A description/proof of for covering map, criterion for lift of continuous map from path-connected locally path-connected topological space to exist
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 knows a definition of path-connected topological space.
- The reader knows a definition of locally path-connected topological space.
- The reader admits the proposition that for any covering map, there is the unique lift of any continuous map from any real closed interval for each initial value.
- 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 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 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 for any covering map, the lift of the reverse of any path is the reverse of the lift of the path.
- The reader admits the proposition that for any covering map, the lift of the product of any 2 paths such that the product exists is the product of the lifts of the paths such that the product (of the lifts) exists.
- 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 a criterion for any covering map, for any lift of any continuous map from any path-connected locally path-connected topological space to exist.
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 path-connected locally path-connected topological space, \(T_3\), and any continuous map, \(f: T_3 \to T_2\), for any point, \(p_0 \in T_3\), and each point \(p'_0 \in \pi^{-1} (f (p_0))\), there is a lift of \(f\), \(\widetilde{f}: T_3 \to T_1\), such that \(\widetilde{f} (p_0) = p'_0\), if and only if \(f_* (\pi_1 (T_3, p_0)) \subseteq \pi_* (\pi_1 (T_1, p'_0))\) where \(f_*\) and \(\pi_*\) are the fundamental group homomorphisms induced by \(f\) and \(\pi\) and \(\pi_1\) s are the fundamental groups.
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.
Let us suppose that \(f_* (\pi_1 (T_3, p_0)) \subseteq \pi_* (\pi_1 (T_1, p'_0))\).
For any point, \(p \in T_3\), as \(T_3\) is path-connected, there is a path, \(\lambda_p: I \to T_3\) such that \(\lambda_p (0) = p_0\) and \(\lambda_p (1) = p\). \(f \circ \lambda_p: I \to T_2\) is a path on \(T_2\) and \(f \circ \lambda_p (0) = f (p_0)\) and \(f \circ \lambda_p (1) = f (p)\). There is the unique lift of \(f \circ \lambda_p\), \(\widetilde{f \circ \lambda_p}: I \to T_1\) such that \(\widetilde{f \circ \lambda_p} (0) = p'_0\), by the proposition that for any covering map, there is the unique lift of any continuous map from any real closed interval for each initial value. Let us define \(f' (p) := \widetilde{f \circ \lambda_p} (1)\).
When \(p = p_0\), \(\lambda_p\) can be taken to be the constant path, \(f \circ \lambda_p\) is the constant path, and the unique lift of \(f \circ \lambda_p\) such that \(\widetilde{f \circ \lambda_p} (0) = p'_0\) is the constant path, and so, \(f' (p_0) = \widetilde{f \circ \lambda_p} (1) = p'_0\).
Let us confirm that \(f' (p)\) does not depend on the choice of \(\lambda_p\).
For any another path, \(\lambda'_p: I \to T_3\) such that \(\lambda'_p (0) = p_0\) and \(\lambda'_p (1) = p\), \(l_p := \lambda_p \overline{\lambda'_p}\), where the over-line denotes the reverse path, is a loop on \(T_3\) that starts at \(p_0\). \(f_* ([l_p]) \in f_* (\pi_1 (T_3, p_0)) \subseteq \pi_* (\pi_1 (T_1, p'_0))\), which means that there is a loop on \(T_1\) that starts at \(p'_0\), \(l'_p: I \to T_1\), such that \(f_* ([l_p]) = \pi_* ([l'_p])\), which means that \([f \circ l_p] = [\pi \circ l'_p]\), which means that \(f \circ l_p \simeq \pi \circ l'_p\) where \(\simeq\) means being path-homotopic.
There is the unique lift, \(\widetilde{\pi \circ l'_p}\) such that \(\widetilde{\pi \circ l'_p} (0) = p'_0\), 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 \(\widetilde{\pi \circ l'_p} = l'_p\), because \(\pi \circ l'_p = \pi \circ l'_p\) and \(l'_p (0) = p'_0\).
There is the unique lift, \(\widetilde{f \circ l_p}\), such that \(\widetilde{f \circ l_p} (0) = p'_0\), 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. \(\widetilde{f \circ l_p} \simeq l'_p\), by the proposition that the lifts, that start at any same point, of any path-homotopic paths are path-homotopic. \(\widetilde{f \circ l_p}\) is a loop. \(f \circ l_p = f \circ (\lambda_p \overline{\lambda'_p}) = (f \circ \lambda_p) (f \circ \overline{\lambda'_p}) = (f \circ \lambda_p) \overline{(f \circ \lambda'_p)}\). So, \(\widetilde{f \circ l_p} = \widetilde{(f \circ \lambda_p) \overline{(f \circ \lambda'_p)}} = \widetilde{(f \circ \lambda_p)} \widetilde{\overline{(f \circ \lambda'_p)}}\) where \(\widetilde{\overline{(f \circ \lambda'_p)}} (0) = \widetilde{(f \circ \lambda_p)} (1)\), by the proposition that for any covering map, the lift of the product of any 2 paths such that the product exists is the product of the lifts of the paths such that the product (of the lifts) exists, but \(\widetilde{\overline{(f \circ \lambda'_p)}} = \overline{\widetilde{(f \circ \lambda'_p)}}\), by the proposition that for any covering map, the lift of the reverse of any path is the reverse of the lift of the path. So, \(\widetilde{(f \circ \lambda_p)} (1) = \widetilde{\overline{(f \circ \lambda'_p)}} (0) = \overline{\widetilde{(f \circ \lambda'_p)}} (0) = \widetilde{(f \circ \lambda'_p)} (1)\).
So, \(f' (p)\) does not depend on the choice of \(\lambda_p\).
\(\pi \circ f' (p) = \pi \circ \widetilde{f \circ \lambda_p} (1) = f \circ \lambda_p (1) = f (p)\).
Let us confirm that \(f'\) is continuous.
For any point, \(p \in T_3\), there are an evenly covered open neighborhood of \(f (p)\), \(U_{f (p)} \subseteq T_2\), and the \({\pi^{-1} (U_{f (p)})}_\alpha\) such that \(f' (p) \in {\pi^{-1} (U_{f (p)})}_\alpha\). As \(f\) is continuous and \(T_3\) is locally path-connected, there is an open path-connected neighborhood, \(U_p \subseteq T_3\), such that \(f (U_p) \subseteq U_{f (p)}\). For any point, \(p' \in U_p\), there is a path, \(\lambda_{p, p'}: I \to U_p\), that connects \(p\) to \(p'\). Let us take the path, \(\lambda_{p'} = \lambda_{p} \lambda_{p, p'}\). \(f' (p') = \widetilde{f \circ \lambda_{p'}} (1) = \widetilde{f \circ (\lambda_{p} \lambda_{p, p'})} (1) = \widetilde{(f \circ \lambda_{p}) (f \circ \lambda_{p, p'})} (1) = \widetilde{f \circ \lambda_{p}} \widetilde{f \circ \lambda_{p, p'}} (1)\) such that \(\widetilde{f \circ \lambda_{p, p'}} (0) = \widetilde{f \circ \lambda_{p}} (1) = f' (p)\), by the proposition that for any covering map, the lift of the product of any 2 paths such that the product exists is the product of the lifts of the paths such that the product (of the lifts) exists.
As \(f \circ \lambda_{p, p'} (I) \subseteq U_{f (p)}\), \(\widetilde{f \circ \lambda_{p, p'}} (I) \subseteq {\pi^{-1} (U_{f (p)})}_\alpha\), because there is no \(r \in I\) such that \(\widetilde{f \circ \lambda_{p, p'}} (r) \notin {\pi^{-1} (U_{f (p)})}\), because otherwise, \(\pi \circ \widetilde{f \circ \lambda_{p, p'}} (r) = f \circ \lambda_{p, p'} (r) \notin U_{f (p)}\); if there was an \(r \in I\) such that \(\widetilde{f \circ \lambda_{p, p'}} (r) \in {\pi^{-1} (U_{f (p)})}_{\alpha'}\) for an \(\alpha' \neq \alpha\), as \(\widetilde{f \circ \lambda_{p, p'}} (I)\) would not be contained in any 1 connected component of \({\pi^{-1} (U_{f (p)})}\), \(\widetilde{f \circ \lambda_{p, p'}} (I)\) would not be connected on \({\pi^{-1} (U_{f (p)})}\), and so, would not be connected on \(T_1\), a contradiction against its being connected as a continuous image of connected \(I\).
So, \(f' (p') = \widetilde{f \circ \lambda_{p}} \widetilde{f \circ \lambda_{p, p'}} (1) = \widetilde{f \circ \lambda_{p, p'}} (1) \in {\pi^{-1} (U_{f (p)})}_\alpha\). So, \(f'\vert_{U_p} = \pi_{p, \alpha}^{-1} \circ f\vert_{U_p}\), continuous. As there is an open neighborhood at any point on \(T_3\) where \(f'\) is continuous, \(f'\) is continuous on \(T_3\), 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.
So, \(f'\) is a lift of \(f\), \(\widetilde{f}\), such that \(\widetilde{f} (p_0) = p'_0\).
Let us suppose that there is a lift, \(\widetilde{f}\), such that \(\widetilde{f} (p_0) = p'_0\).
For any loop, \(l: I \to T_3\), such that \(l (0) = l (1) = p_0\), \(f_* ([l]) = [f \circ l]\). Let us define \(l' := \widetilde{f} \circ l: I \to T_1\). \(l' (0) = l' (1) = \widetilde{f} \circ l (0) = \widetilde{f} (p_0) = p'_0 = \widetilde{f} \circ l (1) = \widetilde{f} (p_0)\). \(\pi \circ l' = \pi \circ \widetilde{f} \circ l = f \circ l\). So, \(\pi_* ([l']) = [\pi \circ l'] = [f \circ l]\). So, \(f_* (\pi_1 (T_3, p_0)) \subseteq \pi_* (\pi_1 (T_1, p'_0))\).
