Locally path-connected, connected covering morphism is a covering
Let
Proof
Let
, and let Λ π₯ β² β Λ π β² be a path from Λ πΌ β² : π β Λ π β² to Λ π₯ β² 0 . Further let Λ π₯ β² and πΌ = π β Λ πΌ β² be the unique lift of Λ πΌ : π β Λ π with πΌ . Then Λ πΌ ( 0 ) = Λ π₯ 0 πΌ = π β Λ πΌ = π β π β Λ πΌ = π β Λ πΌ β² and thus both
and π β Λ πΌ are lifts of Λ πΌ β² over πΌ with π , so π β Λ πΌ ( 0 ) = Λ πΌ β² ( 0 ) = Λ π₯ β² 0 and in particular π β Λ πΌ = Λ πΌ β² . Thus π ( Λ πΌ ( 1 ) ) = Λ π₯ β² is surjective. π Let
. Then π₯ 1 β π has a open neighbourhood π₯ 1 that is evenly covered by both π and π (simply take the intersection of open neighbourhoods with respect to each covering) which we may assume to be connected without loss of generality (otherwise take the connected component containing π ). Now let π₯ 1 and { Λ π π } π β πΌ denote the sheets over { Λ π β² π } π β π½ in π and Λ π respectively. By connectedness it follows that for each Λ π β² , π β πΌ for exactly one π ( Λ π π ) β Λ π β² π . Fix some π β π½ and let π β πΌ as above. It follows π β π½ ( π βΎ Λ π π ) β 1 = ( π βΎ Λ π π ) β 1 β ( π βΎ Λ π β² π ) since
( π βΎ Λ π π ) β 1 β ( π βΎ Λ π β² π ) β ( π βΎ Λ π π ) = ( π βΎ Λ π π ) β 1 β ( π βΎ Λ π π ) = i d Λ π π ( π βΎ Λ π π ) β ( π βΎ Λ π π ) β 1 β ( π βΎ Λ π β² π ) = ( π β Λ π π ) β ( π β π βΎ Λ π π ) β 1 β ( π βΎ Λ π β² π ) = ( π β Λ π π ) β ( ( π βΎ Λ π β² π ) β ( π βΎ Λ π π ) ) β 1 β ( π βΎ Λ π β² π ) = ( π β Λ π π ) β ( π βΎ Λ π π ) β 1 β ( π βΎ Λ π β² π ) β 1 β ( π βΎ Λ π β² π ) = i d Λ π β² π hence
is a homeomorphism. Clearly ( π βΎ Λ π π ) : Λ π π β Λ π β² π , and so from above it follows that the former is some disjoint union of π β 1 ( Λ π β² π ) β π β 1 ( π ) . Therefore Λ π π is a locally path-connected, connected covering. π