I had asked a similar question before.

Functions from an $n$-dimensional hypercube to $\mathbb{R}^m$ when $n >m$.

I am wondering if there are any surjective functions.

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 surjective 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 $\mathbb{R}^m$?**


1 Answer 1


Yes, this is possible, you can even find a $C^\infty$-smooth map with this property.

Let $C\subset R^n$ be a closed subset. A $C^k$-smooth function $F: C\to R^m$ is a function such that for every $a\in C$ there exists a neighborhood $U$ of $a$ in $R^n$ and a $C^k$-smooth extension of $F|_{C\cap U}$ to $U$.

Just to settle the terminology, I will be considering the $n$-dimensional (hyper)cube $Q$ given by the inequalities $$ |x_i|\le 1, i=1,...,n. $$ Then the $2^n$ regions $P_i, i=1,...,2^n$, of $Q$ appearing in your question are the connected components of the complement $Q-D$, where $$ D= \{(x_1,...,x_n): x_1\cdots x_n=0\}. $$ (Each $P_i$ is a unit cube in $R^n$ missing part of its boundary.) In the answer to your previous question you were given a $C^\infty$-function $f: Q\to R^m$ such that the images of the cubes $P_i\subset Q$ are pairwise disjoint. I will construct a $C^\infty$-smooth extension of $f$ to a surjective function $F: R^n\to R^m$.

Consider the $n$-dimensional orthant $$ O_n= \{x\in R^n: x_i\ge 2, i=1,...,n\}. $$

Lemma. For every $m\le n$, there exists a surjective $C^\infty$-smooth map $g: O_n\to R^m$.

Proof. Consider the function $\phi: R\to R$, $$ \phi(t)= t^{k+1}\sin(t). $$ This function is easily seen to be $C^\infty$-smooth and surjective. Moreover, its restriction of $\phi$ to each interval $[T,\infty)$ is also surjective.

Now, define $$ g(x_1,...,x_n)= (\phi(x_1),...,\phi(x_m)). $$ This function $R^n\to R^m$ is also clearly $C^\infty$-smooth and satisfies $g(O_n)=R^m$. qed

Now, for the set $C=Q\cup O_n$ we define a function $h: C\to R^m$, $$ h(x)= \begin{cases} f(x), \quad x\in Q\\ g(x), \quad x\in O_n. \end{cases} $$ By the construction, this function is surjective, $C^\infty$-smooth and the images of cubes $P_i$ are pairwise disjoint.

The last ingredient we need is Whitney extension theorem which is a smooth analogue of Tietze Extension Theorem. I will formulate only a weak version of Whitney's theorem which is much easier to prove (the full Whitney's theorem also prescribes values of partial derivatives).

Theorem. Suppose that $A\subset R^n$ is a closed subset and $f: R^n\to R^m$ is a $C^k$-smooth function, $1\le k\le \infty$. Then $f$ admits a $C^k$-smooth extension $F: R^n\to R^m$.

Remark. Whitney's theorem is usually stated only for $R$-valued functions but applying it to each component $f_i, i=1,...,m$ of the function $f: A\to R^m$, one obtains the theorem stated above.

Lastly, applying Whitney's theorem to the function $h$ defined above, we obtain a surjective $C^\infty$-smooth map $F: R^n\to R^m$ extending the map $h$ and, therefore, satisfying the property that the images of the cubes $P_i$ are pairwise disjoint. qed

Here is what I do not know:

Question. Is there an open surjective smooth (or even continuous) map $F: R^n\to R^m$ such that the images of the cubes $P_i$ are pairwise disjoint?


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