I know covering maps reflect Hausdorffness in the sense that Hausdorff base implies Hausdorff total space. This is proven by cases: if two points are in the same fiber, the discreteness of fibers ensures a separation; if they're in different fibers, their images can be separated by opens in the Hausdorff base, and pulled back.

I'm wondering about the other direction - does a Hausdorff total space imply a Hausdorff base? I don't see why this holds but also can't think of any counterexample.

  • $\begingroup$ The covering of a manifold section of the wikipedia page seems to contain a counterexample. $\endgroup$ Dec 19, 2016 at 22:15
  • $\begingroup$ @MichaelBurr I don't understand that example :\ $\endgroup$
    – Arrow
    Dec 19, 2016 at 22:19
  • $\begingroup$ That's one of the simplest examples you're going to get. The group needs to be infinite. $\endgroup$
    – user98602
    Dec 19, 2016 at 22:45

1 Answer 1


Sketch behind the wikipedia example:

Start with the covering space $\mathbb{R}^2\setminus\{(0,0)\}$. Then, consider the relation where $(x,y)\simeq (2x,y/2)$ and extend it to an equivalence relation on $\mathbb{R}^2\setminus\{(0,0)\}$. Under this relation, $(0,1)$ is equivalent to $(0,2^n)$ for all integers $n$ and $(1,0)$ is equivalent to $(2^m,0)$ for all integers $m$.

Consider the quotient map $q$ under this relation. Our goal is to show that the images $q(0,1)$ and $q(1,0)$ do not have disjoint neighborhoods in the quotient space. Let $U$ be an open set containing $q(0,1)$ and $V$ an open set containing $q(1,0)$. Consider $q^{-1}(U)$ and $q^{-1}(V)$. Our goal is to show that these sets overlap.

Since $q^{-1}(U)$ and $q^{-1}(V)$ are open, $q^{-1}(U)$ contains an open disk of radius $\delta$ containing $(0,1)$ and $q^{-1}(V)$ contains an open disk of radius $\delta$ containing $(1,0)$. Now, look at how this disk distorts under the equivalence relation. These disks will stretch until they overlap with each other.

Let's make this more precise, the horizontal interval $(0,1)-(\delta,1)$ is contained within $q^{-1}(U)$ and the interval $(1,0)-(1,\delta)$ is contained within $q^{-1}(V)$. Applying the relation, we have that the interval $\left(0,\frac{1}{2^n}\right)-\left(2^n\delta,\frac{1}{2^n}\right)$ is also contained within $q^{-1}(U)$. Similarly, the interval $\left(\frac{1}{2^n},0\right)-\left(\frac{1}{2^n},2^n\delta\right)$ is contained within $q^{-1}(V)$. For $n$ sufficiently large, the two segments cross.

Hence $U$ and $V$ intersect and so the quotient is not Hausdorff.


You must log in to answer this question.

Not the answer you're looking for? Browse other questions tagged .