Let $n$ and $m$ be integers such that $n > m$. Suppose there exists a $n$-dimensional hypercube in $\mathbb{R}^n$. Let the hypercube be divided into $2^n$ regions ($n$-dimensional volumes) by perpendicularly bisecting each orthogonal dimension. For example, an analogous idea would be to divide a square into quadrants or a cube into octants.

Does there exist a differentiable function $f: \mathbb{R}^n \to \mathbb{R}^m$ such that the images of the regions of the hypercube are pairwise non-intersecting in $\mathbb{R}^m$?

  • $\begingroup$ Do you mean the images of the regions are pairwise non-intersecting or have empty intersection? $\endgroup$
    – Nick L
    Feb 14, 2019 at 19:29
  • $\begingroup$ I mean pairwise not intersecting. $\endgroup$
    – Halbort
    Feb 14, 2019 at 19:45

1 Answer 1


If $m=1$, no. By the Intermediate Value Theorem, the image of each region must be an interval. By continuity, the image of each region must contain $f(O)$ in its closure, where $O$ is the center. With four or more intervals that have zero either as an endpoint or in their interior, they must overlap.

If $m>1$, yes. Taking the cuts to be the coordinate hyperplanes in $\mathbb{R}^n$, here's a map from $\mathbb{R}^n$ to $\mathbb{R}^2$ that works:
Let $g(x_1,x_2,\dots,x_n)$ be a function from $\mathbb{R}^n$ to the unit circle in $\mathbb{R}^2$ that depends only on the sign of the $x_i$, and takes different values in each "quadrant"; if $x_i$ and $y_i$ have opposite signs for some $i$, $g(x_1,x_2,\dots,x_n)$ and $g(y_1,y_2,\dots,y_n)$ are different.
Then, define $f(x_1,x_2,\dots,x_n)=x_1^2x_2^2\cdots x_n^2g(x_1,x_2,\dots,x_n)$; this $f$ is the function we seek. Each "quadrant" is mapped to a different ray, while the hyperplanes separating them are all mapped to the single point $0$. Inside each "quadrant", $f$ is differentiable because it's a constant multiple of the differentiable function $x_1^2x_2^2\cdots x_n^2$, and $f$ is differentiable on the hyperplanes with derivative zero because we're multiplying something locally bounded by the square of the coordinate that's zero there.

For $m>2$? Pad the function with zeros in the extra coordinates.

This relies on the interpretation that only the interiors of the cut regions have to be separate, and the images of the boundary hyperplanes are allowed to intersect.

  • $\begingroup$ How do you know the function $g$ exists? I don't think this works because the hypercube graph is not planar for $n > 3$. $\endgroup$
    – Halbort
    Feb 14, 2019 at 21:17
  • $\begingroup$ Just choose the $2^n$ values arbitrarily. For example, in $\mathbb{R^4}$, we could choose $g(1,1,1,1)=(1,0)$, $g(1,1,1,-1)=(0.8,0.6)$, $g(1,1,-1,1)=(0.6,0.8)$, and so on. $\endgroup$
    – jmerry
    Feb 14, 2019 at 21:20
  • $\begingroup$ It has to be differentiable. $\endgroup$
    – Halbort
    Feb 14, 2019 at 21:20
  • $\begingroup$ $g$ doesn't. Or, at least, it only has to be differentiable away from the separating hyperplanes, where it's locally constant. $\endgroup$
    – jmerry
    Feb 14, 2019 at 21:22
  • $\begingroup$ If $g$ is not differentiable, $f$ will not be. $g$ does not count as a constant factor. $\endgroup$
    – Halbort
    Feb 14, 2019 at 21:23

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