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I am looking for locally compact Hausdorff spaces $X$ with the following property:

If $f:X\to Y$ is a perfect map onto locally compact Hausdorff space $Y$, then $Y$ is zero-dimensional.

One can see that if $|X|<\mathfrak{c}$, then $Y$ being a Tychonoff space of size $|Y|<\mathfrak{c}$, is necessarily zero-dimensional. In fact in this case we only need $f$ to be a continuous surjective map.

The inspiration for this question is this answer, and it will help me to produce more interesting examples of this type.

If $f:X\to Y$ is an open perfect map and $X$ is zero-dimensional, then $Y$ is zero-dimensional, see e.g. Engelking's General Topology. However, I am allowing $f$ which aren't open.

I am looking for sufficient conditions on the space $X$ so that it has the above property.

Of particular interest would be an example of such space $X$, which contains a compact non-scattered subspace (in particular, is not scattered).

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    $\begingroup$ Here are two simple comments. First, in the case that $|X| < \frak{c}$, it is not even necessary that $f$ be continuous, as long as it is a surjection. Second, a compact space has this property if and only if it is scattered (because a non-scattered compact space admits a continuous map onto $[0,1]$ and a compact scattered space, or even a Lindel\"{o}f scattered space, is functionally countable). $\endgroup$
    – Anonymous
    Commented Nov 9 at 1:36

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Using @Anonymous' second comment one can show that your property characterizes locally compact scattered spaces.

For if $X$ is locally compact Hausdorff and not scattered then it contains a closed dense-in-itself subspace $F$. As $X$ is zero-dimensional we can take a compact clopen set $C$ that meets~$F$, and then $A=C\cap F$ is compact and dense-in-itself. By the second comment let $f:A\to[0,1]$ be continuous and onto. Form the adjunction space $Y=X\cup_f[0,1]$ (or replace $A$ by its quotient $[0,1]$); the quotient map $q:X\to Y$ is perfect and $Y$ is not zero-dimensional. See Theorem 2.4.13 in Engelking's General Topology and the material preceding it for details on the adjunctionn operation.

Conversely if $X$ is scattered and $f:X\to Y$ is perfect and onto then, by @Anonymous' second comment again, if $K$ is a compact subset of $Y$ then $f^{-1}[K]$ is compact as well, as it is scattered this implies that for every continuous $g:K\to[0,1]$ the range is countable and so $K$ is scattered. This implies that $Y$ is scattered too.

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  • $\begingroup$ Thank you, inspired by this answer I've shown that for any locally compact Hausdorff scattered space $Y$ there exists a space $X$ with only zero-dimensional compactifications and $\beta X\setminus X\cong Y$. Moreover, whenever $X$ is a strongly zero-dimensional space and $\beta X\setminus X$ is locally compact, then having only zero-dimensional compactifiactions is equivalent to $\beta X\setminus X$ being scattered. $\endgroup$
    – Jakobian
    Commented Nov 10 at 17:33

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