Quotients of powers of the Sierpinski space Is every space isomorphic to some quotient of a power of the Sierpinski space?
More precisely: Let $(X,\tau)$ be a topological space, and let $\mathbb{S} = (\{0,1\}, \{\emptyset, \{0\},\{0,1\})$ be the Sierpinski space. Is there a set $I$ and an equivalence relation $\sim$ on $\mathbb{S}^I$ such that $X \cong (\mathbb{S}^I/\sim)$?
 A: No, you can't get a nontrivial discrete space this way. Let $X$ be a space with disjoint, nonempty open sets $U$ and $V$ and let $\varphi\colon \mathbb S^I\to X$ be a quotient map. Then $\varphi^{-1}(U)$ and $\varphi^{-1}(V)$ must be disjoint nonempty open sets in $\mathbb S^I$. But the constant function with range $\{0\}$ belongs to every nonempty open subset of $\mathbb S^I$.
EDIT (8/20/15):
This is a response to Andrej Bauer's question
``Is every space a quotient of a T0-space?''.
Yes. Let $X$ be a space, let $I$ be an infinite
set with the indiscrete topology, give $X\times I$
the product topology (say $\pi$), then let $\gamma$
be the cofinite topology on $X\times I$. Let $\pi^+ = \pi\vee \gamma$
be the topology on $X\times I$ that is generated by $\pi$
and $\gamma$.  This topology has a basis consisting of sets
of the form ``$U\times I$ minus a finite set'', where $U$
is open in $X$.
Claim. $(X\times I,\pi^+)$ is $T_1$ and the first
projection $p\colon X\times I\to X$ is a quotient map.
(So an arbitrary space is a quotient of a $T_1$ space.)
Proof:
$\gamma$ is already $T_1$, so $\pi^+$ is. We only
need to argue that $Z\subseteq X$ is open in the original
topology on $X$ iff $Z\times I$ is open in $\pi^+$.
``Only if'' follows from the fact that if $Z$ is open
in $X$, then $Z\times I\in \pi\subseteq \pi^+$.
Conversely, assume that $Z\times I\in \pi^+$.
Choose any $z\in Z$.
For any $i\in I$, we have $(z,i)\in Z\times I$, which we assume is open
in $\pi^+$. There must be a basic open
set $B\in\pi^+$, which is a cofinite subset of some set $V\times I$
with $V$ open in $X$, such that $(z,i)\in B\subseteq Z\times I$.
But then $(z,i)\in V\times I\subseteq Z\times I$, too, so
$z\in V\subseteq Z$ in $X$. This establishes the openness of $Z$
in $X$. \\
