0708-1300/not homeomorphic

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

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