A description/proof of that continuous map from topological space into Hausdorff topological space with continuous left inverse 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 Hausdorff topological space.
- The reader knows a definition of left inverse of map.
- The reader knows a definition of proper map.
- The reader admits the proposition that any compact subset of any Hausdorff topological space is closed.
- The reader admits the proposition that the compactness of any topological subset as a subset equals the compactness as a subspace.
- The reader admits the proposition that any closed subset of any compact topological space is compact.
- 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 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 for any map between sets, the composition of the map after any preimage is contained in the argument set.
Target Context
- The reader will have a description and a proof of the proposition that any continuous map from any topological space into any Hausdorff topological space with any continuous left inverse 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 space, \(T_1\), and any Hausdorff topological space, \(T_2\), any continuous map, \(f: T_1 \rightarrow T_2\), with any continuous left inverse, \(f': T_2 \rightarrow T_1\), which means that \(f' \circ f\) is the identity map, is proper.
2: Proof
\(f\) is injective and \(f'\) is surjective. Let \(S\) be any compact subset of \(T_2\). \(S\) is compact as a subspace, by the proposition that the compactness of any topological subset as a subset equals the compactness as a subspace. \(S\) is closed, by the proposition that any compact subset of any Hausdorff topological space is closed. \(f^{-1} (S)\) is closed. Let \(O := \{U_\alpha \subseteq T_1\vert \alpha \in A\}\) where \(A\) is any possibly uncountable indices set be any open cover of \(f^{-1} (S)\). \(S \cap f'^{-1} \circ f^{-1} (S)\) is closed on \(S\), and so is compact on \(S\), by the proposition that any closed subset of any compact topological space is compact, and so 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.
\(f'^{-1} (U_\alpha)\) is open on \(T_2\), and \(\{f'^{-1} (U_\alpha)\vert \alpha \in A\}\) is an open cover of \(S \cap f'^{-1} \circ f^{-1} (S)\) on \(T_2\), because \(f^{-1} (S) \subseteq \cup_{\alpha \in A} U_\alpha\), \(f'^{-1} (f^{-1} (S)) \subseteq f'^{-1} (\cup_{\alpha \in A} U_\alpha) = \cup_{\alpha \in A} (f'^{-1} (U_\alpha))\), 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 has a finite subcover, \(\{f'^{-1} (U_i)\vert i \in I \subseteq A\}\) where \(I\) is a finite indices set. \(\{U_i\vert i \in I\}\) is a finite subcover of \(O\), because for any \(p \in f^{-1} (S)\), \(f (p) \in S \cap f \circ f^{-1} (S) \subseteq S \cap f'^{-1} \circ f^{-1} (S)\), so, \(f (p) \in f'^{-1} (U_i)\) for an \(i \in I\), \(f' \circ f (p) = p \in f' \circ f'^{-1} (U_i) \subseteq U_i\), by the proposition that for any map between sets, the composition of the map after any preimage is contained in the argument set.
