description/proof of that covering map into simply connected topological space is homeomorphism
Topics
About: topological space
The table of contents of this article
- Starting Context
- Target Context
- Orientation
- Main Body
- 1: Structured Description
- 2: Natural Language Description
- 3: Proof
Starting Context
- The reader knows a definition of covering map.
- The reader knows a definition of lift of continuous map by covering map.
- The reader knows a definition of simply connected topological space.
- The reader admits the proposition that for any covering map, there is the unique lift of any path for each point in the covering map preimage of the path image of any point on the path domain.
- The reader admits the proposition that the lifts, that start at any same point, of any path-homotopic paths are path-homotopic.
- The reader admits the proposition that any path-connected topological component is open and closed on any locally path-connected topological space.
- The reader admits the proposition that any map between topological spaces is continuous if the domain restriction of the map to each open set of a possibly uncountable open cover is continuous.
Target Context
- The reader will have a description and a proof of the proposition that any covering map into any simply connected topological space is a homeomorphism.
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:
\(T_1\): \(\in \{\text{ the connected topological spaces }\} \cap \{\text{ the locally path-connected topological spaces }\}\)
\(T_2\): \(\in \{\text{ the connected topological spaces }\} \cap \{\text{ the locally path-connected topological spaces }\} \cap \{\text{ the simply connected topological spaces }\}\)
\(\pi\): \(:T_1 \to T_2\), \(\in \{\text{ the covering maps }\}\)
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Statements:
\(\pi \in \{\text{ the homeomorphisms }\}\).
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2: Natural Language Description
For any connected and locally path-connected topological spaces, \(T_1, T_2\), where \(T_2\) is also simply connected, any covering map, \(\pi: T_1 \to T_2\), which means that \(\pi\) is continuous and surjective and around any point, \(p \in T_2\), there is a neighborhood, \(N_p \subseteq T_2\), that is evenly covered by \(\pi\), is a homeomorphism.
3: Proof
For any point \(p \in T_2\), let us suppose that there were 2 points, \(p_1, p_2 \in \pi^{-1} (p)\), \(p_1 \neq p_2\). \(p_1\) and \(p_2\) would not be path-connected, because if there was a path, \(\lambda: I \to T_1\), such that \(\lambda (0) = p_1\) and \(\lambda (1) = p_2\), \(\pi \circ \lambda\) would be a loop on \(T_2\) and \(\pi \circ \lambda \simeq c_p\) where \(c_p\) would be the constant path into \(\{p\}\), because \(T_2\) would be simply connected; there would be the unique lifts, \(\widetilde{\pi \circ \lambda}\) such that \(\widetilde{\pi \circ \lambda} (0) = p_1\) and \(\widetilde{c_p}\) such that \(\widetilde{c_p} (0) = p_1\), by the proposition that for any covering map, there is the unique lift of any path for each point in the covering map preimage of the path image of any point on the path domain, and \(\widetilde{\pi \circ \lambda} \simeq \widetilde{c_p}\) and \(\widetilde{\pi \circ \lambda} (1) = \widetilde{c_p} (1)\), by the proposition that the lifts, that start at any same point, of any path-homotopic paths are path-homotopic; but \(\widetilde{\pi \circ \lambda} = \lambda\) and \(\widetilde{c_p} = c_{p_1}\), and \(\widetilde{\pi \circ \lambda} (1) = \lambda (1) = p_2\) and \(\widetilde{c_p} (1) = c_{p_1} (1) = p_1\), a contradiction.
So, there would be the path-connected component around each point, \(p_\alpha \in \pi^{-1} (p)\), \(S_\alpha\). There would be no point on \(T_1\) that did not belong to any \(S_\alpha\), because for any point, \(p' \in T_1\), \(p\) and \(\pi (p')\) would be path-connected (because \(T_2\) would be simply connected, which would imply being path-connected), so, there would be a loop, \(l_{\pi (p'), p}: I \to T_2\), such that \(l_{\pi (p'), p} (0) = l_{\pi (p'), p} (1) = \pi (p')\) and \(l_{\pi (p'), p} (1/2) = p\); there would be the unique lift, \(\widetilde{l_{\pi (p'), p}}\), such that \(\widetilde{l_{\pi (p'), p}} (0) = p'\), by the proposition that for any covering map, there is the unique lift of any path for each point in the covering map preimage of the path image of any point on the path domain, and \(\widetilde{l_{\pi (p'), p}}\) would pass one of \(\pi^{-1} (p)\), so, \(p'\) would be path-connected with one of \(\pi^{-1} (p)\).
So, \(T_1 = \cup_\alpha S_\alpha\), but each \(S_\alpha\) would be open, by the proposition that any path-connected topological component is open and closed on any locally path-connected topological space, and \(\{S_\alpha\}\) would be disjoint, so, if there were multiple points in \(\pi^{-1} (p)\), \(T_1\) would not be connected. So, there is only 1 point in \(\pi^{-1} (p)\). So, \(\pi\) is a bijection.
So, \(\pi^{-1}\) is a map, which is continuous, because around each point, \(p \in T_2\), there is an open neighborhood, \(U_p \subseteq T_2\), such that \(\pi^{-1} \vert_{U_p}: U_p \to \pi^{-1} (U_p)\) is continuous, by the proposition that any map between topological spaces is continuous if the domain restriction of the map to each open set of a possibly uncountable open cover is continuous.
