# Difference between revisions of "0708-1300/not homeomorphic"

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 is a contraction of $S^n$ which is a contradiction with the fact that no such contraction exists.