definition of proper map
Topics
About: topological space
The table of contents of this article
Starting Context
- The reader knows a definition of compact subset of topological space.
- The reader knows a definition of map.
Target Context
- The reader will have a definition of proper map.
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: Structured Description
Here is the rules of Structured Description.
Entities:
\( T_1\): \(\in \{\text{ the topological spaces }\}\)
\( T_2\): \(\in \{\text{ the topological spaces }\}\)
\(*f\): \(: T_1 \to T_2\)
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Conditions:
\(\forall K \in \{\text{ the compact subsets of } T_2\} (f^{-1} (K) \in \{\text{ the compact subsets of } T_1\})\)
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2: Note
There is another narrower definition of 'proper map', which requires \(f\) closed continuous, and for each \(t \in T_2\), \(f^{-1} (t)\) compact.
Let us see that that narrower definition implies this definition.
Let \(K \subseteq T_2\) be any compact subset.
Let \(\{U_j \subseteq T_1 \vert j \in J\}\) be any open cover of \(f^{-1} (K)\).
For each \(k \in K\), \(f^{-1} (k) \subseteq T_1\) is compact and \(f^{-1} (k) \subseteq f^{-1} (K)\).
So, \(\{U_j \subseteq T_1 \vert j \in J\}\) is an open cover of \(f^{-1} (k)\), and there is a finite subcover, \(\{U_j \subseteq T_1 \vert j \in J_k\}\).
\(T_1 \setminus \cup_{j \in J_k} U_j \subseteq T_1\) is closed, so, \(f (T_1 \setminus \cup_{j \in J_k} U_j) \subseteq T_2\) is closed, because \(f\) is closed, so, \(T_2 \setminus f (T_1 \setminus \cup_{j \in J_k} U_j) \subseteq T_2\) is open.
\(k \in T_2 \setminus f (T_1 \setminus \cup_{j \in J_k} U_j)\), because \(k \notin f (T_1 \setminus \cup_{j \in J_k} U_j)\), because as \(f^{-1} (k) \subseteq \cup_{j \in J_k} U_j\), any point in \(T_1 \setminus \cup_{j \in J_k} U_j\) does not map to \(k\) under \(f\).
So, \(K \subseteq \cup_{k \in K} T_2 \setminus f (T_1 \setminus \cup_{j \in J_k} U_j)\), and as \(K\) is compact, there is a finite subcover, \(\{T_2 \setminus f (T_1 \setminus \cup_{j \in J_{k_j}} U_j) \vert k_j \in K^`\}\), where \(K^` = \{k_1, ..., k_n\} \subseteq K\) is a finite subset.
Let us see that \(f^{-1} (K) \subseteq (\cup_{j \in J_{k_1}} U_j) \cup ... \cup (\cup_{j \in J_{k_n}} U_j)\).
Let \(p \in f^{-1} (K)\) be any.
\(f (p) \in K\). So, \(f (p) \in T_2 \setminus f (T_1 \setminus \cup_{j \in J_{k_j}} U_j)\) for a \(k_j\). So, \(f (p) \notin f (T_1 \setminus \cup_{j \in J_{k_j}} U_j)\) for that \(k_j\). That means that \(p \in \cup_{j \in J_{k_j}} U_j\), because otherwise, \(p \in T_1 \setminus \cup_{j \in J_{k_j}} U_j\) and \(f (p) \in f (T_1 \setminus \cup_{j \in J_{k_j}} U_j)\), a contradiction.
So, \(p \in (\cup_{j \in J_{k_1}} U_j) \cup ... \cup (\cup_{j \in J_{k_n}} U_j)\).
So, \(f^{-1} (K) \subseteq (\cup_{j \in J_{k_1}} U_j) \cup ... \cup (\cup_{j \in J_{k_n}} U_j)\).
That means that \(\{U_j \vert j \in J_{k_1} \cup ... \cup J_{k_n}\}\) is a finite subcover of \(f^{-1} (K)\): \(J_{k_1} \cup ... \cup J_{k_n}\) is finite, because each \(J_{k_j}\) is finite.
So, \(f^{-1} (K)\) is compact.
