Countable-infinity-to-one function Are there continuous functions $f:I\to I$ such that $f^{-1}(\{x\})$ is countably infinite for every $x$? Here, $I=[0,1]$.
The question "Infinity-to-one function" answers is similar but without the condition that it be countable. (The range was $S^2$, not $I$, but the accepted answer also worked for $I$.)
I doubt one exists, since I haven't been able to come up with one, but I'm not sure. There's probably some topological reason why this is impossible.
 A: Here is an example.  Start with the Cantor function $g:[0,1]\to[0,1]$, i.e. the function that sends $x=\sum a_n/3^n$ to $\sum a_n/2^{n+1}$ if every $a_n$ is $0$ or $2$ and is locally constant off of the Cantor set $K$.  Note that for each dyadic rational $q\in(0,1)$, there is a unique (nondegenerate) interval $[a_q,b_q]$ with $a_q,b_q\in K$ such that $g(x)=q$ for all $x\in[a_q,b_q]$, and $[0,1]\setminus K$ is the disjoint union of the intervals $(a_q,b_q)$.  Define a function $h:[0,1]\to[0,1]$ by saying $h=g$ on $K$, and on each interval $[a_q,b_q]$, $h$ is a finite-to-one continuous surjection $[a_q,b_q]\to[q,q+1/2^n]$, where $2^n$ is the denominator of $q$ (in lowest terms), and $h(a_q)=h(b_q)=q$.
The function $h$ is clearly continuous when restricted to each interval $[a_q,b_q]$, and in particular is continuous off of $K$.  As you approach a point of $K$ without staying in a single interval $[a_q,b_q]$, you pass through infinitely many such intervals $[a_q,b_q]$, with the denominators of the numbers $q$ getting larger and larger, and so $h$ remains continuous because $g$ is continuous.  Thus $h$ is continuous on all of $[0,1]$.
I claim every point of $[0,1]$ except $0$ has countably infinitely many preimages under $h$ (since $h(x)\geq g(x)$ for all $x$, $0$ is the only preimage of $0$).  It is clear that every point has countably many preimages: $h$ agrees with $g$ on $K$ and $g$ is finite-to-one on $K$, and off of $K$, $h$ is finite-to-one on each of the countably many intervals $[a_q,b_q]$.  Now if $x\in[0,1]$ and $q\in(0,1)$ is any dyadic rational obtained by truncating a binary expansion of $x$ at some point, then by construction, $h$ takes the value $x$ somewhere on the interval $[a_q,b_q]$ (since $q\leq x\leq q+1/2^n$).  If $x\neq0$, then there are infinitely many different dyadic rationals $q\in(0,1)$ that can be obtained by truncating a binary expansion of $x$ (if $x$ is a dyadic rational, use the binary expansion of it that ends in $1$s).  So we find that $x$ must have infinitely many preimages unless $x=0$.
Finally, it is easy to modify $h$ to give $0$ infinitely many preimages.  For instance, define $f(x)=i(x)$ if $x\in[0,1/2]$ and $f(x)=h(2x-1)$ if $x\in[1/2,1]$, where $i:[0,1/2]\to[0,1]$ is any countable-to-one continuous function that achieves the value $0$ infinitely many times and satisfies $i(1/2)=0$ (it is easy to construct such a function by appropriately modifying the function $x\mapsto x\sin^2(1/x)$).  Then $f$ is a continuous function $[0,1]\to[0,1]$ with countably infinitely many preimages for each point.
