For a covering space \(p: \tilde{X} \to X\) and a subspace \(A \subset X\), let \(\tilde{A} = p^{-1}(A)\). Show that the restriction \(p: \tilde{A} \to A\) is a covering space.
Solution
1. Fix \(a\in A\) and choose an evenly covered open neighborhood \(U\subseteq X\) of \(a\) for the covering \(p:\widetilde X\to X\).
Since \(p\) is a covering map, there is an open set \(U\ni a\) and pairwise disjoint open subsets \(V_\lambda\subseteq\widetilde X\) such that \[ p^{-1}(U)=\coprod_{\lambda\in\Lambda}V_\lambda \] and each restriction \[ p|_{V_\lambda}:V_\lambda\longrightarrow U \] is a homeomorphism.
2. The set\[ U_A=U\cap A \]is an open neighborhood of \(a\) in the subspace \(A\).
This is the definition of the subspace topology.
3. Its inverse image under the restricted map is\[ p^{-1}(U_A)\cap\widetilde A = \coprod_{\lambda\in\Lambda}(V_\lambda\cap\widetilde A). \]
By definition, \[ \widetilde A=p^{-1}(A). \] Hence \[ p^{-1}(U_A) =p^{-1}(U\cap A) =p^{-1}(U)\cap p^{-1}(A) =\left(\coprod_\lambda V_\lambda\right)\cap\widetilde A. \] Intersecting preserves the disjointness of the sheets, giving the displayed decomposition. Each \(V_\lambda\cap\widetilde A\) is open in the subspace \(\widetilde A\).
4. For every \(\lambda\), the restriction\[ p:V_\lambda\cap\widetilde A\longrightarrow U_A \]is a homeomorphism.
The homeomorphism \[ p|_{V_\lambda}:V_\lambda\to U \] restricts to a homeomorphism between the inverse image of \(A\cap U\) and \(A\cap U\). But that inverse image is exactly \[ V_\lambda\cap p^{-1}(A)=V_\lambda\cap\widetilde A. \]
5. Therefore every point of \(A\) has an evenly covered neighborhood for\[ p|_{\widetilde A}:\widetilde A\to A, \]so this restriction is a covering map.
The point \(a\) in 1 was arbitrary, and <1>2–<1>4 verify the covering-space definition over \(U_A\).