A description/proof of that restriction of proper map between topological spaces on saturated domain subset and range codomain is proper
Topics
About: topological space
The table of contents of this article
Starting Context
- The reader knows a definition of map.
- The reader knows a definition of topological space.
- The reader knows a definition of saturated subset with respect to map.
- The reader knows a definition of proper map.
- The reader admits the proposition that for any topological space, any compact subset of any subspace is compact on the base space.
- The reader admits the proposition that for any topological space, any subspace subset that is compact on the base space is compact on the subspace.
Target Context
- The reader will have a description and a proof of the proposition that the restriction of any proper map between any topological spaces on any domain saturated subset with respect to the map and the range codomain is proper.
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, \(T_1, T_2\), any not necessarily continuous proper map, \(f: T_1 \rightarrow T_2\), and any saturated subset, \(S \subseteq T_1\), with respect to \(f\), the restriction, \((f\vert_S)': S \rightarrow f (S)\), is proper, where \((f\vert_S)'\) is denoted with the prime because just \(f\vert_S\) is into \(T_2\), not into \(f (S)\).
2: Proof
Let \(S' \subseteq f (S)\) be compact on \(f (S)\). \(S'\) is compact on \(T_2\), by the proposition that for any topological space, any compact subset of any subspace is compact on the base space. \((f\vert_S)'^{-1} (S') = f^{-1} (S')\), because as \(f^{-1} f (S) = S\), \(f^{-1} (S') \subseteq S\). \(f^{-1} (S')\) is compact on \(T_1\) as \(f\) is proper. \((f\vert_S)'^{-1} (S') = f^{-1} (S')\) is compact on \(S\), the proposition that for any topological space, any subspace subset that is compact on the base space is compact on the subspace.
