description/proof of that for \(C^\infty\) 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
Topics
About: topological space
The table of contents of this article
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.
Target Context
- The reader will have a description and a proof of the proposition that for any \(C^\infty\) 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.
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.
Main Body
1: Structured Description
Here is the rules of Structured Description.
Entities:
\((E, T, \pi)\): \(\in \{\text{ the } C^\infty \text{ vectors bundles of rank } k\}\)
\(B\): \(\in \{\text{ the possibly uncountable index sets }\}\)
\(\{U_\beta \vert \beta \in B\}\): \(\in \{\text{ the trivializing open subsets covers of } T\}\)
\(\{\Phi_\beta \vert \beta \in B\}\): \(\Phi_\beta: \pi^{-1} (U_\beta) \to U_\beta \times \mathbb{R}^k \in \{\text{ the trivializations over } U_\beta\}\)
\(\{B_\beta \vert \beta \in B\}\): \(B_\beta \in \{\text{ the bases of } U_\beta\}\)
\(\{B'_\beta \vert \beta \in B\}\): \(B'_\beta \in \{\text{ the bases of } \mathbb{R}^k\}\)
\(\{B''_\beta \vert \beta \in B\}\): \(= \{V \times V' \vert V \in B_\beta, V' \in B'_\beta\}\)
\(D\): \(= \{{\Phi_\beta}^{-1} (U) \vert \beta \in B, U \in B''_\beta\}\)
//
Statements:
\(D \in \{\text{ the bases of } E\}\)
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2: Note
\(B'_\beta\) s can and usually will be chosen to be the same.
3: Proof
Whole Strategy: Step 1: see that \({\Phi_\beta}^{-1} (U) \subseteq E\) 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 \(p \in {\Phi_\beta}^{-1} (U) \subseteq N_p\); Step 3: take an open neighborhood of \(p\), \(U_p \subseteq E\), such that \(U_p \subseteq N_p\) and take \(\Phi_\beta (U_p \cap \pi^{-1} (U_\beta))\) and a \(U\) such that \(\Phi_\beta (p) \in U \subseteq \Phi_\beta (U_p \cap \pi^{-1} (U_\beta))\); Step 4: see that \(p \in {\Phi_\beta}^{-1} (U) \subseteq U_p \subseteq N_p\).
Step 1:
\(U \in B''_\beta \subseteq U_\beta \times \mathbb{R}^k\) is open, by the definition of product topology.
\({\Phi_\beta}^{-1} (U) \subseteq E\) is open, because \(\Phi_\beta\) is a homeomorphism and \({\Phi_\beta}^{-1} (U) \subseteq E\) is open.
Step 2:
Let us take any \(p \in E\) and any neighborhood of \(p\), \(N_p \subseteq E\).
There is a \(U_\beta \in \{U_\beta \vert \beta \in B\}\) 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 \(p \in {\Phi_\beta}^{-1} (U) \subseteq N_p\).
Step 3:
Let us take an open neighborhood of \(p\), \(U_p \subseteq E\), such that \(U_p \subseteq N_p\).
Let us take \(\Phi_\beta (U_p \cap \pi^{-1} (U_\beta)) \subseteq U_\beta \times \mathbb{R}^k\), which is an open neighborhood of \(\Phi_\beta (p)\), because \(U_p \cap \pi^{-1} (U_\beta)\) is open on \(\pi^{-1} (U_\beta)\) and \(\Phi_\beta\) is a homeomorphism.
By the definition of product topology, there are an open subset of \(U_\beta\), \(\tilde{V}\), and an open subset of \(\mathbb{R}^k\), \(\tilde{V'}\), such that \(\Phi_\beta (p) \in \tilde{V} \times \tilde{V'} \subseteq \Phi_\beta (U_p \cap \pi^{-1} (U_\beta))\).
There are a \(V \in B_\beta\) such that \(V \subseteq \tilde{V}\) and a \(V' \in B'_\beta\) such that \(V' \subseteq \tilde{V'}\) such that \(\Phi_\beta (p) \in V \times V'\), inevitably \(V \times V' \subseteq \tilde{V} \times \tilde{V'}\).
\(U := V \times V' \in B''_\beta\).
After all, \(\Phi_\beta (p) \in U \subseteq \tilde{V} \times \tilde{V'} \subseteq \Phi_\beta (U_p \cap \pi^{-1} (U_\beta))\).
Step 4:
\(p = {\Phi_\beta}^{-1} (\Phi_\beta (p)) \in {\Phi_\beta}^{-1} (U) \subseteq {\Phi_\beta}^{-1} (\Phi_\beta (U_p \cap \pi^{-1} (U_\beta))) = U_p \cap \pi^{-1} (U_\beta) \subseteq N_p\).
