# proof of invariance of dimension

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.

Unfortunately neither of these two arguments extends well to the cases where $n,m>1$. Indeed even the case for $n=1$ requires a reasonable amount of work to fill in the details. However, the latter approach does provide the necessary hint for a full solution.

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.htmlhttp://www.math.cornell.edu/ hatcher/AT/ATpage.html

Title proof of invariance of dimension ProofOfInvarianceOfDimension 2013-03-22 15:56:39 2013-03-22 15:56:39 Algeboy (12884) Algeboy (12884) 8 Algeboy (12884) Proof msc 55-00