Covering space Hausdorff implies base space Hausdorff

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 $x,y$.

Solution 1:

Take distinct $y_1 \neq y_2$ in $X$. Then $F_1 = f^{-1}[\{y_1\}]$ and $F_2 = f^{-1}[\{y_2\}]$ are disjoint, finite and non-empty subsets of $\tilde{X}$. By Hausdorffness we can find disjoint open sets $U_1$ and $U_2$ around $F_1$ and $F_2$. Let $O_1$ be an evenly covered neighbourhood of $y_1$, and $O_2$ one of $y_2$.

Now, $f^{-1}[O_1] = \cup_{i=1}^k V_i$, where the $V_i$ are disjoint and open and $f|V_i$ is a homeomorphism between $V_i$ and $O_1$ for all $i$; define $V'_i = V_i \cap U_1$, and define $O'_1 = \cap_{i=1}^k f[V'_i]$, which is open, as a finite intersection of open sets.

Similarly we define $W'_i$ from the inverse image of $O_2$ and $U_2$, and the intersection of their images is called $O'_2$.

These $O'_1$ and $O'_2$ are the required disjoint open neighbourhoods as can be easily checked.

Solution 2:

Take $x \neq y$ in $X$ and choose $x' \in \tilde X$ mapping to $x$. Choose a neighborhood $U'$ of $x'$ and neighborhoods $V_{y'}$ for each $y' \in p^{-1}(y)$ such that none of these neighborhoods intersect. Let $U$ be the image of $U'$ in $X$. Define $V$ by pushing the $V_{y'}$ to $X$ first and then intersecting.

Show that $U$ and $V$ are neighborhoods of $x$ and $y$ that don't intersect.

Edit: Oops, forgot to say you'll want to choose the $V_{y'}$ so that no two intersect. I've edited my answer above to reflect this. You can do this by proving the following little lemma (if you don't already know it) and applying it to $\{x'\} \cup p^{-1}(x)$.

Lemma: Let $Y$ be Hausdorff and $\{v_1, \ldots, v_n\}$ a set of finitely many distinct points in $Y$. Then there exist neighborhoods $V_i$ of $v_i$ such that no two of the $V_i$ intersect.

Solution 3:

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.