Hausdorff Property for a Covering Space of a Manifold $E\to M$. I want to show that if $\pi : E \to M$ is a topological covering map and $M$ is a manifold then $E$ is a manifold. I was reading this post which helped me for the second-countability. The OP says it is simple to show that $E$ is Hausdorff but I don't see it. If I take $p\neq q \in M$.
Either $\pi(p) \neq \pi(q)$. In this case since $M$ is Housdorff then take $U,V$ that separate $\pi(p)$ and $\pi(q)$ and $\pi^{-1}(U)$ and $\pi^{-1}(V)$ will separate $p$ and $q$.
Or $\pi(p) = \pi(q)$. I now consider an open $U \ni \pi(p)=\pi(q)$. Then I was told that the sheets of $\pi^{-1}(U)$ containing $p$ and $q$ are disjoints but I don't understand why.
 A: Like @ArcticChar commented, part of the definition of a covering map $\pi\colon E\to M$ is that if $U$ is an evenly covered neighborhood in $M$, then $\pi^{-1}(U)$ is a disjoint union $\bigsqcup_\alpha V_\alpha$ in $E$ such that for each $\alpha$, $\pi|V_\alpha\colon V_\alpha\to U$ is a homeomorphism. In particular, $\pi|V_\alpha$ is injective.
Do you see how to finish from here?
A: In the definition of a covering space $f:Y\to X$, there is a "local triviality" condition which states that for all $x\in X$, there exists an open neighborhood $U$ such that there exists an isomorphism
$$f^{-1}(U)\xrightarrow{\phi} U\times F,$$ for $F$ a discrete set with cardinality equal to the cardinality of the fibre of $f$ over $x$. Moreover, $\operatorname{pr}_1\circ \:\phi=f.$ It follows that if $p,q\in f^{-1}(x)$ we have that either $p=q$, or $p$ and $q$ lie in disjoint copies of $U$ over the fibre.
If this is unclear, you should think of how $\exp:\mathbb{R}\to S^1$ is a covering space and carry through the argument you are trying to understand.
