Let \(p : (\widetilde{X},y_0) \to (X,x_0)\) be a covering space projection, \(x_0 \in A \subseteq X\), and \(\iota : A \to X\) the inclusion map. Show that \[q = p|_{p^{-1}(A)} : (p^{-1}(A),y_0) \to (A,x_0)\] is also a covering space, and \[\ker(\iota_*) \subseteq \operatorname{im}(q_*) \subseteq \pi_1(A,x_0).\]
Solution
1. The restriction\[ q=p|_{p^{-1}(A)}:p^{-1}(A)\longrightarrow A \]is a covering map.
Let \(a\in A\). Since \(p:\widetilde X\to X\) is a covering map, there is an open neighborhood \(U\subseteq X\) of \(a\) such that \[ p^{-1}(U)=\coprod_{\lambda\in\Lambda}U_\lambda, \] where each \(U_\lambda\) is open in \(\widetilde X\) and \[ p|_{U_\lambda}:U_\lambda\longrightarrow U \] is a homeomorphism.
Set \[ V=U\cap A. \] Then \(V\) is an open neighborhood of \(a\) in the subspace \(A\). Moreover, \[ q^{-1}(V) = p^{-1}(U\cap A) = \coprod_{\lambda\in\Lambda} \left(U_\lambda\cap p^{-1}(A)\right). \] For each \(\lambda\), \[ U_\lambda\cap p^{-1}(A) = (p|_{U_\lambda})^{-1}(V), \] so it is open in the subspace \(p^{-1}(A)\), and restriction of the sheet homeomorphism gives a homeomorphism \[ q|_{U_\lambda\cap p^{-1}(A)}: U_\lambda\cap p^{-1}(A) \xrightarrow{\ \cong\ } V. \] Thus \(V\) is evenly covered by \(q\). Since \(a\) was arbitrary, \(q\) is a covering map.
2. Let\[ [\gamma]\in\ker(\iota_*), \]where \(\gamma:I\to A\) is a loop based at \(x_0\).The lift of \(\gamma\) through \(p\) starting at \(y_0\) is a loop.
The condition \[ [\gamma]\in\ker(\iota_*) \] means that the loop \[ \iota\circ\gamma:I\longrightarrow X \] is nullhomotopic relative to its basepoint \(x_0\).
Let \[ \widetilde\gamma:I\longrightarrow\widetilde X \] be the unique \(p\)-lift of \(\iota\circ\gamma\) satisfying \[ \widetilde\gamma(0)=y_0. \] Choose a based homotopy in \(X\) from \(\iota\circ\gamma\) to the constant loop at \(x_0\). The covering homotopy property lifts this homotopy starting from \(\widetilde\gamma\). The endpoints of the lifted paths remain fixed during the lifted homotopy because they lift the constant basepoint path. At the other end of the homotopy, the lift starting at \(y_0\) of the constant loop at \(x_0\) is the constant loop at \(y_0\). Therefore \[ \widetilde\gamma(1)=y_0. \] Hence \(\widetilde\gamma\) is a loop based at \(y_0\).
3. The lift \(\widetilde\gamma\) from 2 lies entirely in \(p^{-1}(A)\) and is a \(q\)-lift of \(\gamma\).
For every \(t\in I\), \[ p(\widetilde\gamma(t)) = \iota(\gamma(t)) = \gamma(t) \in A. \] Thus \[ \widetilde\gamma(t)\in p^{-1}(A) \] for all \(t\). On this subspace \(q\) is the restriction of \(p\), so \[ q\circ\widetilde\gamma=\gamma. \] By 2, \(\widetilde\gamma\) is a loop based at \(y_0\).
4. We have\[ \ker(\iota_*)\subseteq\operatorname{im}(q_*). \]
Let \([\gamma]\in\ker(\iota_*)\). By 3, there is a loop \(\widetilde\gamma\) in \(p^{-1}(A)\) based at \(y_0\) such that \[ q\circ\widetilde\gamma=\gamma. \] Therefore \[ q_*([\widetilde\gamma])=[\gamma], \] so \([\gamma]\in\operatorname{im}(q_*)\).
5. Finally,\[ \operatorname{im}(q_*)\subseteq\pi_1(A,x_0), \]and hence\[ \ker(\iota_*) \subseteq \operatorname{im}(q_*) \subseteq \pi_1(A,x_0). \]
By definition, \[ q_*: \pi_1(p^{-1}(A),y_0) \longrightarrow \pi_1(A,x_0). \] Thus its image is a subgroup of \(\pi_1(A,x_0)\). Combine this with 4.