# Injective maps $\mathbb{R}^{n} \to \mathbb{R}^{m}$

Let $f: \mathbb{R}^{n} \to \mathbb{R}^{m}$ be an injection for $n>m$. Can $f$ be continuous? Why?

I got this question in mind when I was trying to find a continuous map from $\mathbb{R}^{2}$ to $\mathbb{R}$.

-

If $f$ is injective and continuous from $\mathbb{R}^n$ to $\mathbb{R}^m$ where $n>m$ then $f$ restricts to a continuous bijection from $S^{n-1}$, the unit sphere in $R^n$, to a compact subset $K$ of $\mathbb{R}^m$. Thus you can embed $S^{n-1}$, and a foriori $S^m$ in $\mathbb{R}^m$.

But there are homological obstructions to embedding $S^m$ in $\mathbb{R}^m$. Using the arguments of this excellent paper

Albrecht Dold, A simple proof of the Jordan-Alexander complement theorem, Amer. Math. Monthly 100 (1993), 856-85.

(essentially a cunning use of the Mayer-Vietoris theorem) it would entail the homology of the space $\mathbb{R}^m-K$ being nonzero in negative dimension, which is absurd.

Added As I replied in haste I forgot the sledgehammer that cracks this little nut, namely Alexander duality.

Added later In fact this result also follows from Brouwer's Invariance of Domain. This is Theorem 2B.3 on page 172 of Hatcher's book. This implies that if one has an embedding from $\mathbb{R}^n$ to itself, then its image is open. One gets such an embedding by composing your putative embedding with the natural embedding of $\mathbb{R}^m$ in $\mathbb{R}^n$. Adapting the proof, gives a swift proof that the answer of your original question is no.

If you have a continuous injection from $\mathbb{R}^n$ to $\mathbb{R}^m$, with $n>m$ then you have an embedding of $S^{n-1}$ into a nontrivial hyperplane in $\mathbb{R}^n$. By the Jordan-Brouwer separation theorem the image $K$ of this embedding separates $\mathbb{R}^n$ but it's easy to see that since the image is in a hyperplane any two points of the complement of $K$ can be connected by a path (exercise for reader :-)).

-
There's a simpler argument when m=1. Any such map would induce a continuous map $\mathbb{R}^n\setminus\{p\}\rightarrow \mathbb{R}$ whose image is disconnected, a contradiction. Here $p$ is the preimage of any interior point of the original image. – Kevin Ventullo Aug 2 '10 at 10:18
Dear all, Chandru usually wishes elementary (or at least "ingenious") arguments; cf. mathoverflow.net/questions/34055/transcendence-of-pi . – Wadim Zudilin Aug 2 '10 at 10:52
Set of measure zero, the answer is different. The Baire space $J$ of irrationals is homeomorphic to $J \times J$, which gives us a 1-1 map from $J \times J$ into $\mathbb{R}$. – Gerald Edgar Aug 2 '10 at 12:28
Chandru, is it a simple question to prove that $\mathbf{R}^m$ and $\mathbf{R}^n$ are not homeomorphic for $m\ne n$? – Robin Chapman Aug 2 '10 at 15:22
Chandru, I don't know any way of proving that $\mathbb{R}^m$ and $\mathbb{R}^n$ for $m\ne n$ save by using the methods or results of algebraic topology such as homology theory or the Jordan-Brouwer separation theorem. Your question looks just as hard, and to me at least, it looks unlikely there's a naive proof. – Robin Chapman Aug 2 '10 at 17:20

Alternatively, you may use the Borsuk-Ulam antipodal theorem, in order to prove that such a map cannot be one-to-one.

-

More generally, you might ask whether there is a continuous injective map from a space of dimension $n$ to a space of dimension $m$, where $n > m$. The subject of dimension theory investigates this problem. (Hurewicz & Wallman wrote a classic, though hard to find nowadays, book on the subject; perhaps an expert could recommend a more modern reference?) Here is a representative theorem:

If $f: X \to Y$ is a continuous surjection bewteen separable compact metric spaces, where $X$ has dimension $n$ and $Y$ has dimension $m$, then there is a point $y \in Y$ whose preimage contains at least $n - m + 1$ points.

This applies to your question, as Robin Chapman explained, by restricting your map to a nice compact subspace of dimension $n$, for instance the unit closed ball.

In any case, my point is that the techniques above for Euclidean space and for spheres have appropriate generalizations to separable metric spaces.

As to the "why?", I recommend tracking down a copy of Hurewicz & Wallman from a library. It's very pleasant reading and assumes only a brief encounter with point set topology.

-
For a more modern reference: Engelking, Theory of Dimensions, finite and infinite. For separable metric spaces, also Jan van Mill, "infinite-dimensional topology, prerequisites and an introduction", or the newer "the infinite-dimensional topology of function spaces", both of which have a chapter on this subject. – Henno Brandsma Aug 2 '10 at 18:56

I taught an introductory topology course this last autumn where I covered this theorem from an elementary point of view. The argument just uses the Brouwer fixed point theorem (which itself has a proof via Stokes' theorem which is readily accessible to students with several variable calculus) plus elementary point set topology. In particular no homology or Jordan--Brouwer separation theorems are used. The treatment was based upon that of Hurewicz & Wallman but was also inspired by Larry Guth's ICM-2010 presentation in Hyderabad.

If you're interested the notes are available on