View source for 0708-1300/not homeomorphic

Please, read the following carefully. It can contain some mistake.

Assume  <math>\overline{f} : R^n --> R^m</math> is a homeomorphism. Since <math>\overline{f}</math> is proper we can extend it to a continuous map <math>f : S^n --> S^m</math> which in fact will be
a homeomorphism. Taking inverse if necessary we may assume <math>n < m</math>. Let
<math>F : Sn × [0, 1] --> Sm</math> be a homotopy of <math>f</math> to a smooth map i.e. <math>F</math> is continuous, <math>F(x, 0) = f(x)</math> and <math>F(x, 1)</math> is smooth. Since <math>F(x, 1)</math> is smooth and <math>n < m</math> all of its image points are singular values and by Sard's theorem constitute a set of measure zero. Then there is a point in <math>S^m</math> not in the image of <math>F(x, 1)</math>, but the complement of that point is contractible. This means that we can extend <math>F</math> to <math>\overline{F} : Sn × [0, 2] --> S^m</math> to be a homotopy of <math>f</math> to a constant map. But then <math>f^{－1}\circ\overline{F}</math> is a contraction of <math>S^n</math> which is a contradiction with the fact that no such contraction exists.