A description/proof of that if preimage of closed set under topological spaces map is closed, map 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 knows a definition of closed set.
Target Context
- The reader will have a description and a proof of the proposition that if the preimage of any closed set under a topological spaces map is closed, the map 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 for any close set, \(C \subseteq T_2\), \(f^{-1} (C)\) is closed, then \(f\) is continuous.
2: Proof
Suppose that for any close set, \(C \subseteq T_2\), \(f^{-1} (C)\) is closed. For any open set, \(U \subseteq T_2\), \(T_2 \setminus U\) is closed. \(f^{-1} (T_2 \setminus U)\) is closed. \(f^{-1} (U) = T_1 \setminus f^{-1} (T_2 \setminus U)\), open. So, as the preimage of any open set is open, \(f\) is continuous.
