Regular covering

Correspondence between regular coverings and orbit spaces of their deck transformation groups

Let 𝑝 :Λœπ‘‹ ↠𝑋 be a connected and locally path-connected regular covering and Ξ“ =Autπ–’π—ˆπ—π‘‹β‘(𝑝) be its deck transformation group. Let Λœπ‘‹/Ξ“ be the orbit space with projection π‘ž :Λœπ‘‹ β† Λœπ‘‹/Ξ“. Then there exists an isomorphism Ξ¦ :Λœπ‘‹/Ξ“ →𝑋 such that the following diagram commutes1: #m/thm/homotopy

https://q.uiver.app/#q=WzAsMyxbMCwyLCJYIl0sWzQsMiwiXFx0aWxkZSBYIC8gXFxHYW1tYSJdLFsyLDAsIlxcdGlsZGUgWCJdLFsyLDAsInAiLDIseyJzdHlsZSI6eyJoZWFkIjp7Im5hbWUiOiJlcGkifX19XSxbMiwxLCJxIiwwLHsic3R5bGUiOnsiaGVhZCI6eyJuYW1lIjoiZXBpIn19fV0sWzEsMCwiXFxQaGkiLDAseyJjdXJ2ZSI6LTF9XSxbMCwxLCIiLDEseyJjdXJ2ZSI6LTF9XV0=

Proof

First note that Ξ“ acts properly discontinuously (The deck transformation group acts properly discontinuously) and π‘ž is a regular covering (Orbit space of a properly discontinuous effective group action). Since 𝑝 is clearly constant for each fibre of π‘ž, there exists a function Ξ¦ such that 𝑝 =Ξ¦π‘ž, and by Universal property this is continuous. Since 𝑝 is surjective so is π‘ž, and since 𝑝 is regular and thus Ξ“ is transitive Ξ¦ is injective, because if Ξ¦[˜π‘₯1] =Ξ¦[˜π‘₯2] it follows

𝑝(˜π‘₯1)=Ξ¦π‘ž(˜π‘₯1)=Ξ¦π‘ž(˜π‘₯2)=𝑝(˜π‘₯2)

and thus there exists 𝛾 βˆˆΞ“ with ˜π‘₯2 =𝛾(˜π‘₯1), implying [˜π‘₯2] =[˜π‘₯1]. Since both 𝑝 and π‘ž are local homeomorphisms, so is Ξ¦, in particular it is open. Therefore Ξ¦ is a homeomoprhism.


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Footnotes

  1. 2010, Algebraische Topologie, ΒΆ2.3.38, pp. 96ff ↩