## 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.