0708-1300/not homeomorphic

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

Assume ${\overline {f}}:R^{n}-->R^{m}$ is a homeomorphism. Since ${\overline {f}}$ is proper we can extend it to a continuous map $f:S^{n}-->S^{m}$ which in fact will be a homeomorphism. Taking inverse if necessary we may assume $n<m$. Let Failed to parse (syntax error): {\displaystyle F : Sn × [0, 1] --> Sm} 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 Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle \overline{F} : Sn × [0, 2] --> S^m} to be a homotopy of $f$ to a constant map. But then Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle f^{－1}\circ\overline{F}} is a contraction of $S^{n}$ which is a contradiction with the fact that no such contraction exists.