What does it mean to say that \(p:Y\to X\) is a covering map?
Let \(p:Y\to X\) be a covering map. Prove that, for any \(y\in Y\), the homomorphism \[ p_*:\pi_1(Y,y)\to\pi_1(X,p(y)) \] is injective.
Let \(p:Y\to X\) be a covering map with \(Y\) and \(X\) path-connected. Suppose that \(p_*\) is an isomorphism. Prove that \(p\) is a homeomorphism.
Solution
1. A covering map is a continuous surjection \(p:Y\to X\) such that every \(x\in X\) has an open neighborhood \(U\) for which\[ p^{-1}(U)=\bigsqcup_{\lambda\in\Lambda}V_\lambda, \]where each \(V_\lambda\) is open in \(Y\) and every restriction\[ p|_{V_\lambda}:V_\lambda\longrightarrow U \]is a homeomorphism.
Such a neighborhood \(U\) is called evenly covered, and the sets \(V_\lambda\) are its sheets. This is the definition requested in part (a).
2. For every covering map \(p:Y\to X\) and every \(y\in Y\), the induced homomorphism\[ p_*:\pi_1(Y,y)\longrightarrow\pi_1(X,p(y)) \]is injective.
Let \([\gamma]\in\pi_1(Y,y)\) satisfy \[ p_*[\gamma]=1. \] Then the loop \(p\circ\gamma\) is nullhomotopic in \(X\) relative to its basepoint \(p(y)\). Choose a homotopy \[ H:[0,1]\times[0,1]\to X \] from \(p\circ\gamma\) to the constant loop at \(p(y)\), with the two endpoint edges fixed at \(p(y)\).
By the homotopy lifting property for covering maps, \(H\) has a unique lift \[ \widetilde H:[0,1]\times[0,1]\to Y \] whose initial loop is \(\gamma\). Since the endpoint edges of \(H\) are constant at \(p(y)\) and begin at \(y\), their lifts are constant at \(y\). Hence the terminal loop of \(\widetilde H\) is the lift, based at \(y\), of the constant loop at \(p(y)\), so it is the constant loop at \(y\). Thus \(\gamma\) is nullhomotopic in \(Y\) relative to \(y\). Therefore \([\gamma]=1\), proving that \(p_*\) is injective.
3. In part (c), every fiber of \(p\) contains exactly one point.
Fix \(x\in X\) and suppose \[ y_0,y_1\in p^{-1}(x). \] Because \(Y\) is path-connected, choose a path \[ \gamma:[0,1]\to Y \] from \(y_0\) to \(y_1\). Then \(p\circ\gamma\) is a loop at \(x\).
The map \(p_*\) is surjective, so there is a loop \(\delta\) in \(Y\) based at \(y_0\) such that \[ p_*[\delta]=[p\circ\gamma] \] in \(\pi_1(X,x)\). Thus the loops \(p\circ\delta\) and \(p\circ\gamma\) are homotopic relative to the basepoint.
Lift such a homotopy through \(p\), beginning with the lift \(\delta\) based at \(y_0\). By uniqueness of lifted paths, the other boundary loop of the lifted homotopy is the lift \(\gamma\), since both lift \(p\circ\gamma\) and start at \(y_0\). The endpoints of lifts are constant during a homotopy relative to endpoints, so \(\delta\) and \(\gamma\) have the same terminal point. But \[ \delta(1)=y_0, \qquad \gamma(1)=y_1. \] Hence \(y_0=y_1\).
Since \(p\) is surjective, each fiber is nonempty, so every fiber is a singleton.
4. Therefore the covering map in part (c) is a homeomorphism.
By 3, \(p\) is bijective. A covering map is a local homeomorphism and therefore an open map: if \(W\subseteq Y\) is open and \(y\in W\), choose an evenly covered neighborhood \(U\) of \(p(y)\) and the sheet \(V\) containing \(y\); then \[ p(W\cap V) \] is an open neighborhood of \(p(y)\) contained in \(p(W)\). Thus \(p(W)\) is open.
A bijective continuous open map has continuous inverse, so \(p\) is a homeomorphism. This proves part (c).