A continuous self map on the Cantor space $C = \{0,1\}^\mathbb{N}$ is a mapping $f = (f_i)_{i\in \mathbb{N}}$ such that each $f_i$ is a map from $C$ to $\{0,1\}$ that depends only on a finite number of coordinates.

If for each coordinate there is only a finite number of $f_i$'s that depend on it, it is easily seen that $f$ is also open when restricting its codomain to its range.

If we take the codomain to be the range, what are examples of open continuous maps that are not of this form, if any? Is there a similarly simple way to describe continuous open self maps on Cantor space in general? What about the case where the codomain is not the range but the full Cantor space?

Sorry if this is too elementary for this site.

EDIT: the first question has a trivial answer (simply take a diagonal embedding, see comments). The two other questions remain.

  • $\begingroup$ Here you encode "the Cantor space" as $\{0,1\}^{\mathbf{N}}$. $\endgroup$
    – YCor
    Mar 10, 2019 at 19:44
  • $\begingroup$ yes, with the product topology. Edited the question as there was a typo making it incomprehensible. $\endgroup$
    – alesia
    Mar 10, 2019 at 20:08
  • 2
    $\begingroup$ Write $f_i(x)=x_0$, that is, $f(x)=(x_0,x_0,\dots)$. So the image of $f$ is reduced to the diagonal $\{0,1\}$ (the constant functions) and hence $f$ is clearly open as map to its image, but $f$ does not satisfy "for each coordinate there's only a finite number of $f_i$ that depend on it", since it fails for the 0th coordinate. $\endgroup$
    – YCor
    Mar 10, 2019 at 23:34
  • $\begingroup$ hmm right indeed... Is it reasonable to think open maps (or open maps to their image) have no simple description/parametrization? Or might it make sense to search for one? $\endgroup$
    – alesia
    Mar 11, 2019 at 19:46


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