Does there exist a Borel (or even continuous) function $f:\mathcal{C}\to\mathcal{C}$, where $\mathcal{C}$ is the Cantor set (or Cantor space $2^\omega$) such that for every nonempty closed perfect set $P\subseteq\mathcal{C}$, $f|P$ maps surjectively onto $\mathcal{C}$?


Such functions (on $\mathbb{R}$) are called *perfectly everywhere surjective* here: <https://core.ac.uk/download/pdf/83599431.pdf>, but the maps constructed there rely on, in essence, a well-ordering of $\mathbb{R}$ and are likely far from any kind of measurability.


Hoping that the self-similarity of the Cantor set could be exploited here.



A Borel perfectly everywhere surjective function on the Cantor set
As suggested by the comment above, the answer is no. The relevant fact is that every nonmeager subset of $\mathcal{C}$ with the Baire property (in particular, any Borel set) contains a nonempty closed perfect set.


Suppose that there was a Borel function $f$ with this property. Observe that for every $p\in\mathcal{C}$, $f^{-1}(\{p\})$ is a comeager Borel set: $f^{-1}(\{p\})$ is a Borel set which meets every nonmeager set, since nonmeager subsets of $\mathcal{C}$ must contain perfect sets, so its complement is meager.


But then, if we take $p\in\mathcal{C}$, $f^{-1}(\{p\})$ contains a perfect set which maps only onto $p$, a contradiction.


This argument works for any function which is Baire measurable, a similar argument can be given for Lebesgue measurable.


