Let $\pi: X\to Y$ be a surjective function between the compact, metric and connected spaces $X$, $Y$, and $Y_0 = \{y\in Y: \#\pi^{-1}(y)>1\}$. Suppose that:
- $Y_0$ is dense in $Y$,
- $Y\setminus Y_0$ is a dense $G_\delta$ in $Y$, and
- $\#\pi^{-1}(y) \leq N$ for all $y\in Y$ and an universal constant $N$.
My question is: can a function $\pi$ like this exist? Maybe I need more hypothesis on the topology of $X$ and $Y$.
Observe that the connectedness is crucial: sturmian codings of irrational rotations of the circle are surjective functions $\pi:K\to S^1$ from the Cantor set (a totally disconnected space) and the circle such that $\#\pi^{-1}(y) = 2$ for $y$ in a countable dense subset $Y_0 \subseteq S^1$ and $\#\pi^{-1}(y) = 1$ for $y \in S^1\setminus Y_0$.
Condition (3) is also needed: if $f\colon[0,1]\to[0,1]$ is the Thomae's function, $X = \{(x,y)\in[0,1]^2 : 0\leq y\leq f(x)\}$ is the subgraph of $f$, $\pi: X\to[0,1]$ is the projection onto the first coordinate, and $Y_0 := \{y\in Y: \#\pi^{-1}(y)>1\} = [0,1]\cap\mathbb{Q}$, then $X$ is connected and (1),(2) hold, but $\pi(y)$ is an uncountable set when $y \in Y_0$.
After building these examples I am more convinced than at the beginning that $\pi$ must be injective.
I would appreciate any comment.
Best Answer
I think this example is a surjective non-injective function satisfying your hypothesis. $$\pi\colon[0,1]\to[0,1]$$ $$t\mapsto\left\lbrace\begin{array}{ll} t &\mbox{if $t$ is irrational}\\2\min\{t,1-t\}&\mbox{if $t$ is rational}\end{array}\right.$$
Here $Y_0=(\mathbb{Q}\cap[0,1])\setminus\{\frac{1}{2}\}$, which is dense in $[0,1$]; the complement $[0,1]\setminus(\mathbb{Q}\setminus \{\frac{1}{2}\}) $ is dense as well and form a $G_\delta$ set. And $card(\pi^{-1}(t))$ equals $1$ for irrational $t$ and $t=\frac{1}{2}$ but equals $2$ for rational $t\neq \frac{1}{2}$.