0708-1300/not homeomorphic

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

Assume [math]\displaystyle{ f_0 : R^n \rightarrow R^m }[/math] is a homeomorphism. Since [math]\displaystyle{ f_0 }[/math] is proper we can extend it to a continuous map [math]\displaystyle{ f : S^n \rightarrow S^m }[/math] which in fact will be a homeomorphism. Taking inverse if necessary we may assume [math]\displaystyle{ n \lt m }[/math]. Let [math]\displaystyle{ F:S^n\times[0, 1] \rightarrow S^m }[/math] be a homotopy of [math]\displaystyle{ f }[/math] to a smooth map i.e. [math]\displaystyle{ F }[/math] is continuous, [math]\displaystyle{ F(x, 0) = f(x) }[/math] and [math]\displaystyle{ F(x, 1) }[/math] is smooth. Since [math]\displaystyle{ F(x, 1) }[/math] is smooth and [math]\displaystyle{ n \lt 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]\displaystyle{ S^m }[/math] not in the image of [math]\displaystyle{ F(x, 1) }[/math], but the complement of that point is contractible. This means that we can extend [math]\displaystyle{ F }[/math] to [math]\displaystyle{ F_0:S^n\times[0, 2]\rightarrow S^m }[/math] to be a homotopy of [math]\displaystyle{ f }[/math] to a constant map. But then [math]\displaystyle{ f^{-1}F_{0} }[/math] is a contraction of [math]\displaystyle{ S^n }[/math] which is a contradiction with the fact that no such contraction exists.