# Perfect images of complete Erdős space

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 $$\mathfrak E_{\mathrm{c}}$$?

• What about the original Erdos space? (Does everybody but I know?) – Wlod AA Apr 19 at 1:28
• @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. – D.S. Lipham Apr 19 at 1:37

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). 