Let $q:\widetilde{X} \rightarrow X$ be a covering space. Let $A \subseteq X$ and $\widetilde{A}= q^{-1}(A)$. Show that the restriction of $q$ to $\widetilde{A}$ is a covering space.
I know there is a similar question like this but really didnt get the solution there and my problem is still a bit different so I try again here.
My attempt: Assume $q$ is a covering map and let $B_n$ be evenly covered subsets of $X$. Then the family of all such $B_n$ is an open cover of $X$. We want to show that $q|_A:A\rightarrow \widetilde{A}$ is a covering space. The restriction is a covering map if $A$ is connected and locally path-connected and for every $a \in \widetilde{A}$, $a$ has an evenly covered neighborhood. The later is true since $\widetilde{A}$ is a universal covering space so every element must have an evenly covered neighborhood. Now how do I get from here?
Best Answer
Let $x\in A$ and let $U\subset X$ be open neighborhood of $x$ such that $q^{-1}(U)=\cup_j V_j$ where $V_j\cap V_i=\emptyset$ and $V_j\approx U$ for all $j$. Then $U\cap A$ is open in $A$ and $q^{-1}(U\cap A)=\cup_j (V_j\cap \tilde{A})$ satisfies $V_j\cap \tilde{A}\approx U\cap A$ (check this!) and $(V_j\cap \tilde{A})\cap (V_i\cap \tilde{A})=\emptyset$. The definition of a covering space.