0708-1300/not homeomorphic

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.