Everyone knows that there is a closed equivalence relation $\sim$ on the Cantor set $C$ such that each non-trivial equivalence class has exactly $2$ points and $[0,1]\simeq C/\sim$. Thus a closed quotient of a zero-dimensional space may not be zero-dimensional, even if the equivalence classes are compact. Note that in this case, if we specify that $C/\sim$ must be totally separated, then $C/\sim$ is automatically zero-dimensional, because every totally separated compact metric space is zero-dimensional. *Totally separated* means that for every two points $x$ and $y$ in the space, there is a clopen set containing $x$ and missing $y$. *Zero-dimensional* means that the space has a basis of clopen sets.


This question is about similar quotients of the irrationals $\mathbb P$. 


**Question.** Let $\sim$ be a closed equivalence relation on $\mathbb P$ such that $\mathbb P/\sim$ is Polish and every equivalence class is compact. If $\mathbb P/\sim$ is totally separated, then is $\mathbb P/\sim$ necessarily zero-dimensional? 


Note that every Polish space is a closed quotient of $\mathbb P$; [shown here](https://www.jstor.org/stable/2036424). So the condition that the equivalence classes are compact is critical. I believe my question is equivalent to: *Is every totally disconnected Polish perfect image of $\mathbb P$ zero-dimensional?* A continuous mapping is *perfect* if it is closed has compact point preimages.



Quotients of the irrationals
The answer to this question is negative and follows from the characterization:



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> **Theorem.** A topological space $X$ is an image of the space of irrationals $\mathbb P$ under a perfect map $f:\mathbb P\to X$ if and only if $X$ is Polish and nowhere locally compact.
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*Proof.* To prove the "only if" part, assume that a topological space $X$ is the image of $\mathbb P$ under a perfect map $f:\mathbb P\to X$. By Theorem 3.7.20 of Engelking's "General Topology" (denoted later by [EGT]), the space $X$ is regular, by Theorem 3.7.19 in [EGT], $X$ is second-countable and by the Uryssohn Metrization Theorem, $X$ is metrizable and separable. By Theorem 3.9.10 of [EGT], the space $X$ is Cech-complete and being separable and metrizable is Polish. Assuming that $X$ contains a compact subset $K$ with non-empty interior, we can apply Theorem 3.7.2 of [EGT] and conclude that the preimage $f^{-1}(K)$ is compact and by the continuity of $f$, it has non-empty interior in $\mathbb P$. On the other hand, it is well-known that $\mathbb P$ contains no compact sets with nonempty interior. This contradiction shows that $X$ is nowhere locally compact.


To prove the "if" part, assume that $X$ is a nowhere locally compact Polish space. Let $\bar X$ be any metric compactification of $X$. By Theorem 4.18 in Kechris' "Classical Descriptive Set Theory" (denoted later by [CDST]), $\bar X$ is the image of the Cantor cube under a continuous map $\tilde g:2^\omega\to\bar X$. Using the Kuratowski-Zorn Lemma, choose a minimal closed subset $Z\subseteq 2^\omega$ such that $f(Z)=\bar X$. Let $g=\tilde g{\restriction}Z$. The minimality of $Z$ and closedness of the map $g:Z\to\bar X$ ensure that for any non-empty open set $U\subset Z$ the image $g(U)$ has nonempty interior in $\bar X$. 


Being Polish, the space $X$ is a $G\_\delta$-set in $\bar X$ and its preimage $P=g^{-1}(X)$ is a $G\_\delta$-set in the zero-dimensional compact space $Z$. We claim that $P$ is nowhere locally compact. Assuming that $P$ contains a compact subset $K$ with nonempty interior, we can use the minimality of $Z$ and conclude that $g(K)$ is a compact set with nonempty interior in $X$, which contradicts the nowhere local compactness of $X$. This contradiction shows that the Polish zero-dimensional space $P$ is nowhere locally compact. By the Aleksandrov-Urysohn characterization of $\mathbb P$ (see Theorem 7.7 in [CDST]), the space $P$ is homeomorphic to the space of irraionals $\mathbb P$. It remains to apply Proposition 3.7.6 of [EGT] to see that the restriction $f=g\restriction P:P\to X$ is perfect.


