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

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

Revision as of 12:27, 18 November 2007

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 Failed to parse (lexing error): f^{-1}F_0

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