Let \(p:\widetilde X\to X\) be a covering space. Show that if \(X\) is Hausdorff, then \(\widetilde X\) is also Hausdorff.
Solution
Let \(\widetilde x,\widetilde y\in\widetilde X\) be distinct.
1. If \(p(\widetilde x)\neq p(\widetilde y)\), then \(\widetilde x\) and \(\widetilde y\) have disjoint open neighborhoods.
Since \(X\) is Hausdorff, there are disjoint open \(U\ni p(\widetilde x)\) and \(V\ni p(\widetilde y)\). Then \(p^{-1}(U)\) and \(p^{-1}(V)\) are disjoint open neighborhoods of \(\widetilde x\) and \(\widetilde y\).
2. If \(p(\widetilde x)=p(\widetilde y)=x\), then \(\widetilde x\) and \(\widetilde y\) have disjoint open neighborhoods.
Choose an evenly covered open neighborhood \(U\) of \(x\), with \(p^{-1}(U)=\coprod_{\alpha\in A}V_\alpha\) and each \(p|_{V_\alpha}\colon V_\alpha\to U\) a homeomorphism. Let \(\widetilde x\in V_\alpha\) and \(\widetilde y\in V_\beta\). If \(\alpha=\beta\), injectivity of \(p|_{V_\alpha}\) would give \(\widetilde x=\widetilde y\); so \(\alpha\neq\beta\), and \(V_\alpha\), \(V_\beta\) are disjoint open neighborhoods.
3. Q.E.D.
Steps 1 and 2 cover both cases, so \(\widetilde X\) is Hausdorff.