460: For Covering Map, Criterion for Lift of Continuous Map from Path-Connected Locally Path-Connected Topological Space to Exist
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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
Target Context
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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, , any covering map, , which means that is continuous and surjective and around any point, , there is a neighborhood, , that is evenly covered by , any path-connected locally path-connected topological space, , and any continuous map, , for any point, , and each point , there is a lift of , , such that , if and only if where and are the fundamental group homomorphisms induced by and and s are the fundamental groups.
2: Proof
The subspace, , may consist of multiple connected components, each denoted as where where is a possibly uncountable indices set.
We can take an open neighborhood, , as , because if is not open, there is an open neighborhood, , which is homeomorphic to each by , because is obviously bijective, is continuous with the domain and the codomain regarded as the subspaces of and respectively as a restriction of continuous , 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 , 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 and respectively.
Let us suppose that .
For any point, , as is path-connected, there is a path, such that and . is a path on and and . There is the unique lift of , such that , 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 .
When , can be taken to be the constant path, is the constant path, and the unique lift of such that is the constant path, and so, .
Let us confirm that does not depend on the choice of .
For any another path, such that and , , where the over-line denotes the reverse path, is a loop on that starts at . , which means that there is a loop on that starts at , , such that , which means that , which means that where means being path-homotopic.
There is the unique lift, such that , 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 , because and .
There is the unique lift, , such that , 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. , by the proposition that the lifts, that start at any same point, of any path-homotopic paths are path-homotopic. is a loop. . So, where , 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 , 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, .
So, does not depend on the choice of .
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Let us confirm that is continuous.
For any point, , there are an evenly covered open neighborhood of , , and the such that . As is continuous and is locally path-connected, there is an open path-connected neighborhood, , such that . For any point, , there is a path, , that connects to . Let us take the path, . such that , 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 , , because there is no such that , because otherwise, ; if there was an such that for an , as would not be contained in any 1 connected component of , would not be connected on , and so, would not be connected on , a contradiction against its being connected as a continuous image of connected .
So, . So, , continuous. As there is an open neighborhood at any point on where is continuous, is continuous on , 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, is a lift of , , such that .
Let us suppose that there is a lift, , such that .
For any loop, , such that , . Let us define . . . So, . So, .
References
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