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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.

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    $\begingroup$ 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 $S^2$. $\endgroup$
    – Tyler Lawson
    Commented Dec 22, 2018 at 9:25
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    $\begingroup$ 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? $\endgroup$
    – YCor
    Commented Dec 23, 2018 at 9:52
  • $\begingroup$ @YCor The answer to your follow-up question (a') is that it is possible. For every pair of antipodal points $(a,b)$ define its twist to be the mapping $z\mapsto z^2$ after we identify the sphere with the complex projective space $\mathbb{C}P^1$ by projecting it from the point $b$ to the complex plane that is tangent to the sphere at point $a$ so that $a$ is paired with $0$ and $b$ is paired with $\infty$. Then the composition of twists for two different pairs of antipodal points is a mapping that is 2-to-1 thru 4-to-1. There is a similar but more complicated construction for 2-to-1 and 3-to-1. $\endgroup$
    – cnikbesku
    Commented Dec 12, 2022 at 11:47

1 Answer 1

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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.

Remark. The journal issue with the quoted paper by Kerékjártó is also available here. A modern treatment can be found here.

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  • $\begingroup$ Could this technique answer the analogue of the question where $f$ is a 3-to-1 function? $\endgroup$
    – cnikbesku
    Commented Dec 9, 2022 at 19:12
  • $\begingroup$ @donaastor I don't know, and I have no time to think about it. The same question was asked by YCor in a comment below my post. $\endgroup$
    – GH from MO
    Commented Dec 11, 2022 at 15:21
  • $\begingroup$ Could you at least let us know where to find the works of Civin and Kerékjártó you mentioned in your post, please? Because I wanted to know if there are some generalizations in the papers, that was the point of my question. $\endgroup$
    – cnikbesku
    Commented Dec 11, 2022 at 15:39
  • $\begingroup$ @donaastor I added links and a formal remark to my post. $\endgroup$
    – GH from MO
    Commented Dec 11, 2022 at 17:32
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    $\begingroup$ Great, thank you! $\endgroup$
    – cnikbesku
    Commented Dec 11, 2022 at 19:04

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