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

  • $\begingroup$ I don't know much topology, but here's an idea: if you take $f^{-1}$ to be continuous as well, then $f$ would be a homeomorphism from some connected subset of the sphere to itself. However, a sphere partitions $3$-space into two parts, and no connected subset of a sphere partitions $3$-space into two parts (because, non-rigorously speaking, it must have a hole). $\endgroup$ – Frpzzd Dec 18 '18 at 0:00
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    $\begingroup$ We know that $f$ cannot be a covering map. This leaves the question whether your function $f$ must be a covering map. The answers by Florentin MB and SmileyCraft are have gaps. Florentin MB cannot show that $f$ is locally injective and SmileyCraft assumes that $f$ is open. The arguments used in these answers cannot be sufficient because they are of a very general nature and do not invoke special features of $S^2$. This is shown by yoyo's comment where you can find a continuous function $f : S^1 \to S^1$ which is not a covering although preimages of points contain two points $\endgroup$ – Paul Frost Dec 18 '18 at 14:11
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    $\begingroup$ Related: mathoverflow.net/questions/17707/… $\endgroup$ – Dap Dec 20 '18 at 14:19
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    $\begingroup$ See my answer at MathOverflow. $\endgroup$ – GH from MO Dec 23 '18 at 2:17

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.

  • $\begingroup$ Could you please explain how compactness lead us to the fact $d(x,y) \ge 2 \varepsilon$? $\endgroup$ – yoyo Dec 11 '18 at 13:47
  • $\begingroup$ @yoyo In fact, a uniform $\varepsilon$ is not needed (but it is true). I edited the answer. $\endgroup$ – Florentin MB Dec 11 '18 at 14:14
  • $\begingroup$ Sorry that I still can't understand how you deduced the fact that the restriction of $f$ to $U$ is injective. $\endgroup$ – yoyo Dec 11 '18 at 14:43
  • $\begingroup$ @yoyo I believe the idea is that the diameter of the preimages of $f$ is a positive continuous function on $S^2$, and since that's a compact domain, it is bounded below by some uniform positive constant. That is the chosen $\epsilon$ $\endgroup$ – Balarka Sen Dec 11 '18 at 14:53
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    $\begingroup$ EX:$g:S^1\to S^1$, $g(\theta)=2\theta$ if $0 \le \theta \le \pi$ and $g(\theta)=-2 (\pi + \theta)$ if $-\pi \le \theta \le 0$, which satisfies $|g^{-1}(\{\theta\})|=2$ for all $\theta \in S^1$ but $S^1$ is compact and the $inf$ of the diameters of the preimages of $g$ is $0$. $\endgroup$ – yoyo Dec 11 '18 at 15:26

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 $x\in S^2$ and $f(x)=f(y)$ for $x\neq y$. Then let $U$ and $V$ be disjoint opens such that $x\in U$ and $y\in V$. If $f$ is open, then $f(U)$ and $f(V)$ are open, so $W:=f(U)\cap f(V)$ is open. Now for every $w\in W$ there exist $u\in U$ and $v\in V$ such that $f(u)=f(v)=w$. Because $U$ and $V$ are disjoint, we find $u\neq v$, so because $|f^{-1}(\{w\})|=2$, we find $f^{-1}(\{w\})=\{u,v\}$. Hence, we find $f^{-1}(W)\subset U\cup V$. Because $f$ is continuous, $f^{-1}(W)$ is open, so $N:=f^{-1}(W)\cap U$ is open. Notice that $x\in N$, so $N$ is a neighborhood of $x$. Finally, $f$ is injective in $N$, because for all $n\in N$ we have $f(n)\in W$. This gives $f^{-1}(\{f(n)\})=\{n,v\}$ for some $v\in V$, and $U$ and $V$ are disjoint, so $v\not\in U$.

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    $\begingroup$ If it should be possible to prove that $f$ is open, then it must be based on special features of $S^2$. See yoyo's example of a function $f : S^1 \to S^1$ which is not open. $\endgroup$ – Paul Frost Dec 18 '18 at 14:18
  • $\begingroup$ Right, so its probably not making things any easier... $\endgroup$ – SmileyCraft Dec 18 '18 at 16:40

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