2024-01-21

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


  • 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, T1,T2, any covering map, π:T1T2, which means that π is continuous and surjective and around any point, pT2, there is a neighborhood, NpT2, that is evenly covered by π, any path-connected locally path-connected topological space, T3, and any continuous map, f:T3T2, for any point, p0T3, and each point p0π1(f(p0)), there is a lift of f, f~:T3T1, such that f~(p0)=p0, if and only if f(π1(T3,p0))π(π1(T1,p0)) where f and π are the fundamental group homomorphisms induced by f and π and π1 s are the fundamental groups.


2: Proof


The subspace, π1(Np), may consist of multiple connected components, each denoted as π1(Np)α where αAp where Ap is a possibly uncountable indices set.

We can take an open neighborhood, UpT2, as Np, because if Np is not open, there is an open neighborhood, UpNp, which is homeomorphic to each π1(Up)α by πp,α:=π|π1(Up)α:π1(Up)αUp, because πp,α is obviously bijective, is continuous with the domain and the codomain regarded as the subspaces of π1(Np)α and Np respectively as a restriction of continuous π|π1(Np)α:π1(Np)αNp, 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 π|π1(Np)α1, 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 T1 and T2 respectively.

Let us suppose that f(π1(T3,p0))π(π1(T1,p0)).

For any point, pT3, as T3 is path-connected, there is a path, λp:IT3 such that λp(0)=p0 and λp(1)=p. fλp:IT2 is a path on T2 and fλp(0)=f(p0) and fλp(1)=f(p). There is the unique lift of fλp, fλp~:IT1 such that fλp~(0)=p0, 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):=fλp~(1).

When p=p0, λp can be taken to be the constant path, fλp is the constant path, and the unique lift of fλp such that fλp~(0)=p0 is the constant path, and so, f(p0)=fλp~(1)=p0.

Let us confirm that f(p) does not depend on the choice of λp.

For any another path, λp:IT3 such that λp(0)=p0 and λp(1)=p, lp:=λpλp, where the over-line denotes the reverse path, is a loop on T3 that starts at p0. f([lp])f(π1(T3,p0))π(π1(T1,p0)), which means that there is a loop on T1 that starts at p0, lp:IT1, such that f([lp])=π([lp]), which means that [flp]=[πlp], which means that flpπlp where means being path-homotopic.

There is the unique lift, πlp~ such that πlp~(0)=p0, 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 πlp~=lp, because πlp=πlp and lp(0)=p0.

There is the unique lift, flp~, such that flp~(0)=p0, 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. flp~lp, by the proposition that the lifts, that start at any same point, of any path-homotopic paths are path-homotopic. flp~ is a loop. flp=f(λpλp)=(fλp)(fλp)=(fλp)(fλp). So, flp~=(fλp)(fλp)~=(fλp)~(fλp)~ where (fλp)~(0)=(fλp)~(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 (fλp)~=(fλp)~, 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, (fλp)~(1)=(fλp)~(0)=(fλp)~(0)=(fλp)~(1).

So, f(p) does not depend on the choice of λp.

πf(p)=πfλp~(1)=fλp(1)=f(p).

Let us confirm that f is continuous.

For any point, pT3, there are an evenly covered open neighborhood of f(p), Uf(p)T2, and the π1(Uf(p))α such that f(p)π1(Uf(p))α. As f is continuous and T3 is locally path-connected, there is an open path-connected neighborhood, UpT3, such that f(Up)Uf(p). For any point, pUp, there is a path, λp,p:IUp, that connects p to p. Let us take the path, λp=λpλp,p. f(p)=fλp~(1)=f(λpλp,p)~(1)=(fλp)(fλp,p)~(1)=fλp~fλp,p~(1) such that fλp,p~(0)=fλp~(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 fλp,p(I)Uf(p), fλp,p~(I)π1(Uf(p))α, because there is no rI such that fλp,p~(r)π1(Uf(p)), because otherwise, πfλp,p~(r)=fλp,p(r)Uf(p); if there was an rI such that fλp,p~(r)π1(Uf(p))α for an αα, as fλp,p~(I) would not be contained in any 1 connected component of π1(Uf(p)), fλp,p~(I) would not be connected on π1(Uf(p)), and so, would not be connected on T1, a contradiction against its being connected as a continuous image of connected I.

So, f(p)=fλp~fλp,p~(1)=fλp,p~(1)π1(Uf(p))α. So, f|Up=πp,α1f|Up, continuous. As there is an open neighborhood at any point on T3 where f is continuous, f is continuous on T3, 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, f~, such that f~(p0)=p0.

Let us suppose that there is a lift, f~, such that f~(p0)=p0.

For any loop, l:IT3, such that l(0)=l(1)=p0, f([l])=[fl]. Let us define l:=f~l:IT1. l(0)=l(1)=f~l(0)=f~(p0)=p0=f~l(1)=f~(p0). πl=πf~l=fl. So, π([l])=[πl]=[fl]. So, f(π1(T3,p0))π(π1(T1,p0)).


References


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