closed quotient maps and $T_{k}$ spaces with $k \in \{1,2,3\}$ Let X be a topological space with $q: X \rightarrow X\setminus{\sim}$ the resulting quotient map.
(i) q is a closed map $\implies$ for all $x \in X$ and for any open $U \subset X$ with $q^{-1}(\{q(x)\}) \subset U$ there exists an open $V \subset Y$ with $q^{-1}(\{q(x)\}) \subset q^{-1}(V) \subset U$
(ii) X is regular ($T_{1}$ and $T_{3}$ hold) and q is both open and closed $\implies$ Y is a Hausdorff space (:= $T_{2}$)
X is a $T_{1}$ space if for $x, y \in X$ with $x \neq y$ there exists open neighborhoods $U,V \subset X$ of $x,y$ with $x \notin V$ and $y \notin U$.
X is a $T_{3}$ space if for $C \subset X$ closed and $x 
\in X \setminus C$ there exists open neighborhoods $U,V \subset X$ of x and C with $U \cap C = \varnothing$.
(i) Assume q is closed. Take some $x \in X$ and let $U \subset X$ be open with $q^{-1}(\{q(x)\}) \subset U$. Then $X \setminus U \subset X$ is closed and thus is $q (X \setminus U) \subset Y$ closed. Then $Y \setminus q(X \setminus U) \subset Y$ is open. So there exists an open $V \subset Y$, namely $V := Y \setminus q(X \setminus U) \subset Y$.
Now I want to show that $q^{-1}(\{q(x)\}) \subset q^{-1}(V)$ and $q^{-1}(V) \subset U$ hold for this $V \subset Y$.
$q^{-1}(V) = q^{-1}(Y \setminus q(X \setminus U)) \subset U \iff q(q^{-1}(Y \setminus q(X \setminus U))) \subset q(U) \iff Y \setminus q(X \setminus U) \subset q(U)$ (since q is a quotient map and thus surjective) $\iff q(X \setminus U) \subset Y \setminus q(U)$. I don't know how to continue now, since I think I can't say that $q(X \setminus U) = q(X) \setminus q(U)$, right?
$q^{-1}(\{q(x)\}) \subset q^{-1}(V) \iff q(q^{-1}(\{q(x)\})) \subset q(q^{-1}(V)) \iff \{q(x)\} \subset V \iff q(x) \in V := Y \setminus q(X \setminus U)$
But I think that $q(x) \in q(X \setminus U)$ is also possible for $x \in X$. So then, not for all $x \in X$ we have that $q(x) \in V$. So maybe this V is not possible, I don't know.
(ii) I know that q is closed. So I think that I have to use (i). With (i) I can get an open $V \subset Y$. And maybe I can use that X is $T_{3}$. But I don't know how and in which order.
 A: (i) holds for all functions $f: X \to Y$; it's a standard characterisation of closed maps. Suppose $x \in X$ and $U$ is open with $f^{-1}[f\{x\}] \subseteq U$.
As $f$ is closed, $f[X\setminus U]$ is open in $Y$ so indeed define $V = Y\setminus f[X\setminus U]$, which is then open in $Y$. Note that $f(x) \in V$ in particular, which is useful to know.
Claim: $f^{-1}[V] \subseteq U$. Proof: let $p \in f^{-1}[V]$. Suppose $p \notin U$, then $p \in X\setminus U$ and so $f(p) \in f[X\setminus U]$ and $f(p) \notin V$ contradicting that $p \in f^{-1}[V]$. So $p \in U$ and the inclusion holds.
This shows (i), and we need no property of a quotient map. It's just a lemma for the remainder I suppose.

If $q$ is open-and-closed and $X$ is $T_3$ we need to show $Y$ is $T_2$, so let $y_1 \neq y_2$ in $Y$. We can find $x_1,x_2\in X$  so that $q(x_i)=y_i, i =1,2$. Let $F_i = q^{-1}[\{y_i\}]$ which are closed in $X$ (as $X$ is $T_1$ and $f$ is closed, $\{y_i\} = q[\{x_i\}]$ are closed in $Y$ and $q$ is continuous in particular, being quotient). Of course, $x_1 \notin F_2$
So we can find $U$ open containing $x_1$ and $V$ containing $F_2$ that are disjoint by $T_3$-ness, and (i) gives us an open $V'$ containing $y_2$
such that $q^{-1}[V'] \subseteq V$.
Note that $q[U]$ and $V'$ are disjoint: if $x \in U$ existed with $q(x) \in V'$ then $x \in U \cap q^{-1}[V'] \subseteq U \cap V = \emptyset$, contradiction.
As $q$ is open this shows that $f[U]$ and $V'$ are the required disjoint open neighbourhoods of $y_1$ and $y_2$ resp., and we're done.
