If $F:\mathbb R^m\to \mathbb R^m$ is continuous with $|F(x)-F(y)|\geq \lambda|x-y|$, then $F$ is surjective.

Question

I know that this question has been asked for several times (such as this post and the one-dimensional case). However, I still couldn't find an answer without using invariance of domain. The original question is stated as:

If $F:\mathbb R^m\to \mathbb R^m$ is continuous with $|F(x)-F(y)|\geq \lambda |x-y|$ for some $\lambda>0$, then $F$ is a homeomorphism.

Here is my attempt: It is clear that $F$ is injective and $F^{-1}:F(\mathbb R^m)\to \mathbb R^m$ is continuous. Hence, it remains to show that $F$ is surjective. As $\mathbb R^m$ is connected, we can infer that $F(\mathbb R^m)=\mathbb R^m$ by proving that $F(\mathbb R^m)$ is a clopen subset. The closedness of $F(\mathbb R^m)$ is easy: Suppose $q_n$ is a sequence in $F(R^m)$ with $q_n\to q$. Then there exists $p_n\in \mathbb R^m$ such that $F(p_n)=q_n$. Since $|q_n-q_m|\geq \lambda |p_n-p_m|$ and $\{q_n\}$ is Cauchy, $p_n$ is Cauchy, which implies that $p_n\to p$ for some $p\in \mathbb R^m$. Them by the continuity of $F$, $q=F(p)\in F(\mathbb R^m)$.

However, I still need show that $F(\mathbb R^m)$ is open and that is where I get stumped. Since it is a question given in our introductory real analysis courses, invariance of domain is completely out of scope of this curriculum. Thus, I seek for an elementary solution.

Answer 1

Here is my take. If you were to prove surjectivity, you will need to use (or to prove) some nontrivial facts about topology of $\mathbb R^n$ which are not covered in a typical undergraduate real analysis/general topology class, such as the invariance of domain theorem. Here is an example of a metric space $(X,d)$ which is path-connected, locally compact, complete and a continuous expanding map $F: X\to X$. (Moreover, it satisfies the Heine-Borel property: a subset of $X$ is compact if and only if it is closed and bounded.) The image of the map $F$ will be closed but not open.

Consider the following version of Hawaiian Earrings:

For $n\ge 3$ take $X$ to be the union of round spheres $S(k {\mathbf e}_1, k)\subset \mathbb R^n, k\in \mathbb N$, ($\mathbb R^n$ is equipped with the standard metric), centered at $k {\mathbb e}_1$ and of the radius $k$, together with the coordinate hyperplane orthogonal to the vector $\mathbf e_1$. Use the restriction of the standard metric on $\mathbb R^n$ to get a metric $d$ on $X$. The resulting metric space $(X,d)$ is complete, path-connected, locally compact, hence, shares all the basic properties with $\mathbb R^n$. However, it is not a topological manifold.

Take the map $F(\mathbf x)=2\mathbf x$; $F: X\to X$ is continuous and satisfies the inequality: $$ d(F(\mathbf x), F(\mathbf y))=2 d(\mathbf x, \mathbf y) $$ for all pairs of points in $X$. But this map is not surjective.

Why is $X$ not a manifold? Because it has a "cut-point", a point $p$ such that $X\setminus \{p\}$ is disconnected. (This point is the origin.) By multiplying this example by the real line you get a similar example which has no cut-points. It is still not a manifold but it has a "cut-line." Why a connected topological manifold of dimension $\ge 3$ cannot have a cut-line (i.e. a subset $A$ homeomorphic to $\mathbb R$ such that removing $A$ disconnects the manifold)? I do not know of any easy proof of this result which avoids some algebraic topology or algebraic topology in disguise. (In contrast, it is elementary to see that a connected manifold of dimension $\ge 2$ cannot have cut-points.)

By further modifying the above example, one gets a similar example where the metric $d$ is uniquely geodesic: Every two points are connected by a length-minimizing path, such a path is unique (up to a reparameterization) and depends continuously on the end-points. (This is another property one has in $\mathbb R^n$ with the standard metric.)

Lastly, I think it is irresponsible to give this problem in introductory analysis courses without discussing first the invariance of domain theorem or assuming $C^1$-smoothness of $F$.