1136: Covering Map Is Proper iff Cardinality of Sheets Is Finite

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description/proof of that covering map is proper iff cardinality of sheets is finite

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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: Structured Description
• 2: Proof

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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.

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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.

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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: Structured Description

Here is the rules of Structured Description.

Entities:
$T_1$: $\in \{\text{ the topological spaces }\}$
$T_2$: $\in \{\text{ the topological spaces }\}$
$\pi$: $:T_1\to T_2$, $\in \{\text{ the covering maps }\}$
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Statements:
$\pi \in \{\text{ the proper maps }\}$
$⟺$
$\text{ the cardinality of sheets }\in \mathbb{N}\setminus \{0\}$
//

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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 $t_2$, $V_{t_2}\subseteq T_2$.

${\pi }^{-1}(\{t_2\})=\{{\pi }^{-1}(\{t_2\}{)}_j|j\in J\}\subseteq {\pi }^{-1}(V_{t_2})={\cup }_{j\in J}{\pi }^{-1}(V_{t_2}{)}_j$, which means that $\{{\pi }^{-1}(V_{t_2}{)}_j|j\in J\}$ 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 $\{U_j|j\in J\}$ 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, $\pi {|}_{W_{{\pi }^{-1}(\{k\}{)}_1}}:W_{{\pi }^{-1}(\{k\}{)}_1}\to \pi (W_{{\pi }^{-1}(\{k\}{)}_1})\subseteq V_k$ 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, $\pi {|}_{W_{{\pi }^{-1}(\{k\}{)}_2}}:W_{{\pi }^{-1}(\{k\}{)}_2}\to \pi (W_{{\pi }^{-1}(\{k\}{)}_2})\subseteq V_k$ 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 $\{V_k|k\in K\}$ 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.

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References

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