A description/proof of that closed continuous map between topological spaces with compact fibers is proper
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 map.
- The reader knows a definition of compact subset.
- The reader knows a definition of fiber of map.
- The reader knows a definition of proper map.
- The reader admits the proposition that for any map, the map preimage of any union of sets is the union of the map preimages of the sets.
- The reader admits the proposition that the preimage of the codomain minus any codomain subset under any map is the domain minus the preimage of the subset.
- The reader admits the proposition that for any map, the composition of the preimage after the map of any subset contains the argument set.
Target Context
- The reader will have a description and a proof of the proposition that any closed continuous map between topological spaces with compact fibers 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 closed continuous map with compact fibers, \(f: T_1 \rightarrow T_2\), is proper.
2: Proof
Let \(S \subseteq T_2\) be any compact subset. Let \(O := \{U_\alpha \subseteq T_1\vert \alpha \in A\}\) where \(A\) is any possibly uncountable indices set be any open cover of \(f^{-1} (S)\). For any point, \(p \in S\), \(f^{-1} (p)\) is compact. \(O\) is an open cover of \(f^{-1} (p)\) and has a finite subcover, \(O_p := \{U_i \in O\vert i \in I_p\}\) where \(I_p \subseteq A\) is a finite indices set. Let us define \(U'_p := \cup_{i \in I_p} U_i\). \(T_1 \setminus U'_p\) is closed on \(T_1\). \(f (T_1 \setminus U'_p)\) is closed on \(T_2\). \(T_2 \setminus f (T_1 \setminus U'_p)\) is open on \(T_2\). \(p \in T_2 \setminus f (T_1 \setminus U'_p)\), because for any \(p' \in T_1 \setminus U'_p\), \(f (p') \neq p\). \(\{T_2 \setminus f (T_1 \setminus U'_p)\vert p \in S\}\) is an open cover of \(S\) and has a finite subcover, \(\{T_2 \setminus f (T_1 \setminus U'_{p_i})\vert p_i \in S' \subseteq S\}\) where \(S'\) is a set of finite number of points.
\(f^{-1} (S) \subseteq f^{-1} (\cup_{p_i \in S'} (T_2 \setminus f (T_1 \setminus U'_{p_i}))) = \cup_{p_i \in S'} f^{-1} (T_2 \setminus f (T_1 \setminus U'_{p_i})) = \cup_{p_i \in S'} (T_1 \setminus f^{-1} f (T_1 \setminus U'_{p_i}))\), by the proposition that for any map, the map preimage of any union of sets is the union of the map preimages of the sets and by the proposition that the preimage of the codomain minus any codomain subset under any map is the domain minus the preimage of the subset. \(T_1 \setminus U'_{p_i} \subseteq f^{-1} f (T_1 \setminus U'_{p_i})\), by the proposition that for any map, the composition of the preimage after the map of any subset contains the argument set. \(T_1 \setminus f^{-1} f (T_1 \setminus U'_{p_i}) \subseteq U'_{p_i}\). So, \(f^{-1} (S) \subseteq \cup_{p_i \in S'} U'_{p_i}\), which means that \(\{U_i \subseteq O\vert i \in I_{p_j}, p_j \in S'\}\) is a finite subcover of \(O\).
