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Is it true that quotient space of a Hausdorff space is necessarily Hausdorff?

In the book "Algebraic Curves and Riemann Surfaces", by Miranda, the author writes:

"$\mathbb{}P^2$ can be viewed as the quotient space of $\mathbb{C}^3-\{0\}$ by the multiplicative action of $\mathbb{C}^*$. In this way, $\mathbb{}P^2$ inherits a Hausdorff topology, which is the quotient topology from the natural map from $\mathbb{C}^3-\{0\}$ to $\mathbb{}P^2$"

It is true that the complex projective plane $\mathbb{}P^2$ is Hausdorff, but the above reasoning by Miranda will be true if the statement in the question is true.

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You can get the Mobius strip as a (topological) quotient, by identifying the edges of a square in the right way (and, of course, each point is in its own class), but I don't know if this topological quotient can also be made into a group quotient. So at least in the case when the quotient is a topological one, the answer is no. –  gary Jul 7 '11 at 7:34
@Theo, Zev: but I think the article Zev links to refers to topological quotient, and not group quotients, i.e., spaces that result from group actions. –  gary Jul 7 '11 at 7:38
@gary: The Möbius strip is pretty Hausdorff... A better example would be the line with two origins (let $\mathbb{Z}/2$ act on the two lines by switching $(0,t)$ with $(1,t)$ except if $t = 0$) or, more drastically: consider $\mathbb{R}/\mathbb{Q}$. –  t.b. Jul 7 '11 at 7:38
@Zev: Yes, but the objection is (I think): given that you know that quotient's aren't Hausdorff in general, you cannot infer that in the special situation of group actions the statement doesn't hold. –  t.b. Jul 7 '11 at 7:43
@Zev: sure, I gave two examples in my second comment. I'm out of this discussion now. –  t.b. Jul 7 '11 at 7:48

1 Answer 1

I thought I'd better make my comments into an answer.

Short answer: No, the quotient space of a Hausdorff space need not be Hausdorff.

Here are some of the canonical examples (in view of the comments I formulate them using group actions):

  1. Let $\mathbb{Z}/2$ act on $\mathbb{R} \times \{0\} \cup \mathbb{R} \times \{1\}$ by $g\cdot(t,0) = (t,1)$ and $g\cdot (t,1) = (t,0)$ if $t \neq 0$ and $g\cdot(0,0) = (0,0)$ and $g\cdot(1,1) = 1$. Then the quotient space is the line with two origins which is certainly not Hausdorff.

  2. One could object that this is not a particularly good example because the action is not by homeomorphisms, and I'd have to agree with that. So here's a better example: Let $\mathbb{Z}$ act on $\mathbb{R}^2\smallsetminus\{0\}$ via the matrix $\begin{bmatrix}2&0\\0&1/2\end{bmatrix}$ (more precisely, define $n \cdot \begin{bmatrix}x\\y\end{bmatrix} = A^{n}\begin{bmatrix}x\\y\end{bmatrix}$). Then it is easy to see that the images of $\begin{bmatrix}1\\0\end{bmatrix}$ and $\begin{bmatrix}0\\1\end{bmatrix}$ in the quotient don't have disjoint neighborhoods.

  3. A more drastic example is the action of the additive subgroup $\mathbb{Q}$ on $\mathbb{R}$. The quotient $\mathbb{R}/\mathbb{Q}$ carries the trivial topology (because $\mathbb{Q}$ is dense in $\mathbb{R}$).

In a positive direction, I'd like to make the following remarks:

  1. If $G$ acts by homeomorphisms the quotient map $p: X \to X / G$ is always open (contrary to general quotient maps): this is because $V \subset X/G$ is open if and only if $p^{-1}(V) \subset X$ is open and $p^{-1}(p(U)) = \bigcup_{g \in G}gU$ is a union of open sets if $U \subset X$ is open. Therefore $X/G$ is Hausdorff if and only if the orbit equivalence relation is a closed subset of $X \times X$.

  2. If a group $G$ acts properly on the Hausdorff space $X$ then $X/G$ is Hausdorff. See e.g. my post on MO for some quick facts on proper actions and some references.

  3. If $G$ is a Lie group acting smoothly, properly and freely on a manifold $M$ then $M/G$ is a manifold. This can be found e.g. in Duistermaat-Kolk Lie groups, or Montgomery-Zippin, Topological transformation groups. Unfortunately, I can't give you more precise references, as Google doesn't let me look at the relevant pages. Update: Olivier Bégassat recommends J.M. Lee, Introduction to smooth manifolds for this (which I can only second, thanks!)

The last fact is pretty difficult to prove in this generality (and I hope I haven't forgotten a hypothesis).

Of course, your question about $P^2 = (\mathbb{C}^3 \smallsetminus \{0\}) / \mathbb{C}^{\ast}$ being a Hausdorff space is covered by remark 2, while remark 3 shows that $P^2$ is even a manifold.

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+1 This answer deserves far more upvotes that it has received. Unfortunately, the site is less active at this time of the day. I would give it many more votes if I could! –  Amitesh Datta Jul 7 '11 at 9:42
Thank you! I don't worry too much about the votes I get, else I'd lose lots of hair... –  t.b. Jul 7 '11 at 9:57
Another excellent reference for quotient manifolds by a Lie group action as described by you is Lee's Introduction to Smooth Manifolds. –  Olivier Bégassat Jul 7 '11 at 11:49
@Oliver: Thanks! Added. –  t.b. Jul 7 '11 at 12:01
Good answer! And nice example 2. I am wondering if the quotient space in the example is (or homeomorphic to) something known or useful in some applications? Or it is just an abstract construction useful as a counter-example only? –  Vadim Apr 24 '12 at 2:56

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