# 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.

Currently, all I have is that the map cannot be locally 1-to-1. For if $$f$$ is locally 1-to-1, then it must be a covering and from there you can obtain contradictions in numerous ways.

Thanks for any help.

Quick note: I've already asked this question here on MSE. However, the question has sat for a week with no definitive answers, so I figured it was okay to ask it here as well.

• If a 2-to-1 map $X \to Y$ is sufficiently regular (eg if there are compatible triangulations or CW decompositions of the source and target), then you get an identity of Euler characteristics $2 \chi(Y) = \chi(X)$. This would give a contradiction in your case because both the source and target have Euler characteristic 2. I haven't been able to think of a sufficiently clever argument that the Euler characteristic formula does or does not hold for a general 2-to-1 map of $S^2$. – Tyler Lawson Dec 22 '18 at 9:25
• Follow-up questions: (a) same question with 2 replaced by $n\ge 2$ (a') same with the assumption that for every $x$, $f^{-1}(\{x\})$ is finite of cardinal $\ge 2$ (b) original question on higher-dimensional spheres (d) which closed 3-manifolds admit a continuous self-map in which every preimage has cardinal 2? – YCor Dec 23 '18 at 9:52

## 1 Answer

There is no such function as I will prove below using the work of Civin (1943) and Kerékjártó (1919).

Let me review what we need from the work of Civin (1943). Let $$f$$ be a continuous 2-to-1 function defined on a compact manifold $$M$$. For each $$x\in M$$, the preimage $$f^{-1}(f(x))$$ consists of $$x$$ and another point $$s(x)\in M$$. Let $$K\subset M$$ be the set of points where $$s$$ is continuous. For $$x\in M$$, let $$t(x)=s(x)$$ when $$x\in K$$, and let $$t(x)=x$$ when $$x\not\in K$$. Then $$t:M\to M$$ is a homeomorphism of order 2 (i.e. $$t^2=1$$); I will call $$t$$ the homeomorphism associated to $$f$$. It is also known that $$K$$ is dense and open in $$M$$, and it is invariant under $$s$$. Therefore, $$F:=M\setminus K$$ is a nowhere dense compact subset of $$M$$, which is invariant under $$s$$, and the restriction of $$f$$ to $$F$$ is also 2-to-1. Note that $$F$$ is the set of fixed points of $$t$$.

Now assume that $$f:S^2\to S^2$$ is a continuous 2-to-1 function, and use the notations of the previous paragraph. By the theorem of Kerékjártó (1919), $$t$$ is conjugate within the group of homoeomorphisms of $$S^2$$ to a rotation of angle $$\pi$$ or a reflection. Therefore, we can assume without loss of generality (namely after composing $$f$$ from inside by a suitable homeomorphism of $$S^2$$), that $$F$$ is an antipodal pair of points or a great circle. In the first case, $$f$$ induces a homeomorphism from the annulus $$S^2\setminus F$$ divided by a rotation of angle $$\pi$$ (which is still an annulus) to the punctured sphere $$S^2\setminus f(F)$$. This is clearly absurd. Hence $$F$$ is a great circle, and $$f$$ induces a homeomorphism from either (hemisphere) connected component of $$S^2\setminus F$$ to $$S^2\setminus f(F)$$. Consider the restriction $$g:=f_{\mid F}:F\to S^2$$, which is a continuous 2-to-1 function, and the (order two) homeomorphism $$u:F\to F$$ associated to $$g$$. Similarly as before, $$u$$ is conjugate within the group of homeomorphisms of $$F$$ to a rotation of angle $$\pi$$ or a reflection. In either case, we can see that $$f(F)$$ is homeomorphic to $$S^1$$. Therefore, by the Jordan curve theorem, $$S^2\setminus f(F)$$ has two connected components, contradicting our earlier finding that it is homeomorphic to an open hemisphere.