description/proof of that for 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 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:
\((E, T, \pi)\): \(\in \{\text{ the vectors bundles of rank } k\}\)
\(B\): \(\in \{\text{ the possibly uncountable index sets }\}\)
\(\{U_\beta \vert \beta \in B\}\): \(\in \{\text{ the trivializing open subsets covers of } T\}\)
\(\{\Phi_\beta \vert \beta \in B\}\): \(\Phi_\beta: \pi^{-1} (U_\beta) \to U_\beta \times \mathbb{R}^k \in \{\text{ the trivializations over } U_\beta\}\)
\(\{B_\beta \vert \beta \in B\}\): \(B_\beta \in \{\text{ the bases of } U_\beta\}\)
\(\{B'_\beta \vert \beta \in B\}\): \(B'_\beta \in \{\text{ the bases of } \mathbb{R}^k\}\)
\(\{B''_\beta \vert \beta \in B\}\): \(= \{V \times V' \vert V \in B_\beta, V' \in B'_\beta\}\)
\(D\): \(= \{{\Phi_\beta}^{-1} (U) \vert \beta \in B, U \in B''_\beta\}\)
//
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:
\(p = {\Phi_\beta}^{-1} (\Phi_\beta (p)) \in {\Phi_\beta}^{-1} (U) \subseteq {\Phi_\beta}^{-1} (\Phi_\beta (U_p \cap \pi^{-1} (U_\beta))) = U_p \cap \pi^{-1} (U_\beta) \subseteq N_p\).
