TITLE: Every topological space can be realized as the quotient of some Hausdorff space.
QUESTION [8 upvotes]: Prove that every topological space can be realized as the quotient of some Hausdorff space.

I tried to show this by using the intersection of two open sets in $x$ (for $f:z\to x$).

REPLY [3 votes]: Let me elaborate on the comment of mine that Slade mentioned above.  Using convergence of ultrafilters, there is a very conceptually simple way to do this.  First, let us suppose $X$ is a $T_1$ space.  Then the topology on $X$ is completely determined by the convergence of nonprincipal ultrafilters on $X$: for each nonprincipal ultrafilter $F$ on $X$, there is some subset $L(F)\subseteq X$ of points that $F$ converges to, and the topology on $X$ is the finest topology such that $F$ converges to $x$ for each $F$ and each $x\in L(F)$.  Given a nonprincipal ultrafilter $F$ on $X$ and a point $x\in X$, we can let $X_{F,x}$ be $X$ equipped with the finest topology for which $F$ converges to $x$.  Explicitly, every point in $X_{F,x}$ is open except $x$, and the neighborhoods of $x$ are the sets in $F$.  Since $F$ is nonprincipal, this topology is Hausdorff.  We can form the disjoint union $Y=\bigsqcup X_{F,x}$ where $(F,x)$ ranges over all pairs of a nonprincipal ultrafilter $F$ and a point $x\in L(F)$.  There is then a canonical continuous map $p:Y\to X$ which is the identity map on each $X_{F,x}$.  Since $p$ is continuous with respect to a topology on $X$ iff $F$ converges to $x$ for each $(F,x)$, we conclude that the given topology on $X$ is the finest topology that makes $p$ continuous.  That is, $p$ is a quotient map.
What can we do if $X$ is not $T_1$?  Well, we just have to detect the convergence of principal ultrafilters as well.  For instance, for every pair $(x,y)\in X\times X$ such that the principal ultrafilter at $x$ converges to $y$, we could take the space $\mathbb{N}\cup\{\infty\}$ and map each element of $\mathbb{N}$ to $x$ and $\infty$ to $y$.  This map is continuous with respect to a topology on $X$ iff the principal ultrafilter at $x$ converges to $y$.  So we can let $Z$ be the disjoint union of the $Y$ constructed above and a copy of $\mathbb{N}\cup\{\infty\}$ for each such $(x,y)$, and define a map $q:Z\to X$ as the joining of of $p$ and the maps defined above.  Then by a similar argument as in the $T_1$ case, $q$ is a quotient map.
Let me further remark that the space $Y$ used here is not just Hausdorff but normal.  It suffices to check that each $X_{F,x}$ is normal.  But this is trivial: given two disjoint closed sets, one of them does not contain $x$, and so is clopen.
(Note that the use of ultrafilters (and thus of the axiom of choice) can be eliminated by using all convergent filters containing the cofinite filter, rather than just nonprincipal ultrafilters.  In fact, it suffices to just consider for each point $x\in X$ the pair $(F,x)$, where $F$ is the filter generated by the neighborhoods of $x$ and the cofinite sets.)