I am interested in the following question:
Does there exist a continuous function $f:S^2\to S^2$ such that, for any $p\in S^2$, $|f^{-1}(\{p\})|=2$?
I suspect the answer is no, but I don't know how to prove it. All I know right now is that $f$ cannot be a covering map.
For if $f$ is a covering map, take any $p\in S^2$. Then $f$ restricts to a covering map from $S^2\backslash f^{-1}(p)$ to $S^2\backslash \{p\}$. However, the fundamental group of $S^2\backslash f^{-1}(p)$ is $\mathbb{Z}$ and the fundamental group of $S^2\backslash \{p\}$ is trivial. That $f$ is a covering then gives that there is an injective homomorphism from $\mathbb{Z}$ to the trivial group, a contradiction. Thus $f$ cannot be a covering map.
That's all I've got so far. Any more progress is greatly appreciated.
Update (Dec 21, 2018): I've posted this question on MO
Best Answer
EDIT: the argument is incomplete. As pointed in the comments by yoyo, the separation of the pre-images is not guaranteed by the compactness.
There is no such map.
For is there is, I prove below that it has to be a covering map and you proved that it is not possible (or simply remark that from the fact that $S^2$ is simply connected, it admits no non-trivial covering).
Let's assume $f$ satisfies your hypothesis. By compactness of $S^2$, there exists a $\varepsilon>0$ such that for all $p \in S^2$, for all $x \neq y \in f^{-1}(p)$ we have $d(x,y) \ge 2 \varepsilon$ (where $d$ is any compatible metric on the sphere). Consequently, for any $x \in S^2$, if $U$ is the closed ball centered at $x$ and radius $\varepsilon$ then the restriction of $f$ to $U$ is injective, and by compactness, an homeomorphism onto its image.
We've shown that $f$ is a local homeomorphism, and by hypothesis it is surjective. As its domain is compact and its range connected, it is a covering.