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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 $T_2$??

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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.

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  • $\begingroup$ 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? $\endgroup$ Commented Jan 11 at 6:59
  • $\begingroup$ @SigmaAlgebra: If $X/\sim$ is Hausdorff, then $\sim$ is closed in $X\times X$. $\endgroup$
    – Mittens
    Commented Jan 11 at 7:58
  • $\begingroup$ Ok, but the viceversa is true? If $\sim $ is closed in $X\times X$, then $X/\sim$ is Hausdorff with that hypotesis on $X$? $\endgroup$ Commented Jan 11 at 13:58
  • $\begingroup$ @SigmaAlgebra: ... yes, provided that $p$ is open. $\endgroup$
    – Mittens
    Commented Jan 11 at 14:26

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