395: Closed Continuous Map Between Topological Spaces with Compact Fibers Is Proper

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A description/proof of that closed continuous map between topological spaces with compact fibers 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 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.

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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.

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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 spaces, $T_1,T_2$, any closed continuous map with compact fibers, $f:T_1\to T_2$, is proper.

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2: Proof

Let $S\subseteq T_2$ be any compact subset. Let $O:=\{U_{\alpha }\subseteq T_1|\alpha \in A\}$ where $A$ is any possibly uncountable indices set be any open cover of $f^{-1}(S)$. 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, $O_p:=\{U_i\in O|i\in I_p\}$ where $I_p\subseteq A$ is a finite indices set. Let us define $U_p^{\prime }:={\cup }_{i\in I_p}{U}_i$. $T_1\setminus U_p^{\prime }$ is closed on $T_1$. $f(T_1\setminus U_p^{\prime })$ is closed on $T_2$. $T_2\setminus f(T_1\setminus U_p^{\prime })$ is open on $T_2$. $p\in T_2\setminus f(T_1\setminus U_p^{\prime })$, because for any $p^{\prime }\in T_1\setminus U_p^{\prime }$, $f(p^{\prime })\ne p$. $\{T_2\setminus f(T_1\setminus U_p^{\prime })|p\in S\}$ is an open cover of $S$ and has a finite subcover, $\{T_2\setminus f(T_1\setminus U_{p_i}^{\prime })|p_i\in S^{\prime }\subseteq S\}$ where $S^{\prime }$ is a set of finite number of points.

$f^{-1}(S)\subseteq f^{-1}({\cup }_{p_i\in S^{\prime }}(T_2\setminus f(T_1\setminus U_{p_i}^{\prime })))={\cup }_{p_i\in S^{\prime }}{f}^{-1}(T_2\setminus f(T_1\setminus U_{p_i}^{\prime }))={\cup }_{p_i\in S^{\prime }}(T_1\setminus f^{-1}f(T_1\setminus U_{p_i}^{\prime }))$, 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. $T_1\setminus U_{p_i}^{\prime }\subseteq f^{-1}f(T_1\setminus U_{p_i}^{\prime })$, by the proposition that for any map, the composition of the preimage after the map of any subset contains the argument set. $T_1\setminus f^{-1}f(T_1\setminus U_{p_i}^{\prime })\subseteq U_{p_i}^{\prime }$. So, $f^{-1}(S)\subseteq {\cup }_{p_i\in S^{\prime }}{U}_{p_i}^{\prime }$, which means that $\{U_i\subseteq O|i\in I_{p_j},p_j\in S^{\prime }\}$ is a finite subcover of $O$.

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References

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