime }\}$
$D$: $=\{{{\mathrm{\Phi }}_{\beta }}^{-1}(U)|\beta \in B,U\in B_{\beta }^{″}\}$
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Statements:
$D\in \{\text{ the bases of }E\}$
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2: Note

$B_{\beta }^{\prime }$ s can and usually will be chosen to be the same.

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3: Proof

Whole Strategy: Step 1: see that ${{\mathrm{\Phi }}_{\beta }}^{-1}(U)\subseteq E$ is open; Step 2: take any $p\in E$ and any neighborhood of $p$, $N_p\subseteq E$, and set the goal of getting a $U$ such that $p\in {{\mathrm{\Phi }}_{\beta }}^{-1}(U)\subseteq N_p$; Step 3: take an open neighborhood of $p$, $U_p\subseteq E$, such that $U_p\subseteq N_p$ and take ${\mathrm{\Phi }}_{\beta }(U_p\cap {\pi }^{-1}(U_{\beta }))$ and a $U$ such that ${\mathrm{\Phi }}_{\beta }(p)\in U\subseteq {\mathrm{\Phi }}_{\beta }(U_p\cap {\pi }^{-1}(U_{\beta }))$; Step 4: see that $p\in {{\mathrm{\Phi }}_{\beta }}^{-1}(U)\subseteq U_p\subseteq N_p$.

Step 1:

$U\in B_{\beta }^{″}\subseteq U_{\beta }\times {\mathbb{R}}^k$ is open, by the definition of product topology.

${{\mathrm{\Phi }}_{\beta }}^{-1}(U)\subseteq E$ is open, because ${\mathrm{\Phi }}_{\beta }$ is a homeomorphism and ${{\mathrm{\Phi }}_{\beta }}^{-1}(U)\subseteq E$ is open.

Step 2:

Let us take any $p\in E$ and any neighborhood of $p$, $N_p\subseteq E$.

There is a $U_{\beta }\in \{U_{\beta }|\beta \in B\}$ such that $\pi (p)\in U_{\beta }$.

$p\in {\pi }^{-1}(U_{\beta })$.

Our goal is to get a $U\in B_{\beta }^{″}$ such that $p\in {{\mathrm{\Phi }}_{\beta }}^{-1}(U)\subseteq N_p$.

Step 3:

Let us take an open neighborhood of $p$, $U_p\subseteq E$, such that $U_p\subseteq N_p$.

Let us take ${\mathrm{\Phi }}_{\beta }(U_p\cap {\pi }^{-1}(U_{\beta }))\subseteq U_{\beta }\times {\mathbb{R}}^k$, which is an open neighborhood of ${\mathrm{\Phi }}_{\beta }(p)$, because $U_p\cap {\pi }^{-1}(U_{\beta })$ is open on ${\pi }^{-1}(U_{\beta })$ and ${\mathrm{\Phi }}_{\beta }$ is a homeomorphism.

By the definition of product topology, there are an open subset of $U_{\beta }$, $\tilde{V}$, and an open subset of ${\mathbb{R}}^k$, $\widetilde{V^{\prime }}$, such that ${\mathrm{\Phi }}_{\beta }(p)\in \tilde{V}\times \widetilde{V^{\prime }}\subseteq {\mathrm{\Phi }}_{\beta }(U_p\cap {\pi }^{-1}(U_{\beta }))$.

There are a $V\in B_{\beta }$ such that $V\subseteq \tilde{V}$ and a $V^{\prime }\in B_{\beta }^{\prime }$ such that $V^{\prime }\subseteq \widetilde{V^{\prime }}$ such that ${\mathrm{\Phi }}_{\beta }(p)\in V\times V^{\prime }$, inevitably $V\times V^{\prime }\subseteq \tilde{V}\times \widetilde{V^{\prime }}$.

$U:=V\times V^{\prime }\in B_{\beta }^{″}$.

After all, ${\mathrm{\Phi }}_{\beta }(p)\in U\subseteq \tilde{V}\times \widetilde{V^{\prime }}\subseteq {\mathrm{\Phi }}_{\beta }(U_p\cap {\pi }^{-1}(U_{\beta }))$.

Step 4:

$p={{\mathrm{\Phi }}_{\beta }}^{-1}({\mathrm{\Phi }}_{\beta }(p))\in {{\mathrm{\Phi }}_{\beta }}^{-1}(U)\subseteq {{\mathrm{\Phi }}_{\beta }}^{-1}({\mathrm{\Phi }}_{\beta }(U_p\cap {\pi }^{-1}(U_{\beta })))=U_p\cap {\pi }^{-1}(U_{\beta })\subseteq N_p$.

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References

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