(May 2019) Let \(p:\widetilde X\to X\) and \(q:\widetilde Y\to Y\) be covering maps such that both \(\widetilde X\) and \(\widetilde Y\) are connected and simply connected. Show that for every map \(f:X\to Y\) there is a map \(\widetilde f:\widetilde X\to\widetilde Y\) such that \(f\circ p=q\circ\widetilde f\).
Solution
Choose \(\widetilde x_0\in\widetilde X\), let \(x_0=p(\widetilde x_0)\), and choose \(\widetilde y_0\in q^{-1}(f(x_0))\). Consider \[ f\circ p:\widetilde X\to Y. \] Since \(\widetilde X\) is simply connected, \[ (f\circ p)_*\pi_1(\widetilde X,\widetilde x_0)=1. \] Also \(q_*\pi_1(\widetilde Y,\widetilde y_0)=1\) because \(\widetilde Y\) is simply connected. Hence the covering-space lifting criterion is satisfied, so there is a lift \[ \widetilde f:\widetilde X\to\widetilde Y \] with \(\widetilde f(\widetilde x_0)=\widetilde y_0\) and \[ q\circ\widetilde f=f\circ p. \] Thus every map \(f:X\to Y\) lifts after precomposition with the simply connected cover of \(X\).