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

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Please, read the following carefully. It can contain some mistake.
 
Please, read the following carefully. It can contain some mistake.
  
Assume  <math>\~{f} : R^n --> R^m</math> is a homeomorphism. Since <math>\~{f}</math> is proper
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Assume  <math>\overline{f} : R^n --> R^m</math> is a homeomorphism. Since <math>\overline{f}</math> is proper we can extend it to a continuous map <math>f : S^n --> S^m</math> which in fact will be
we can extend it to a continuous map <math>f : S^n --> S^m</math> which in fact will be
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a homeomorphism. Taking inverse if necessary we may assume <math>n < m</math>. Let
 
a homeomorphism. Taking inverse if necessary we may assume <math>n < m</math>. Let
<math>F : Sn × [0, 1] --> Sm</math> be a homotopy of <math>f</math> to a smooth map i.e. <math>F</math> is continuous,
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<math>F : Sn × [0, 1] --> Sm</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>\overline{F} : Sn × [0, 2] --> S^m</math> to be a homotopy of <math>f</math> to a constant map. But then <math>f^{-1}\circ\overline{F}</math> is a contraction of <math>S^n</math> which is a contradiction with the fact that no such contraction exists.
<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
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of its image points are singular values and by Sard's theorem constitute a set
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of measure zero. Then there is a point in <math>S^m</math> not in the image of <math>F(x, 1)</math>, but
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the complement of that point is contractible. This means that we can extend
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<math>F</math> to <math>\~{F} : Sn × [0, 2] --> S^m</math> to be a homotopy of <math>f</math> to a constant map. But then
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<math>f^{-1}\circ\~{F}</math> is a contraction of <math>S^n</math> which is a contradiction with the fact that no
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such contraction exists.
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Revision as of 12:20, 18 November 2007

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 (lexing error): 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 (lexing error): \overline{F} : Sn × [0, 2] --> S^m
to be a homotopy of f to a constant map. But then Failed to parse (lexing error): f^{-1}\circ\overline{F}
is a contraction of S^n which is a contradiction with the fact that no such contraction exists.