Image under covering map of Hausdorff space is Hausdorff? Let $\overline{X} \rightarrow X$ be a covering space, where $\overline{X}$ is a Hausdorff space. Does $X$ have to be Hausdorff?
I assume there is a counterexample, but I couldn't find one yet.
 A: Here's a counterexample.  Let $X=\mathbb{N}\cup\{\infty,\infty'\}$ where a set is open iff it is either contained in $\mathbb{N}$ or cofinite.  Let $\overline{X}=X\times\mathbb{Z}$ with the following topology (not the product topology).  A set $U\subseteq\overline{X}$ is open iff it satisfies the following two properties:


*

*For all $n\in\mathbb{Z}$ such that $(\infty,n)\in U$, $(m,n+m)\in U$ for all but finitely many $m\in\mathbb{N}$.

*For all $n\in\mathbb{Z}$ such that $(\infty',n)\in U$, $(m,n-m)\in U$ for all but finitely many $m\in\mathbb{N}$.


I now claim that the projection map $p:\overline{X}\to X$ is a covering map.  First, it is easy to check that $p$ is continuous.  Any $m\in\mathbb{N}$ has an evenly covered neighborhood, namely $\{m\}$ (since $p^{-1}(\{m\})=\{m\}\times\mathbb{Z}$ is discrete in $\overline{X}$).  Finally, I claim $\infty$ and $\infty'$ have evenly covered neighborhoods.  Indeed, if $V=X\setminus\{\infty'\}$, then $p^{-1}(V)$ is homeomorphic to $V\times\mathbb{Z}$ (with the product topology) via the map sending $(m,n)$ to $(m,n-m)$ and $(\infty,n)$ to $(\infty,n)$, so $V$ is an evenly covered neighborhood of $\infty$.  Similarly, $W=X\setminus\{\infty\}$ is evenly covered since $p^{-1}(W)\cong W\times \mathbb{Z}$ by mapping $(m,n)$ to $(m,n+m)$.
Finally, $X$ is not Hausdorff ($\infty$ and $\infty'$ cannot be separated by open sets) but $\overline{X}$ is Hausdorff (we can get disjoint neighborhoods of $(\infty,n)$ and $(\infty',n')$ since the collections of points of the form $(m,n+m)$ and $(m,n'-m)$ only intersect at at most one point).
