This is the problem we want to solve:

Can $f\colon \mathbb{R}^k \to \mathbb{R}^n$ such that $ \forall y \in \operatorname{im}(f)$, $ f^{-1}(y) = \{a_y,b_y\}, a_y \neq b_y $ be continuous?

Originally I've seen this question on an exam but it was stated only for the case $ k = n = 1 $ and $f$ surjective, which made it really easy to show $f$ can't be continuous, by using the Weierstrass extreme value theorem. A very similar argument seem to work for any $k$, as long as $n=1$. However, for general $k$ and $n$ this seems much harder. I don't see how surjectivity affects this problem, so I've dropped this assumption for now. Edit:Slup commented below, showing the relevance of surjectivity for this question.

Induction on $n$ and looking at projections of $f$ onto individual coordinates seemed tempting at first, but the composition of $f$ with a projection seems to lose any traces of the property that the inverse image of a point = exactly two points, so I don't see how this could be useful.

Trying to visualise this for $k=n=2$, it intuitively seems that in order to transform the space in this way, we would have to 'tear' it along some curve. For bigger $k = n$, that becomes 'tearing' along some $n-1$ dimensional manifold, but that's obviously completely informal, sort of useless and I completely have no idea how this idea could be translated into a formal proof.

Bonus question: Does the answer or the proof change in a significant way if we limit the domain to $ f:\overline{\mathbb{B}^k} \to \mathbb{R}^n $? We operate on a compact ball now, so that's fairly different from $\mathbb{R}^k$.

  • 3
    $\begingroup$ Note: it is impossible to get a two-to-one map with a polynomial from $\Bbb C$ to $\Bbb C$, since every polynomial equation has a discriminant that must become zero for some input value. $\endgroup$ Jun 29, 2016 at 15:41
  • $\begingroup$ @fermesomme no, they of course depend on y, otherwise the question would be kind of stupid. Good point though, I've edited the question to make this clearer. $\endgroup$
    – Ormi
    Jun 29, 2016 at 15:42
  • 4
    $\begingroup$ Suppose that such continuous map exists. Then it should be a two-fold covering of its image. So your question is about subspaces of $\mathbb{R}^n$ having $\mathbb{Z}_2$ as a fundamental group and $\mathbb{R}^k$ as a universal covering. In particular, such a map could not be surjective. $\endgroup$
    – Slup
    Jun 29, 2016 at 15:44
  • $\begingroup$ @slup Thank you for the comment. I'm not familiar with the theorems you're refering to, but I will try to read up on this. However, in general the question is stated without the assumption of surjectivity, so I'm still very much interested in what happens then. $\endgroup$
    – Ormi
    Jun 29, 2016 at 15:47
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    $\begingroup$ Please see here- mathoverflow.net/questions/17707/… $\endgroup$
    – Hmm.
    Jul 1, 2016 at 16:56

3 Answers 3


I did not read the paper of Mioduszewski mentioned in one of the posts. I only have some partial answers when $f$ is more regular than continuous.

If $k\leq n$ and $f:\overline{B}\rightarrow\mathbb{R}^{n}$ (the bonus question) is Lipschitz continuous, then using the area formula for Lipschitz functions (it's in the book of Evans and Gariepy "Measure theory and fine properties of functions") $$ \infty>L\mathcal{L}^{k}(\overline{B})\geq\int_{\overline{B}}Jf\,dx=\int _{\mathbb{R}^{n}}\mathcal{H}^{0}(\overline{B}\cap f^{-1}(\{y\})\,dy=\int _{\mathbb{R}^{n}}2\,dy=\infty, $$ which is a contradiction. Here, $Jf$ is the Jacobian of $f$, $L$ is the bound of $Jf$, and $\mathcal{H}^{0}$ is the counting measure. If $f:\mathbb{R}% ^{k}\rightarrow\mathbb{R}^{n}$ is Lipschitz continuous and $Jf$ has finite integral, you get the same contradiction.

If $k>n$ and $f\in C^{k+1-n}(\mathbb{R}^{k})$, then using Sard's theorem https://en.wikipedia.org/wiki/Sard's_theorem the set of points $\{x\in \mathbb{R}^{k}:\,Jf(x)=0\}\cap f^{-1}(\{y\})$ is empty for $\mathcal{L}^{n}$ a.e. $y\in\mathbb{R}^{n}$. Hence, if $f$ is onto or if $f(\mathbb{R}^{k})$ has positive measure, then taking $y\in f(\mathbb{R}^{k})$ such that $\{x\in\mathbb{R}^{k}:\,Jf(x)=0\}\cap f^{-1}(\{y\})=\emptyset$, we get that $Jf(x)$ has rank $n$ and so we can apply the implicit function theorem to conclude that $f^{-1}(\{y\}$ is locally the graph of a function. In particular $f^{-1}(\{y\})$ cannot consists of two points.


A negative answer for the case $k>n=1$.

Let $a,b\in\mathbb R^k $ be twin points. Let $c$ and $d $ other points. Take a path from $a $ to $c $ to $d $ to $b $. By the mean value theorem the twin points are extremal on that path. Say they are minima. This remains true even if we change the point $d $ to be any point; therefore $a $and $b $ are global minima. Since this is true for any pair of twins, any such pair are extremal. But this impossible because $ \mathbb R^k$ contains more than 4 points.

The same argument holds for the closed ball.

  • $\begingroup$ Edit: a solution to the bonus problem $\endgroup$
    – Ron
    Aug 31, 2017 at 20:35

(I'm assuming $f$ is surjective). Let $n=k=2$, $y \in \mathbb{R}^2, a_y=a, b_y=b.$ Consider the line between $a$ and $b$, $r$. This splits $\mathbb{R}^2$ in $2$ connected components, $P_1$ and $P_2$. Since these are connected, if $f$ was continous then their images through $f, Q_1$ and $Q_2$, would be connected as well, and their intersection would be empty. Since $r$ is also connected, its image $s$ is connected as well, and it contains $y$. Furthermore, since $f(a)=f(b)=y, s$ is not injective, in fact it splits $\mathbb{R}^2$ in at least $3$ connected components. Since continous functions can only decrease the number of connected components, we have a contradiction.

For a general case with $k=n$ i think it suffices to take an adequate number of points in order to split the domain and codomain and use a similar reasoning (eg for $k=n=3$ take two points such that their preimages are coplanar and split $\mathbb{R}^3$ using a plane and a weird surface). I also think this reasoning, if a bit modified, works for $k>n$ as well (eg if $k>n=1$, it's even simpler: $\mathbb{R}^k - \{a, b\}$ is connected but its image $\mathbb{R} - \{y\}$ is not, so $f$ is not continous).


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