# Surjective functions from a $n$-dimensional hypercube to $\mathbb{R}^m$ when $n > m$

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

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?