Covering map is an open map

We prove a lemma that the covering map p:\tilde{X}\to X is an open map.

Let U be open in \tilde{X}. Let y\in p(U), then y has an evenly covered open neighborhood V, such that p^{-1}(V)=\coprod A_i, where the A_i are disjoint open sets in \tilde{X}, and p|_{A_i}:A_i\to V is a homeomorphism. A_i\cap U is open in \tilde{X}, and open in A_i, so p(A_i\cap U) is open in U, thus open in X.

There exists x\in U such that y=p(x). Thus x\in p^{-1}(y)\subseteq p^{-1}(V) so x\in A_i for some i. Thus x\in A_i\cap U and thus y\in p(A_i\cap U)\subseteq p(U). This shows y is an interior point of p(U). Hence p(U) is open, thus p is an open map.

Author: mathtuition88

Leave a Reply

Fill in your details below or click an icon to log in: Logo

You are commenting using your account. Log Out /  Change )

Google photo

You are commenting using your Google account. Log Out /  Change )

Twitter picture

You are commenting using your Twitter account. Log Out /  Change )

Facebook photo

You are commenting using your Facebook account. Log Out /  Change )

Connecting to %s

This site uses Akismet to reduce spam. Learn how your comment data is processed.