Correspondence between regular coverings and orbit spaces of their deck transformation groups
Let
Proof
First note that
acts properly discontinuously (The deck transformation group acts properly discontinuously) and Ξ is a regular covering (Orbit space of a properly discontinuous effective group action). Since π is clearly constant for each fibre of π , there exists a function π such that Ξ¦ , and by Universal property this is continuous. Since π = Ξ¦ π is surjective so is π , and since π is regular and thus π is transitive Ξ is injective, because if Ξ¦ it follows Ξ¦ [ Λ π₯ 1 ] = Ξ¦ [ Λ π₯ 2 ] π ( Λ π₯ 1 ) = Ξ¦ π ( Λ π₯ 1 ) = Ξ¦ π ( Λ π₯ 2 ) = π ( Λ π₯ 2 ) and thus there exists
with πΎ β Ξ , implying Λ π₯ 2 = πΎ ( Λ π₯ 1 ) . Since both [ Λ π₯ 2 ] = [ Λ π₯ 1 ] and π are local homeomorphisms, so is π , in particular it is open. Therefore Ξ¦ is a homeomoprhism. Ξ¦
Footnotes
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2010, Algebraische Topologie, ΒΆ2.3.38, pp. 96ff β©