# Quotient Space of Hausdorff space

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.

• 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
Commented Jul 7, 2011 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
Commented Jul 7, 2011 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.
Commented Jul 7, 2011 at 7:38
• In fact, it is easy to see that in all non-trivial cases, we can find an equivalence relation on a topological space $X$ such that the quotient topology is not Hausdorff. For example, let $X$ be a non-discrete topological space and choose a subset $A\subseteq X$ such that $A$ is not open in $X$. Define an equivalence relation on $X$ such that the equivalence classes are precisely $A$ and $X\setminus A$. The resulting quotient space is clearly not Hausdorff since the point corresponding to the equivalence class $X\setminus A$ is not a closed point. Commented Jul 7, 2011 at 7:42
• @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.
Commented Jul 7, 2011 at 7:43

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 $$g\cdot(1,1) =(1,1)$$. 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.

• 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? Commented Apr 24, 2012 at 2:56
• @Vadim: thanks. I am not aware of any name for the quotient space nor of any direct use in applications. It is an example showing the erratic behavior of the quotient near a hyperbolic fixed point of a dynamical system. See also Arnold's Cat Map for some nice pictures of a related system.
– t.b.
Commented Apr 24, 2012 at 8:16
• @t.b.: I guess there is a typo in the first example. Shouldn't it be so that the generator of $\mathbb{Z}_2$ sends $(t,k)$ to $(t,|1-k|)$ for nonzero $t$ and $(0,k)$ to $(0,k)$? Commented Nov 7, 2015 at 15:06
• Could you please suggest me reference for first remark you made. Regarding action of topological group Commented Feb 4, 2020 at 5:14
• So the first remark shows that open quotients (quotient map is open) of Hausdorff spaces are Hausdorff? Commented Oct 29, 2021 at 19:46