2024-03-31

520: Covering Map into Simply Connected Topological Space Is Homeomorphism

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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


  • 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:
T1: { the connected topological spaces }{ the locally path-connected topological spaces }
T2: { the connected topological spaces }{ the locally path-connected topological spaces }{ the simply connected topological spaces }
π: :T1T2, { the covering maps }
//

Statements:
π{ the homeomorphisms }.
//


2: Natural Language Description


For any connected and locally path-connected topological spaces, T1,T2, where T2 is also simply connected, 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, is a homeomorphism.


3: Proof


For any point pT2, let us suppose that there were 2 points, p1,p2π1(p), p1p2. p1 and p2 would not be path-connected, because if there was a path, λ:IT1, such that λ(0)=p1 and λ(1)=p2, πλ would be a loop on T2 and πλcp where cp would be the constant path into {p}, because T2 would be simply connected; there would be the unique lifts, πλ~ such that πλ~(0)=p1 and cp~ such that cp~(0)=p1, 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 πλ~cp~ and πλ~(1)=cp~(1), by the proposition that the lifts, that start at any same point, of any path-homotopic paths are path-homotopic; but πλ~=λ and cp~=cp1, and πλ~(1)=λ(1)=p2 and cp~(1)=cp1(1)=p1, a contradiction.

So, there would be the path-connected component around each point, pαπ1(p), Sα. There would be no point on T1 that did not belong to any Sα, because for any point, pT1, p and π(p) would be path-connected (because T2 would be simply connected, which would imply being path-connected), so, there would be a loop, lπ(p),p:IT2, such that lπ(p),p(0)=lπ(p),p(1)=π(p) and lπ(p),p(1/2)=p; there would be the unique lift, lπ(p),p~, such that lπ(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 lπ(p),p~ would pass one of π1(p), so, p would be path-connected with one of π1(p).

So, T1=αSα, but each Sα would be open, by the proposition that any path-connected topological component is open and closed on any locally path-connected topological space, and {Sα} would be disjoint, so, if there were multiple points in π1(p), T1 would not be connected. So, there is only 1 point in π1(p). So, π is a bijection.

So, π1 is a map, which is continuous, because around each point, pT2, there is an open neighborhood, UpT2, such that π1|Up:Upπ1(Up) 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.


References


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