An application of the invariance of dimension theorem shows that $\mathbb{R}^n$ is homeomorphic to $\mathbb{R}^m$ if and only if $m=n$ . Already this is a difficult question. (We will assume $n\leq m$ throughout this article.)

Simple arguments suffice for small dimensions.

• If $n=0$ cardinality is sufficient: there can be no bijection between $\mathbb{R}^0=\{0\}$ and $\mathbb{R}^n$ , $m>0$ , as the latter is uncountable.
• If $n=1$ , then suppose $f:\mathbb{R}\rightarrow\mathbb{R}^m$ is a homeomorphism with $m>1$ . Then certainly the following restriction of $f$ $$ f:\mathbb{R}-\{0\}\rightarrow \mathbb{R}^m-\{f(0)\ $$ is also a homeomorphism. Yet, as $m>1$ , $\mathbb{R}^m-\{f(0)\}$ is (path) connected but $\mathbb{R}-\{0\}$ is not connected. Thus this restriction of $f$ cannot be a homeomorphism so indeed the original $f$ could not be a homeomorphism.

To solve the problem outright depends on algebraic invariants from homology, a surprisingly big hammer for such a basic topological question. But the conceptual steps are still basic, and we will attempt to highlight them in our exposition of the proof.

Let $U$ and $V$ be non-empty open subsets of $\mathbb{R}^n$ and $\mathbb{R}^m$ respectively. Assume that $f:U\rightarrow V$ is a homeomorphism.

Choose a point $x\in U$ (akin to the point we removed when $n=1$ .) Then consider the relative homology groups $H_i(U,U-\{x\})$ , $i\in \mathbb{N}$ . As $U$ is open we may apply the Excision Theorem (axiom) to claim $H_i(U,U-\{x\})\cong H_i(\mathbb{R}^n,\mathbb{R}^n-\{x\})$ - basically, to look at a punctured open disk it to look at a punctured $\mathbb{R}^n$ . Now we look at the induced long exact sequence from the relative pair $(\mathbb{R}^n,\mathbb{R}^n-\{x\})$ and find $H_i(\mathbb{R}^n,\mathbb{R}^n-\{x\})$ is isomorphic to the reduced homology $\tilde{H}_i(\mathbb{R}^n-\{x\})$ . But $\mathbb{R}^n-\{x\}$ contracts to the sphere $S^{n-1}$ - and homologoy preserves homotopy type - so we now have $H_i(U,U-\{x\})\cong H_i(S^{n-1})$ . (Puncture a disk, and it deflates to a sphere of lower dimension.)

Now it is an exercise in homology to prove that $H_i(S^{n-1})=0$ if $i\neq 0,n-1$ and $\mathbb{Z}$ otherwise. In particular we are using the fact that the invariance of dimension of spheres is (more) easily established by the homology groups.

We now repeat the process with $V$ . If $U$ and $V$ are indeed homeomorphic, then this process will result in isomorphic homology groups for every $i\in\mathbb{N}$ . In particular, $$ \mathbb{Z}\cong H_{m-1}(S^{m-1})\cong H_{m-1}(V,V-\{f(x)\})\cong H_{m-1}(U,U-\{x\})\cong H_{m-1}(S^{n-1}) $$ Thus either $m=1$ which implies $n=0,1$ as $n\leq m$ , or $m=n$ . If $n=0$ we have already seen $m=n$ . So the result stands for all $m,n$ .

For a detailed accounting of this theorem together with the necessary lemmas refer to:

Allen Hatcher, Algebraic Topology, Cambridge University Press, Cambridge, 2002. Available on-line at: http://www.math.cornell.edu/~hatcher/AT/ATpage.html