33
$\begingroup$

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.

$\endgroup$
  • 9
    $\begingroup$ 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 Dec 22 '18 at 9:25
  • 3
    $\begingroup$ 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 Dec 23 '18 at 9:52
27
$\begingroup$

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.

$\endgroup$

Your Answer

By clicking "Post Your Answer", you acknowledge that you have read our updated terms of service, privacy policy and cookie policy, and that your continued use of the website is subject to these policies.

Not the answer you're looking for? Browse other questions tagged or ask your own question.