Mental block in the last step in showing $p$ is an open map

Question

Let $p:\tilde{X}\to X$ be a covering map.

Let $U$ be open in $\tilde X$.

Let $y\in P(U)$. $y$ has an evenly covered open neighborhood $V$, such that $p^{-1}(V)=\coprod A_i$, where $A_i$ 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 $X$.

I am stuck at the last step: how to show: $y\in p(A_i\cap U)$?

Answer 1

You know that there is at least some $x \in U \subseteq \tilde{X}$ with $p(x) = y$. This $x$ is in some $A_{i_0}$, because any point that maps to $y$ is in one of the $A_i$. Then $x \in U \cap A_{i_0}$, so $y \in p[U \cap A_{i_0}] \subseteq p[U]$ So $y$ is an interior point of $p[U]$.