# When the quotient of a Hausdorff space is also Hausdorff?

I have to prove the following proposition.

Proposition 1.1. (quotient projections out of compact Hausdorff spaces are closed precisely if the codomain is Hausdorff)

Let $$\pi:(X,\tau_X)\to (Y, \tau_Y)$$ be a continuous function between topological spaces such that

1. $$(X,\tau)$$ is a compact Hausdorff topological space;

2. $$\pi$$ is a surjection and $$\tau_Y$$ is the corresponding quotient topology.

Then, $$(Y,\tau_Y)$$ is an Hausdorff topological space $$\Longleftrightarrow$$ $$\pi$$ is a closed map.

Proof.

The implicaton $$(Y,\tau_Y)$$ Hausdorff $$\Rightarrow \pi$$ is closed it's easy since maps from compact spaces to Hausdorff spaces are closed.

So, assume that $$\pi$$ is a closed map. We need to show that for every pair of distinct points $$y_1,y_2\in Y$$ there exist open neighbourhoods $$U_{y_1}, U_{y_2}$$ which are disjoint.

First notice that the singleton subsets $$\{x\},\{y\}\subseteq Y$$ are closed. This is because they are images of singleton subsets in $$X$$, by surjectivity of $$f$$, and because singletons in a Hausdorff space are closed, and because images under $$f$$ of closed subsets are closed, by the assumption that $$f$$ is a closed map.

It follows that the pre images

$$C_1=\pi^{-1}(\{y_1\}), \quad C_2=\pi^{-1}(\{y_2\})$$

are closed subsets of $$X$$.

Now since compact Hausdorff spaces are normal it follows that we may find disjoint open subset $$U_1, U_2\in \tau_X$$ tali che $$C_1\subseteq U_1$$ and $$C_2\subseteq U_2$$.

By this lemma

Lemma 2.1. Let $$f:X\to Y$$ a function. Then, a subset $$S\subseteq X$$ is $$f$$-saturated precisely if its complement $$X\smallsetminus S$$ is so.

we can find these $$U_1, U_2$$ such that they are both saturated subsets.

Now, by this other lemma

Lemma 2.3. Let

1. $$f:X\to Y$$ a closed map
2. $$C\subseteq X$$ a closed subset which is $$f$$-saturated
3. $$U$$ an open set that contains $$C$$. Then, there exists a smaller open set $$V$$ such that $$U\supset V\supset C$$ and $$V$$ is $$f$$- saturated.

we can conclude because $$\pi(U_i)$$ are open in $$(Y,\tau_Y)$$.

Is it correct? Can I use this fact to characterize $$T_2$$ quotient of $$T_2$$ compact spaces?

Any suggestions to show that if $$X$$ is $$T_2$$ and the quotient projection is open, then the quotient space is $$T_2$$??

It is a well known result (see Dugundji, J., Topology., Wm C Brown Publishers, 1989, pp. 140 (Theorem 1.6) that

Theorem A: if $$(X,\tau)$$ is an arbitrary topological space, $$R\subset X\times X$$ is an equivalence relation in $$X$$ and $$p:X\rightarrow X/R$$ is the identification map (which gives the quotient topology in $$X/R$$), then if $$R$$ is closed in $$X\times X$$ and $$p$$ is an open map, then $$X/R$$ is Hausdorff.

To see this, suppose $$p(x)\neq p(y)$$. This means that $$(x,y)\notin R$$. As $$R$$ is closed, there are open neighborhoods $$U$$ and $$V$$ of $$x$$ and $$y$$ respectively such that $$U\times V\cap R=\emptyset$$. Hence $$p(U)\cap p(V)=\emptyset$$ The assumption that $$p$$ is open implies that $$p(U)$$ and $$p(V)$$ are disjoint open neighborhoods of $$p(x)$$ and $$q(x)$$ respectively.

Comment: If $$X/R$$ is Hausdorff, then $$R$$ must be closed in $$X\times X$$: consider the map $$P:X\times X\rightarrow X/R \times X/R$$ defined as $$(x,y)\mapsto (p(x),p(y))$$. The diagonal $$D=\{(p(x),p(x):x\in X\}$$ is closed in $$X/R \times X/R$$ and $$R=P^{-1}(D)$$.

Another related results are the following:

Theorem B: If $$X$$ and $$Y$$ are topological spaces, $$Y$$ is Hausdorff, and $$f:X\rightarrow Y$$ is an injective and continuous map, then $$X$$ is Hausdorff.

To se this, notice that the inverse map $$f^{-1}:f(X)\rightarrow Y$$ is a closed bijection from the Hausdorff space $$f(X)$$ onto $$X$$.

Theorem C: If $$X$$ and $$Y$$ are topological spaces, $$Y$$ is Hausdorff, and $$f:X\rightarrow Y$$ is continuous, then under the equivalence relation $$K(f):=\{(x,x')\in X\times X: f(x)=f(x')\}$$, the space $$X/K(f)$$ is Hausdorff.

Theorem C follows from Thoerem B any using the map $$fp^{-1}:X/X(f)\rightarrow Y$$, where $$p:X\rightarrow X/K(f)$$ is the quotient map and $$fp^{-1}$$ is the map such that $$(fp^{-1})\circ p=f$$. Notice that $$fg^{-1}$$ is continuous and injective.

Hope this helps.

• Thanks a lot! So, if X is Hausdorff and the quotient map is open I need that the equivalence relation is closed to say that the quotient $X/\sim$ is Hausdorff, without this last hypotesis, can the quotient not to be Hausdorff? Commented Jan 11 at 6:59
• @SigmaAlgebra: If $X/\sim$ is Hausdorff, then $\sim$ is closed in $X\times X$. Commented Jan 11 at 7:58
• Ok, but the viceversa is true? If $\sim$ is closed in $X\times X$, then $X/\sim$ is Hausdorff with that hypotesis on $X$? Commented Jan 11 at 13:58
• @SigmaAlgebra: ... yes, provided that $p$ is open. Commented Jan 11 at 14:26