description/proof of that for covering map, cardinalities of sheets are same
Topics
About: topological space
The table of contents of this article
Starting Context
- The reader knows a definition of covering map.
- The reader knows finite-open-sets-sequence-connected pair of open sets.
- The reader admits the proposition that in any nest of topological subspaces, the connected-ness of any subspace does not depend on the superspace of which the subspace is regarded to be a subspace.
- The reader admits the proposition that for any topological space and any point on any subspace, the intersection of any neighborhood of the point on the base space and the subspace is a neighborhood on the subspace.
- The reader admits the proposition that for any topological space, any point, and any neighborhood of the point, any neighborhood of the point on the neighborhood is a neighborhood of the point on the base space.
- The reader admits the proposition that any continuous image of any connected space is connected.
- The reader admits the proposition that any restriction of any continuous map on the domain and the codomain is continuous.
- The reader admits the proposition that any pair of elements of any open cover of any connected topological space is finite-open-sets-sequence-connected via some elements of the open cover.
Target Context
- The reader will have a description and a proof of the proposition that for any covering map, the cardinalities of the sheets are the same.
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:
\(T_1\): \(\in \{\text{ the connected and locally path-connected topological spaces }\}\)
\(T_2\): \(\in \{\text{ the connected and locally path-connected topological spaces }\}\)
\(\pi\): \(:T_1 \to T_2\), \(\in \{\text{ the covering maps }\}\)
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Statements:
\(\forall N_p \subseteq T_2 \in \{\text{ the evenly covered neighborhoods of } p \text{ by } \pi\}, \forall N_{p'} \subseteq T_2 \in \{\text{ the evenly covered neighborhoods of } p' \text{ by } \pi\} (\vert \{\pi^{-1} (N_p)_\beta \vert \beta \in B\} \vert = \vert \{\pi^{-1} (N_{p'})_{\beta'} \vert \beta' \in B'\} \vert)\)
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2: Proof
Whole Strategy: Step 1: see that for each connected neighborhood of \(p''\), \(N_{p''} \subseteq N_p\), \(\vert \{\pi^{-1} (N_p)_\beta \vert \beta \in B\} \vert = \vert \{\pi^{-1} (N_{p''})_{\beta} \vert \beta \in B\} \vert\); Step 2: determine to think of evenly covered open neighborhoods, \(U_p\) s; Step 3: see that when \(U_p \cap U_{p'} \neq \emptyset\), \(\vert \{\pi^{-1} (U_p)_\beta \vert \beta \in B\} \vert = \vert \{\pi^{-1} (U_{p'})_{\beta'} \vert \beta' \in B'\} \vert\); Step 4: see that each \(U_p\) and \(U_{p'}\) are finite-open-sets-sequence-connected via evenly covered open neighborhoods; Step 5: conclude the proposition.
Step 1:
Let us see that for each connected neighborhood of \(p''\), \(N_{p''} \subseteq N_p\), \(\vert \{\pi^{-1} (N_p)_\beta \vert \beta \in B\} \vert = \vert \{\pi^{-1} (N_{p''})_{\beta} \vert \beta \in B\} \vert\).
Note that when we say "connected neighborhood", whether \(N_{p''}\) is regarded to be the subspace of \(N_p\) or be the subspace of \(T_2\) does not matter by the proposition that in any nest of topological subspaces, the connected-ness of any subspace does not depend on the superspace of which the subspace is regarded to be a subspace, and whether \(N_{p''}\) is regarded to be the neighborhood on \(N_p\) or be the neighborhood on \(T_2\) does not matter by the proposition that for any topological space and any point on any subspace, the intersection of any neighborhood of the point on the base space and the subspace is a neighborhood on the subspace and the proposition that for any topological space, any point, and any neighborhood of the point, any neighborhood of the point on the neighborhood is a neighborhood of the point on the base space.
Let us see that \(N_{p''}\) is evenly covered.
\(\pi^{-1} (N_{p''}) \subseteq \pi^{-1} (N_p) = \cup_{\beta \in B} \pi^{-1} (N_p)_\beta\).
Let \(\pi^{-1} (N_{p''})_\beta := \pi^{-1} (N_{p''}) \cap \pi^{-1} (N_p)_\beta\).
Let \(g_{p, \beta} := \pi \vert_{\pi^{-1} (N_p)_\beta}: \pi^{-1} (N_p)_\beta \to N_p\) be the homeomorphism.
\(\pi^{-1} (N_{p''})_\beta = {g_{p, \beta}}^{-1} (N_{p''})\), which is connected as the subspace of \(\pi^{-1} (N_p)_\beta\), by the proposition that any continuous image of any connected space is connected.
\(\pi^{-1} (N_{p''})_\beta\) is connected as the subspace of \(T_1\), by the proposition that in any nest of topological subspaces, the connected-ness of any subspace does not depend on the superspace of which the subspace is regarded to be a subspace.
All the \(\pi^{-1} (N_{p''})_\beta\) s are disjoint, because they are contained in the disjoint \(\pi^{-1} (N_{p})_\beta\) s.
So, \(\{\pi^{-1} (N_{p''})_{\beta} \vert \beta \in B\}\) is nothing but the connected components of \(\pi^{-1} (N_{p''})\).
Each \(\pi \vert_{\pi^{-1} (N_{p''})_\beta}: \pi^{-1} (N_{p''})_\beta \to N_{p''}\) is a homeomorphism, because it is the restriction of homeomorphic \(g_{p, \beta}\), by the proposition that any restriction of any continuous map on the domain and the codomain is continuous.
So, \(\vert \{\pi^{-1} (N_p)_\beta \vert \beta \in B\} \vert = \vert \{\pi^{-1} (N_{p''})_{\beta} \vert \beta \in B\} \vert\).
Step 2:
Although we are talking about not-necessarily open \(N_p\) s, for each \(N_p\), there is an open neighborhood of \(p\), such that \(U_p \subseteq N_p\), and by Step 1, we know that the cardinality for \(N_p\) equals the cardinality for \(U_p\).
So, if the cardinalities for all the \(U_p\) s are the same, the cardinalities for all the \(N_p\) s will be the same.
So, let us concentrate on \(U_p\) s.
Step 3:
Let us suppose that \(U_p \cap U_{p'} \neq \emptyset\).
Let us see that \(\vert \{\pi^{-1} (U_p)_\beta \vert \beta \in B\} \vert = \vert \{\pi^{-1} (U_{p'})_{\beta'} \vert \beta' \in B'\} \vert\).
\(U_p \cap U_{p'}\) is an evenly covered open neighborhood of any \(p'' \in U_p \cap U_{p'}\).
By Step 1, \(\vert \{\pi^{-1} (U_p)_\beta \vert \beta \in B\} \vert = \vert \{\pi^{-1} (U_p \cap U_{p'})_\beta \vert \beta \in B\} \vert\) and \(\vert \{\pi^{-1} (U_{p'})_\beta \vert \beta \in B\} \vert = \vert \{\pi^{-1} (U_p \cap U_{p'})_\beta \vert \beta \in B\} \vert\).
So, \(\vert \{\pi^{-1} (U_p)_\beta \vert \beta \in B\} \vert = \vert \{\pi^{-1} (U_{p'})_{\beta'} \vert \beta' \in B'\} \vert\): \(B'\) can be taken to be \(B\).
Step 4:
Let us see that each \(U_p\) and \(U_{p'}\) are finite-open-sets-sequence-connected via evenly covered open neighborhoods: refer to the definition of finite-open-sets-sequence-connected pair of open sets.
\(T_2\) is covered by some evenly covered open neighborhoods, \(\{U_p \vert p \in T_2\}\).
By the proposition that any pair of elements of any open cover of any connected topological space is finite-open-sets-sequence-connected via some elements of the open cover, \(U_p\) is connected to \(U_{p'}\) by a finite sequence, \((U_p, U_{p_1}, ..., U_{p_k}, U_{p'})\).
Step 5:
As \(U_p \cap U_{p_1} \neq \emptyset\), by Step 3, \(\vert \{\pi^{-1} (U_p)_\beta \vert \beta \in B\} \vert = \vert \{\pi^{-1} (U_{p_1})_{\beta} \vert \beta \in B\} \vert\).
Likewise, \(\vert \{\pi^{-1} (U_{p_1})_{\beta} \vert \beta \in B\} \vert = \vert \{\pi^{-1} (U_{p_2})_{\beta} \vert \beta \in B\} \vert\).
...
Likewise, \(\vert \{\pi^{-1} (U_{p_k})_{\beta} \vert \beta \in B\} \vert = \vert \{\pi^{-1} (U_{p'})_{\beta} \vert \beta \in B\} \vert\).
So, \(\vert \{\pi^{-1} (U_p)_\beta \vert \beta \in B\} \vert = ... = \vert \{\pi^{-1} (U_{p'})_{\beta} \vert \beta \in B\} \vert\).
