# Does there exist a continuous 2-to-1 function from the sphere to itself?

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

• 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). – Frpzzd Dec 18 '18 at 0:00
• 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 – Paul Frost Dec 18 '18 at 14:11
• – Dap Dec 20 '18 at 14:19
• See my answer at MathOverflow. – 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.

• Could you please explain how compactness lead us to the fact $d(x,y) \ge 2 \varepsilon$? – yoyo Dec 11 '18 at 13:47
• @yoyo In fact, a uniform $\varepsilon$ is not needed (but it is true). I edited the answer. – Florentin MB Dec 11 '18 at 14:14
• Sorry that I still can't understand how you deduced the fact that the restriction of $f$ to $U$ is injective. – yoyo Dec 11 '18 at 14:43
• @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 $\epsilon$ – Balarka Sen Dec 11 '18 at 14:53
• 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 $0$. – 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 $$v\not\in U$$.

• 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. – Paul Frost Dec 18 '18 at 14:18
• Right, so its probably not making things any easier... – SmileyCraft Dec 18 '18 at 16:40