(May 2016) Let \(p\colon\widetilde X\to X\) and \(q\colon\widetilde Y\to Y\) be covering spaces of path-connected, locally path-connected spaces \(X\) and \(Y\) with \(\widetilde X\) and \(\widetilde Y\) locally path-connected and simply connected. Show that if \(X\) and \(Y\) are homeomorphic, then \(\widetilde X\) and \(\widetilde Y\) are homeomorphic.
Solution
Let \(h:X\to Y\) be a homeomorphism. Choose \(\widetilde x_0\in\widetilde X\), put \(x_0=p(\widetilde x_0)\), choose \(\widetilde y_0\in q^{-1}(h(x_0))\), and consider \[ h\circ p:\widetilde X\to Y. \] Since \(\widetilde X\) is simply connected, the lifting criterion gives a lift \[ F:\widetilde X\to\widetilde Y \] of \(h\circ p\) with \(F(\widetilde x_0)=\widetilde y_0\): \[ q\circ F=h\circ p. \]
Similarly, since \(\widetilde Y\) is simply connected, \(h^{-1}\circ q\) has a lift \[ G:\widetilde Y\to\widetilde X \] chosen so that \(G(\widetilde y_0)=\widetilde x_0\): \[ p\circ G=h^{-1}\circ q. \] Then \[ p\circ(GF)=h^{-1}qF=h^{-1}hp=p. \] Thus \(GF\) and the identity map of \(\widetilde X\) are two lifts of \(p:\widetilde X\to X\) that agree at \(\widetilde x_0\). Since \(\widetilde X\) is connected, uniqueness of lifts gives \[ GF=\operatorname{id}_{\widetilde X}. \] Likewise \(FG=\operatorname{id}_{\widetilde Y}\). Hence \(F\) and \(G\) are inverse homeomorphisms, so \[ \widetilde X\cong\widetilde Y. \]