A description/proof of that for covering map, 2 lifts of continuous map from connected topological space totally agree or totally disagree
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 lift of continuous map with respect to covering map.
- The reader admits the proposition that in any nest of topological subspaces, the openness of any subset on any subspace does not depend on the superspace of which the subspace is regarded to be a subspace.
- 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 open subset of any locally path-connected topological space is locally path-connected.
- The reader admits the proposition that any connected component on any locally path-connected topological space is open.
- The reader admits the proposition that any open set on any open topological subspace is open on the base space.
- The reader admits the proposition that any 2 continuous maps from any connected topological space into any topological space such that, for any point, if they (the maps) agree at the point, they agree on a neighborhood and if disagree at the point they disagree on a neighborhood, totally agree or totally disagree on the whole domain.
Target Context
- The reader will have a description and a proof of the proposition that for any covering map, any 2 lifts of any continuous map from any connected topological space totally agree or totally disagree.
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: Description
For any connected and locally path-connected topological spaces, \(T_1, T_2\), 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 \(\pi\), any connected topological space, \(T_3\), and any continuous map, \(f: T_3 \to T_2\), any 2 lifts of \(f\), \(f', f'': T_3 \to T_1\), agree at each point of \(T_3\) or disagree at each point of \(T_3\).
2: Proof
The subspace, \(\pi^{-1} (N_p)\), may consist of multiple connected components, each denoted as \({\pi^{-1} (N_p)}_\alpha\) where \(\alpha \in A_p\) where \(A_p\) is a possibly uncountable indices set.
We can take an open neighborhood, \(U_p \subseteq T_2\), as \(N_p\), because if \(N_p\) is not open, there is an open neighborhood, \(U_p \subseteq N_p\), which is homeomorphic to each \({\pi^{-1} (U_p)}_\alpha\) by \(\pi_{p, \alpha} := \pi\vert_{{\pi^{-1} (U_p)}_\alpha}: {\pi^{-1} (U_p)}_\alpha \to U_p\), because \(\pi_{p, \alpha}\) is obviously bijective, is continuous with the domain and the codomain regarded as the subspaces of \({\pi^{-1} (N_p)}_\alpha\) and \(N_p\) respectively as a restriction of continuous \(\pi\vert_{{\pi^{-1} (N_p)}_\alpha}: {\pi^{-1} (N_p)}_\alpha \to N_p\), by the proposition that any restriction of any continuous map on the domain and the codomain is continuous, and its inverse is continuous with the domain and the codomain regarded likewise as a restriction of continuous \({\pi\vert_{{\pi^{-1} (N_p)}_\alpha}}^{-1}\), likewise, but by the proposition that in any nest of topological subspaces, the openness of any subset on any subspace does not depend on the superspace of which the subspace is regarded to be a subspace, those maps are continuous also with the domain and the codomain regarded as subspaces of \(T_1\) and \(T_2\) respectively.
\({\pi^{-1} (U_p)}_\alpha\) is open on \(T_1\), because \(\pi^{-1} (U_p)\) is open on \(T_1\) and is locally path-connected, by the proposition that any open subset of any locally path-connected topological space is locally path-connected, \({\pi^{-1} (U_p)}_\alpha\) is open on \(\pi^{-1} (U_p)\), by the proposition that any connected component on any locally path-connected topological space is open, and \({\pi^{-1} (U_p)}_\alpha\) is open on \(T_1\), by the proposition that any open set on any open topological subspace is open on the base space.
\(\pi \circ f' = \pi \circ f'' = f\) by the definition of lift of continuous map with respect to covering map.
For any point, \(p \in T_3\), there are an \(U_{f (p)} \subseteq T_2\) and \(\{{\pi^{-1} (U_{f (p)})}_\alpha \subseteq T_1\}\). There are 2 possibilities: 1) \(f' (p) = f'' (p) \in {\pi^{-1} (U_{f (p)})}_\alpha\) for an \(\alpha\); 2) \(f' (p) \neq f'' (p)\) where \(f' (p) \in {\pi^{-1} (U_{f (p)})}_{\alpha'}\) and \(f'' (p) \in {\pi^{-1} (U_{f (p)})}_{\alpha''}\) for some \(\alpha' \neq \alpha''\).
Let us suppose the possibility 1). As \({\pi^{-1} (U_{f (p)})}_\alpha\) is open on \(T_1\) and \(f'\) is continuous, there is an open neighborhood, \(U'_p \subseteq T_3\), such that \(f' (U'_p) \subseteq {\pi^{-1} (U_{f (p)})}_\alpha\). Likewise, there is an open neighborhood, \(U''_p \subseteq T_3\), such that \(f'' (U''_p) \subseteq {\pi^{-1} (U_{f (p)})}_\alpha\). Let us define \(U_p := U'_p \cap U''_p\). For any \(p' \in U_p\), \(f' (p'), f'' (p') \in {\pi^{-1} (U_{f (p)})}_\alpha\), \(\pi_{p, \alpha} \circ f' (p') = \pi_{p, \alpha} \circ f'' (p')\), but as \(\pi_{p, \alpha}\) is bijective, \(f' (p') = f'' (p')\). So, for any point, \(p \in T_3\), if \(f' (p) = f'' (p)\), \(f'\) and \(f''\) agree on a neighborhood of \(p\).
Let us suppose the possibility 2). As \({\pi^{-1} (U_{f (p)})}_{\alpha'}\) is open on \(T_1\) and \(f'\) is continuous, there is an open neighborhood, \(U'_p \subseteq T_3\), such that \(f' (U'_p) \subseteq {\pi^{-1} (U_{f (p)})}_{\alpha'}\). Likewise, there is an open neighborhood, \(U''_p \subseteq T_3\), such that \(f'' (U''_p) \subseteq {\pi^{-1} (U_{f (p)})}_{\alpha''}\). Let us define \(U_p := U'_p \cap U''_p\). For any \(p' \in U_p\), \(f' (p') \in {\pi^{-1} (U_{f (p)})}_{\alpha'}\) and \(f'' (p') \in {\pi^{-1} (U_{f (p)})}_{\alpha''}\). As \({\pi^{-1} (U_{f (p)})}_{\alpha'} \cap {\pi^{-1} (U_{f (p)})}_{\alpha''} = \emptyset\), \(f' (p') \neq f'' (p')\). So, for any point, \(p \in T_3\), if \(f' (p) \neq f'' (p)\), \(f'\) and \(f''\) disagree on a neighborhood of \(p\).
So, \(f'\) and \(f''\) agree at each point of \(T_3\) or disagree at each point of \(T_3\), by the proposition that any 2 continuous maps from any connected topological space into any topological space such that, for any point, if they (the maps) agree at the point, they agree on a neighborhood and if disagree at the point they disagree on a neighborhood, totally agree or totally disagree on the whole domain.
