Prove that \(\RR^2\) is not homeomorphic to \(\RR^n\) for \(n > 2\).
Solution
1. Suppose, for contradiction, that there is a homeomorphism\[ h:\mathbb R^2\to\mathbb R^n \]with \(n>2\).
We will derive a contradiction from the homology of the complements of corresponding points.
2. Restricting \(h\) gives a homeomorphism\[ \mathbb R^2\setminus\{0\} \cong \mathbb R^n\setminus\{h(0)\}. \]
Since \(h\) is bijective, \[ h\bigl(\mathbb R^2\setminus\{0\}\bigr) = \mathbb R^n\setminus\{h(0)\}. \] The restriction of a homeomorphism to a subspace is a homeomorphism onto its image.
3. The punctured plane deformation retracts onto \(S^1\), so\[ H_1(\mathbb R^2\setminus\{0\};\mathbb Z) \cong \mathbb Z. \]
The radial map \[ r(x)=\frac{x}{\|x\|} \] is a deformation retraction of \(\mathbb R^2\setminus\{0\}\) onto the unit circle \(S^1\). Homotopy invariance of singular homology therefore gives \[ H_1(\mathbb R^2\setminus\{0\}) \cong H_1(S^1) \cong \mathbb Z. \]
4. The punctured \(n\)-space deformation retracts onto \(S^{n-1}\), and for \(n>2\),\[ H_1(\mathbb R^n\setminus\{h(0)\};\mathbb Z)=0. \]
Translation by \(-h(0)\) is a homeomorphism \[ \mathbb R^n\setminus\{h(0)\} \cong \mathbb R^n\setminus\{0\}. \] Radial deformation retraction gives \[ \mathbb R^n\setminus\{0\}\simeq S^{n-1}. \] Since \(n>2\), one has \(n-1\ge2\), and the homology of a sphere gives \[ H_1(S^{n-1};\mathbb Z)=0. \] Hence the displayed punctured-space group vanishes.
5. This contradicts 2, so\[ \boxed{\mathbb R^2\not\cong\mathbb R^n\quad(n>2)}. \]
2. Therefore no homeomorphism \(\mathbb R^2\to\mathbb R^n\) exists for \(n>2\).