Htykuut

I am interested in the following question:

Does there exist a continuous function $f:S^2to S^2$ such that, for any $pin S^2$, $|f^{-1}({p})|=2$?

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 $pin S^2$. Then $f$ restricts to a covering map from $S^2backslash f^{-1}(p)$ to $S^2backslash {p}$. However, the fundamental group of $S^2backslash f^{-1}(p)$ is $mathbb{Z}$ and the fundamental group of $S^2backslash {p}$ 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

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.

If I understand correctly, Florentin MB's answer will be complete if we can prove that $f$ is locally injective. Feel free to tell me if I'm wrong here, but I think we can at least show that $f$ is locally injective if $f$ is open. It seems to me that $f$ should be open, but I don't know how to prove that.

Let $xin S^2$ and $f(x)=f(y)$ for $xneq y$. Then let $U$ and $V$ be disjoint opens such that $xin U$ and $yin V$. If $f$ is open, then $f(U)$ and $f(V)$ are open, so $W:=f(U)cap f(V)$ is open. Now for every $win W$ there exist $uin U$ and $vin V$ such that $f(u)=f(v)=w$. Because $U$ and $V$ are disjoint, we find $uneq v$, so because $|f^{-1}({w})|=2$, we find $f^{-1}({w})={u,v}$. Hence, we find $f^{-1}(W)subset Ucup V$. Because $f$ is continuous, $f^{-1}(W)$ is open, so $N:=f^{-1}(W)cap U$ is open. Notice that $xin N$, so $N$ is a neighborhood of $x$. Finally, $f$ is injective in $N$, because for all $nin N$ we have $f(n)in W$. This gives $f^{-1}({f(n)})={n,v}$ for some $vin V$, and $U$ and $V$ are disjoint, so $vnotin U$.