A description/proof of that for vectors bundle, there is chart trivializing open cover
Topics
About: \(C^\infty\) manifold
The table of contents of this article
Starting Context
- The reader knows a definition of \(C^\infty\) vectors bundle.
- The reader admits the proposition that for any vectors bundle, a trivializing open subset is not necessarily a chart open subset, but there is a possibly smaller chart trivializing open subset at any point on any trivializing open subset.
Target Context
- The reader will have a description and a proof of the proposition that for any vectors bundle, there is a chart trivializing open cover.
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: Description
For any \(C^\infty\) manifold, \(M\), and any vectors bundle, \(\pi: E \rightarrow M\), there is a chart trivializing open cover, which means that each trivializing open subset is a chart open subset.
2: Proof
There is a trivializing open cover, \(\{U_\alpha\vert \alpha \in A\}\) where \(A\) is any possibly uncountable indices set, by the definition of vectors bundle, but \(U_\alpha\) is not necessarily a chart open subset but there is a chart trivializing open set around any point, \(p \in U_\alpha\), \(U_{\alpha, p}\), by the proposition that for any vectors bundle, a trivializing open subset is not necessarily a chart open subset, but there is a possibly smaller chart trivializing open subset at any point on any trivializing open subset. \(\{U_{\alpha, p}\vert p \in U_\alpha \land \alpha \in A\}\) is a chart trivializing open cover. Note that \(U_{\alpha, p}\) and \(U_{\alpha, p'}\) for \(p \neq p'\) may be the same, but that is not any problem, and the chart trivializing open cover does not contain any duplication, because the definition of set automatically eliminates any duplications.
