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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 $X \cong (\mathbb{S}^I/\sim)$?

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  • $\begingroup$ No, you can't get a nontrivial discrete space this way. $\endgroup$
    – Keith Kearnes
    Aug 20, 2015 at 7:50
  • $\begingroup$ If you can quickly post an argument for your statement as an answer, we can close this question $\endgroup$
    – Dominic van der Zypen
    Aug 20, 2015 at 7:52
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    $\begingroup$ You probably know this, but it might be worth mentioning that every $T_0$-space is a subspace of a power of Sierpinski space. Hmm, is every space a quotient of a $T_0$-space? If so, then we can "fix" your question by pointing out that every space is a subquotient of a power of Sierpinski space. $\endgroup$
    – Andrej Bauer
    Aug 20, 2015 at 8:36
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    $\begingroup$ As an abstract homotopy theorist, I really like this question and I caution against closure. The Sierpinski space is a cool (counter)example and the comment of Andrej is saying something interesting about the category of topological spaces. I see no reason at all this question should be closed. $\endgroup$
    – David White
    Aug 20, 2015 at 13:06
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    $\begingroup$ Any quotient of a power of a Sierpinski space must be quasicompact. $\endgroup$
    – Joseph Van Name
    Aug 20, 2015 at 18:20

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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 $X$. \\

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