If \(p: (\tilde{X}, \tilde{A}, \tilde{x}_0) \to (X, A, x_0)\) is a covering space with \(\tilde{A} = p^{-1}(A)\), show that the map \(p_*: \pi_n(\tilde{X}, \tilde{A}, \tilde{x}_0) \to \pi_n(X, A, x_0)\) is an isomorphism for all \(n > 1\).
Solution
Represent an element of \(\pi_n(X,A,x_0)\), \(n>1\), by a map of triples \[ f:(D^n,S^{n-1},s_0)\longrightarrow(X,A,x_0). \] Since \(D^n\) is simply connected, \(f\) has a unique lift \[ \widetilde f:(D^n,s_0)\longrightarrow(\widetilde X,\widetilde x_0). \] Because \(f(S^{n-1})\subset A\), we have \[ \widetilde f(S^{n-1})\subset p^{-1}(A)=\widetilde A. \] Thus \(\widetilde f\) represents a class in \(\pi_n(\widetilde X,\widetilde A,\widetilde x_0)\), proving surjectivity of \(p_*\).
If two lifted maps become relatively homotopic after applying \(p\), the relative homotopy \[ F:(D^n\times I,S^{n-1}\times I,\{s_0\}\times I)\to(X,A,x_0) \] has a unique lift after fixing its value at \((s_0,0)\). Uniqueness of lifting makes its two ends the prescribed lifted maps, and its boundary remains in \(p^{-1}(A)\). Hence the two lifted maps were already relatively homotopic.
Therefore \[ \boxed{p_*:\pi_n(\widetilde X,\widetilde A,\widetilde x_0)\xrightarrow{\cong}\pi_n(X,A,x_0)} \qquad(n>1). \]