description/proof of that covering map is proper iff cardinality of sheets is finite
Topics
About: topological space
The table of contents of this article
Starting Context
- The reader knows a definition of covering map.
- The reader knows a definition of proper map.
- The reader admits the proposition that for any covering map, the cardinalities of the sheets are the same.
- 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.
Target Context
- The reader will have a description and a proof of the proposition that any covering map is proper if and only if the cardinality of sheets is finite.
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:
\(T_1\): \(\in \{\text{ the topological spaces }\}\)
\(T_2\): \(\in \{\text{ the topological spaces }\}\)
\(\pi\): \(: T_1 \to T_2\), \(\in \{\text{ the covering maps }\}\)
//
Statements:
\(\pi \in \{\text{ the proper maps }\}\)
\(\iff\)
\(\text{ the cardinality of sheets } \in \mathbb{N} \setminus \{0\}\)
//
2: Proof
Whole Strategy: Step 1: see that 'the cardinality of sheets' is a valid concept; Step 2: suppose that \(\pi\) is proper; Step 3: see that the cardinality of sheets is finite; Step 4: suppose that the cardinality of sheets is finite, \(n\); Step 5: take any compact subset of \(T_2\), \(K\), and any open cover of \(\pi^{-1} (K)\), \(\{U_j\}\), and for each \(k \in K\), take any evenly-covered \(V_k\) such that \(\pi^{-1} (V_k) = \{\pi^{-1} (V_k)_1, ..., \pi^{-1} (V_k)_n\}\), shrink \(V_k\) if necessary such that \(\pi^{-1} (V_k)_1 \subseteq U_{j (k, 1)}\), ..., shrink \(V_k\) if necessary such that \(\pi^{-1} (V_k)_n \subseteq U_{j (k, n)}\); Step 6: take a finite subcover of \(K\), \(\{V_{k_1}, ..., V_{k_m}\}\), and see that \(\{U_{j (k_1, 1)}, ..., U_{j (k_1, n)}, ..., U_{j (k_m, 1)}, ..., U_{j (k_m, n)}\}\) is a finite subcover of \(\pi^{-1} (K)\).
Step 1:
Talked about "the cardinality of sheets", do some evenly-covered open neighborhoods not have some different cardinalities?
No, all the evenly-covered open neighborhoods have the same cardinality, by the proposition that for any covering map, the cardinalities of the sheets are the same, so, 'the cardinality of sheets' is a valid concept.
Step 2:
Let us suppose that \(\pi\) is proper.
Step 3:
Let \(t_2 \in T_2\) be any.
\(\{t_2\} \subseteq T_2\) is a compact subset.
Let us take any evenly-covered open neighborhood of \(t_2\), \(V_{t_2} \subseteq T_2\).
\(\pi^{-1} (\{t_2\}) = \{\pi^{-1} (\{t_2\})_j \vert j \in J\} \subseteq \pi^{-1} (V_{t_2}) = \cup_{j \in J} \pi^{-1} (V_{t_2})_j\), which means that \(\{\pi^{-1} (V_{t_2})_j \vert j \in J\}\) is an open cover of \(\pi^{-1} (\{t_2\})\).
If \(J\) was an infinite set, there would be no finite subcover, because each \(\pi^{-1} (V_{t_2})_j\) would not be able to be removed, because that would make \(\pi^{-1} (\{t_2\})_j\) uncovered.
So, \(J\) is finite.
Step 4:
Let us suppose that the cardinality of sheets is a finite \(n\).
Step 5:
Let \(K \subseteq T_2\) be any compact subset.
Let \(\{U_j \vert j \in J\}\) be any open cover of \(\pi^{-1} (K)\).
Let \(k \in K\) be any.
There is an evenly-covered open neighborhood of \(k\), \(V_k \subseteq T_2\).
\(\pi^{-1} (V_k) = \{\pi^{-1} (V_k)_1, ..., \pi^{-1} (V_k)_n\}\).
\(\pi^{-1} (\{k\}) = \{\pi^{-1} (\{k\})_1, ..., \pi^{-1} (\{k\})_n\}\) where \(\pi^{-1} (\{k\})_j \in \pi^{-1} (V_k)_j\).
1st, let us think of \(\pi^{-1} (V_k)_1\).
As \(\pi^{-1} (\{k\})_1 \in \pi^{-1} (K)\), \(\pi^{-1} (\{k\})_1 \in U_{j (k, 1)}\), where \(j (k, 1)\) means that \(j (k, 1) \in J\) is chosen depending on \((k, 1)\).
There is an open neighborhood of \(\pi^{-1} (\{k\})_1\), \(W_{\pi^{-1} (\{k\})_1} \subseteq T_1\), such that \(W_{\pi^{-1} (\{k\})_1} \subseteq \pi^{-1} (V_k)_1 \cap U_{j (k, 1)}\).
Then, \(\pi \vert_{W_{\pi^{-1} (\{k\})_1}}: W_{\pi^{-1} (\{k\})_1} \to \pi (W_{\pi^{-1} (\{k\})_1}) \subseteq V_k\) is a homeomorphism, and \(\pi (W_{\pi^{-1} (\{k\})_1})\) is an evenly covered open neighborhood of \(k\), which will be denoted \(V_k\) again.
Now we have \(\pi^{-1} (V_k)_1 \subseteq U_{j (k, 1)}\).
Next, let us think of \(\pi^{-1} (V_k)_2\).
As \(\pi^{-1} (\{k\})_2 \in \pi^{-1} (K)\), \(\pi^{-1} (\{k\})_2 \in U_{j (k, 2)}\).
There is an open neighborhood of \(\pi^{-1} (\{k\})_2\), \(W_{\pi^{-1} (\{k\})_2} \subseteq T_1\), such that \(W_{\pi^{-1} (\{k\})_2} \subseteq \pi^{-1} (V_k)_2 \cap U_{j (k, 2)}\).
Then, \(\pi \vert_{W_{\pi^{-1} (\{k\})_2}}: W_{\pi^{-1} (\{k\})_2} \to \pi (W_{\pi^{-1} (\{k\})_2}) \subseteq V_k\) is a homeomorphism, and \(\pi (W_{\pi^{-1} (\{k\})_2})\) is an evenly covered open neighborhood of \(k\), which will be denoted \(V_k\) again.
Now we have \(\pi^{-1} (V_k)_2 \subseteq U_{j (k, 2)}\), while \(\pi^{-1} (V_k)_1 \subseteq U_{j (k, 1)}\) still holds, because shrinking \(V_k\) preserves the relation.
And so on, and after all, we have \(V_k\) such that \(\pi^{-1} (V_k)_j \subseteq U_{j (k, j)}\) for each \(j \in \{1, ..., n\}\), which means that \(\pi^{-1} (V_k) = \pi^{-1} (V_k)_1 \cup ... \cup \pi^{-1} (V_k)_n \subseteq U_{j (k, 1)} \cup ... \cup U_{j (k, n)}\).
Step 6:
Such \(\{V_k \vert k \in K\}\) is an open cover of \(K\), and has a finite subcover, \(\{V_{k_1}, ... V_{k_m}\}\), which means that \(K \subseteq V_{k_1} \cup ... \cup V_{k_m}\).
\(\pi^{-1} (K) \subseteq \pi^{-1} (V_{k_1} \cup ... \cup V_{k_m}) = \pi^{-1} (V_{k_1}) \cup ... \cup \pi^{-1} (V_{k_m})\), 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.
But \(\pi^{-1} (V_{k_1}) \cup ... \cup \pi^{-1} (V_{k_m}) \subseteq U_{j (k_1, 1)} \cup ... \cup U_{j (k_1, n)} \cup ... \cup U_{j (k_m, 1)} \cup ... \cup U_{j (k_m, n)}\).
So, \(\{U_{j (k_1, 1)}, ... U_{j (k_1, n)}, ..., U_{j (k_m, 1)}, ... U_{j (k_m, n)}\}\) is a finite subcover of \(\pi^{-1} (K)\).
So, \(\pi^{-1} (K)\) is compact.
So, \(\pi\) is proper.
