Let $X$ be a compact Hausdorff space and $f:X \rightarrow Y$ be a quotient map. Show that the following are equivalent:
(a)$Y$ is an Hausdorff space,
(b)$f$ is closed map,
(c)The set $\lbrace (x_1,x_2) \in X\times X:f(x_1)=f(x_2) \rbrace$ is closed in $X\times X$
My attempt at solution:
** $(a\Rightarrow b)$ **
Let $A\subset X$ is closed set in $X$. Since $X$ is compact. $A$ is compact in $X$. As f is continuous $f(A)$ is compact. Since $Y$ is Hausdorff $f(A)$ closed. Then we obtain f is closed map.
** $(a\Rightarrow c)$**
Let $\Delta = \lbrace (x_1,x_2) \in X\times X:f(x_1)=f(x_2) \rbrace $,
$(a,b )\in X\times X \setminus \Delta$. Then $f(a)\neq f(b)$. Since $Y$ is Hausdorff.
There is neighbourhood U,V such that $f(a) \in U$ $f(b) \in V$,
$U\cap V= \emptyset$
$a \in f^{-1}(U)$,$b \in f^{-1}(V)$
$f^{-1}(U)$, $ f^{-1}(V)$ are open in X since f is continuous.
Then $f^{-1}(U) \times f^{-1}(V)$ open in $X\times X$
Claim : $f^{-1}(U) \times f^{-1}(V)\cap A = \emptyset$,
Let $(x,y) \in f^{-1}(U) \times f^{-1}(V)$,
Then $x \in f^{-1}(U)$ $y \in f^{-1}(V)$,
$f(x) \in U$, $f(y) \in V$, $(x,y) \in A $ , $f(x)=f(y)$,
$U \cap V= \emptyset$
It is a contradiction. Then
$(a,b) \in f^{-1}(U) \times f^{-1}(V) \subset X \times X \setminus \Delta$
$ X \times X \setminus \Delta$ is open. $\Delta$ is closed.
** $(c\Rightarrow a)$**
Let $\Delta$ is closed. Then $ X \times X \setminus \Delta$ is open. Let $(a,b) \in X \times X \setminus \Delta$, $f(a) \neq f(b)$. Since $ X \times X \setminus \Delta$ is open there is open neighbourhood $U \times V$ of $(a,b)$
$(a,b) \in U \times V \subset X \times X \setminus \Delta $
$U \times V \cap \Delta= \emptyset$...(*)
$f(a), f(b)\in Y$, $f(a) \neq f(b)$
$a \in U$,$f(a) \in f(U)$, $b \in V$, $f(b) \in f(V)$
$f(U),f(V)$ are open since $f$ is quotient.
Claim: $f(U)\cap f(V)= \emptyset$
Let $x \in f(U)\cap f(V)$, $x \in f(U)$,$f^{-1}(x) \in U$ $x \in f(V)$, $f^{-1}(x) \in V$
$f^{-1}(x) \in U \cap V$, $U \times V \cap \Delta \neq \emptyset$.
It is contradiction with (*).Then distinct points in $Y$, have disjoint neighbourhoods.
I must prove $(a\Rightarrow b \Rightarrow c \Rightarrow a)$ But I couldn't $b \Rightarrow c$ or $c \Rightarrow b$
We show $b\Rightarrow a$.
Since $X$ is Hausdorff, single points in $X$ are closed subsets. Since $f$ is closed and onto, for every $p\in Y$ there is $a\in X$ with $f(a)=p$, thus single points in $Y$ are also closed. Now let $p,q\in Y$, $p\neq q$, so the two subsets $f^{-1}(p),f^{-1}(q)$ are disjoint and closed in $X$. $X$ is compact and Hausdorff, thus normal, so there are two disjoint open $U,V\subset X$, with $f^{-1}(p)\subset U,f^{-1}(q)\subset V$. Note that $X\setminus U,X\setminus V$ are closed, and their union is the whole of $X$. Since $f$ is closed and onto, the subsets $f(X\setminus U),f(X\setminus V)$ are closed in $Y$, and their union is the whole of $Y$. Also, by construction, $p\not\in f(X\setminus U),q\not\in f(X\setminus V)$, and it follows that $Y$ is Hausdorff.