I am interested in the following question:



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> Does there exist a continuous function $f:S^2\to S^2$ such that, for any $p\in S^2$, $|f^{-1}(\{p\})|=2$?
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> 
> 


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](https://math.stackexchange.com/questions/3033601/does-there-exist-a-continuous-2-to-1-function-from-the-sphere-to-itself) 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.



Does there exist a continuous 2-to-1 function from the sphere to itself?
There is no such function as I will prove below using the work of [Civin (1943)](https://doi.org/10.1215/S0012-7094-43-01006-3) and [Kerékjártó (1919)](https://doi.org/10.1007/BF01463232).


Let me review what we need from the work of [Civin (1943)](https://doi.org/10.1215/S0012-7094-43-01006-3). 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)](https://doi.org/10.1007/BF01463232), $t$ 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](http://resolver.sub.uni-goettingen.de/purl?PPN235181684_0080). A modern treatment can be found [here](http://doi.org/10.5169/seals-61111).


