# 0708-1300/not homeomorphic

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