First note that Ξ acts properly discontinuously (The deck transformation group acts properly discontinuously) and π 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 Ξ¦ such that π =Ξ¦π,
and by Universal property this is continuous.
Since π is surjective so is π,
and since π is regular and thus Ξ is transitive Ξ¦ is injective,
because if Ξ¦[Λπ₯1] =Ξ¦[Λπ₯2] it follows
π(Λπ₯1)=Ξ¦π(Λπ₯1)=Ξ¦π(Λπ₯2)=π(Λπ₯2)and thus there exists πΎ βΞ with Λπ₯2 =πΎ(Λπ₯1), implying [Λπ₯2] =[Λπ₯1].
Since both π and π are local homeomorphisms, so is Ξ¦, in particular it is open.
Therefore Ξ¦ is a homeomoprhism.