TITLE: Covering space Hausdorff implies base space Hausdorff
QUESTION [11 upvotes]: There is an exercise problem in Hatcher's Algebraic Topology book asking to show that if $p:\tilde{X}\rightarrow X$ is a covering space with $p^{-1}(x)$ finite and nonempty for all $x\in X$, then $\tilde{X}$ is compact Hausdorff iff $X$ is compact Hausdorff.
I've managed to show each part of the statement except for $\tilde{X}$ Hausdorff $\Rightarrow X$ Hausdorff. The problem I encountered was that taking $x\neq y$ in $X$ and $\tilde{U},\tilde{V}$ disjoint open sets in $\tilde{X}$ with $p^{-1}(x)\subset\tilde{U}$ and $p^{-1}(y)\subset \tilde{V}$ (I'm using the assumption that $\tilde{X}$ is compact Hausdorff, thus normal), I could not use these to get disjoint open sets in $X$ separating $x,y$.

REPLY [3 votes]: A space $X$ is Hausdorff if and only if the diagonal is closed in $X \times X$. Now suppose $\tilde X$ is Hausdorff, so the diagonal $\Delta_{\tilde X}$ is closed in $\tilde X \times \tilde X$. Remark that $p \times p : \tilde X \times \tilde X \to X \times X$ is again a finite covering map, hence it is a closed map; and the surjectivity of $p$ implies that $(p\times p) (\Delta_{\tilde X}) = \Delta_{X}$, so $\Delta_X$ is closed in $X \times X$ and therefore $X$ is Hausdorff.