397: Continuous Map from Topological Space into Hausdorff Topological Space with Continuous Left Inverse Is Proper

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A description/proof of that continuous map from topological space into Hausdorff topological space with continuous left inverse is proper

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Topics

About: topological space

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The table of contents of this article

• Starting Context
• Target Context
• Orientation
• Main Body
• 1: Description
• 2: Proof

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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.

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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.

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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.

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Main Body

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1: Description

For any topological space, $T_1$, and any Hausdorff topological space, $T_2$, any continuous map, $f:T_1\to T_2$, with any continuous left inverse, $f^{\prime }:T_2\to T_1$, which means that $f^{\prime }\circ f$ is the identity map, is proper.

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2: Proof

$f$ is injective and $f^{\prime }$ 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 $O:=\{U_{\alpha }\subseteq T_1|\alpha \in A\}$ where $A$ is any possibly uncountable indices set be any open cover of $f^{-1}(S)$. $S\cap f^{\prime -1}\circ f^{-1}(S)$ 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.

$f^{\prime -1}(U_{\alpha })$ is open on $T_2$, and $\{f^{\prime -1}(U_{\alpha })|\alpha \in A\}$ is an open cover of $S\cap f^{\prime -1}\circ f^{-1}(S)$ on $T_2$, because $f^{-1}(S)\subseteq {\cup }_{\alpha \in A}{U}_{\alpha }$, $f^{\prime -1}(f^{-1}(S))\subseteq f^{\prime -1}({\cup }_{\alpha \in A}{U}_{\alpha })={\cup }_{\alpha \in A}(f^{\prime -1}(U_{\alpha }))$, 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, $\{f^{\prime -1}(U_i)|i\in I\subseteq A\}$ where $I$ is a finite indices set. $\{U_i|i\in I\}$ is a finite subcover of $O$, because for any $p\in f^{-1}(S)$, $f(p)\in S\cap f\circ f^{-1}(S)\subseteq S\cap f^{\prime -1}\circ f^{-1}(S)$, so, $f(p)\in f^{\prime -1}(U_i)$ for an $i\in I$, $f^{\prime }\circ f(p)=p\in f^{\prime }\circ f^{\prime -1}(U_i)\subseteq U_i$, by the proposition that for any map between sets, the composition of the map after any preimage is contained in the argument set.

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References

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