Set of odd points of continuous maps $S^1\to \mathbb{R}$ Does there exist a continuous map $f:S^1\to \mathbb{R}$ which for an odd number $o\in \mathbb{N}$ there be infinitely many points in $f(S^1)$ with the property that the preimage of each one has exactly $o$ points in $S^1$ ?
 A: Fix two points $p,q\in S^1$ and start with a function $g:S^1\to\mathbb{R}$ which has $g(p)=0$ and $g(q)=1$, and $g$ interpolates monotonically between them on each of the arcs from $p$ to $q$.  Now modify $g$ on one of those arcs to have infinitely many brief dips downward which accumulate at $q$.  If we arrange that the images of these dips downward do not overlap, then the local minimum at the end of each dip downward will have exactly $3$ preimages: once while going up in each arc, and a third time at the bottom of the dip downward.  So there will be infinitely many points with exactly $3$ preimages.
Note though that the set of values with an odd number of preimages must always be countable.  Indeed, if a value $c$ has only finitely many preimages, then we can split $S^1\setminus f^{-1}(\{c\})$ into $|f^{-1}(c)|$ arcs such that on each arc, $f$ stays on one side of $c$.  Moreover, unless $c$ is a local maximum or minimum value of $f$, then $f$ alternates which side of $c$ it is on these arcs.  Since the arcs form a loop, there must be an even number of such alternations and thus an even number of arcs. That is, $|f^{-1}(c)|$ must be even unless $c$ is a local maxiumum or minimum value of $f$.  But $f$ can have only countably many local maxima or minima, because each one must be the global maximum or minimum on some basic open set in any countable basis for the topology of $S^1$.
A: Yes, there does exist such a function.
[EDIT: Based on Eric Wofsey's comments I realized that this example does not work. Leaving it up since it is an instructive mistake.]
I played around with giving some explicit examples before realizing there is a much more elegant [EDIT: almost] solution: just consider a sample path of Brownian motion on the circle. (More precisely, let $(B_t\colon t\in[0,1])$ be a Brownian bridge with $B_0=B_1=0$ and set $f(e^{2\pi i t})=B_t$ for $t\in[0,1]$.) With probability $1$, the sample path will be a continuous function with local maxima on a countable dense set with pairwise distinct values. The set of these distinct values has the property you are looking for: each level set will be crossed a finite, odd [EDIT: or infinite] number of times.
To phrase what we have done without invoking probability, we have constructed a measure on the space of continuous functions $\mathbb S^1\to\mathbb R$ and we have specified a subset of these functions, each of which constitutes a function with [EDIT: some of] the properties you are seeking, and then we used sample path properties to argue that the measure of this subset of functions is positive - and as a consequence, the subset is non-empty.

The problem with this example can be summarized as "when it rains, it pours" regarding the cardinality of the level sets. My idea was to consider a sufficiently oscillatory function whose local maxima all lie on distinct level sets, since then each of those level sets will have an odd number of elements provided they are finite. However, the example I picked was too oscillatory, and in fact all local maxima but the global max will be crossed infinitely many times. 
