A description/proof of that map between topological spaces is continuous if domain restriction of map to each open set of open cover is continuous
Topics
About: topological space
The table of contents of this article
Starting Context
- The reader knows a definition of continuous map.
- The reader admits the proposition that any open set on any open topological subspace is open on the base space.
Target Context
- The reader will have a description and a proof of the proposition that any map between topological spaces is continuous if the domain restriction of the map to each open set of a possibly uncountable open cover is continuous.
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 topological spaces, \(T_1, T_2\), and any map, \(f: T_1 \rightarrow T_2\), if there is an open cover of \(T_1\), \(\{U_\alpha \subseteq T_1\}, \cup_{\alpha} U_\alpha = T_1\), such that each \(f|_{U_\alpha}: U_\alpha \rightarrow T_2\) is continuous, \(f\) is continuous.
2: Proof
For any open set, \(U \subseteq T_2\), \({f|_{U_\alpha}}^{-1} (U)\) is open on \(U_\alpha\), and on \(T_1\), by the proposition that any open set on any open subspace topological space is open on the base topological space. \(f^{-1} (U) = \cup_\alpha {f|_{U_\alpha}}^{-1} (U)\), because for any \(p \in f^{-1} (U)\), \(f (p) \in U\), but \(p \in \cup_\alpha U_\alpha\), so, \(f|_{U_\alpha} (p) \in U\) for an \(\alpha\), \(p \in {f|_{U_\alpha}}^{-1} (U)\); for any \(p \in \cup_\alpha {f|_{U_\alpha}}^{-1} (U)\), \(p \in {f|_{U_\alpha}}^{-1} (U)\) for an \(\alpha\), so, \(f|_{U_\alpha} (p) \in U\), so, \(f (p) \in U\), so, \(p \in f^{-1} (U)\). So, \(f^{-1} (U)\) is open as the union of open sets.
