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Let $\mathbb P$ denote the space of irrational numbers. In an answer to this question, Taras Banakh showed that the perfect images of $\mathbb P$ are precisely the Polish spaces with no compact neighborhoods. Here, perfect means a continuous, closed, surjective mapping with compact point pre-images.

Increasing the dimension slightly, we go from $\mathbb P$ to complete Erdős space $$\mathfrak E_{\mathrm{c}}=\{x\in \ell^2:x_n\in \mathbb P\text{ for all }n<\omega\}.$$ Here, $\ell^2$ is the Hilbert space of square-summable sequences of real numbers.

Question 1. Is every perfect image of $\mathfrak E_{\mathrm{c}}$ homeomorphic to $\mathfrak E_{\mathrm{c}}$?

Question 2. Is $\mathbb P$ a perfect image of $\mathfrak E_{\mathrm{c}}$?

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  • $\begingroup$ What about the original Erdos space? (Does everybody but I know?) $\endgroup$
    – Wlod AA
    Commented Apr 19, 2020 at 1:28
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    $\begingroup$ @WlodAA That one is usually more difficult to work with because it does not have as many representations, is not Polish, etc. But that may be a good follow-up question. $\endgroup$
    – D.S. Lipham
    Commented Apr 19, 2020 at 1:37

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It seems that the answer to Question 1 is "no".

According to this paper, the Julia set of $f(z)=\pi\sinh(z)$ is equal to the entire complex plane $\mathbb C$, and is the perfect image of a "Cantor bouquet". The endpoint set of any Cantor bouquet is homeomorphic to $\mathfrak E_{\mathrm{c}}$. But according to the image below (from the same paper), these endpoints are mapped to a dendritic connected set (see the dark lines including the imaginary axis). enter image description here

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