Quotient Space of Hausdorff space

Solution 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.