0708-1300/not homeomorphic: Difference between revisions
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Please, read the following carefully. It can contain some mistake. |
Please, read the following carefully. It can contain some mistake. |
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Assume <math> |
Assume <math>f_0 : R^n --> 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 --> 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 |
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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> |
<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>F_0 : Sn × [0, 2] --> S^m</math> to be a homotopy of <math>f</math> to a constant map. But then <math>f^{-1}\circ 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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Revision as of 12:21, 18 November 2007
Please, read the following carefully. It can contain some mistake.
Assume [math]\displaystyle{ f_0 : R^n --\gt R^m }[/math] is a homeomorphism. Since [math]\displaystyle{ f_0 }[/math] is proper we can extend it to a continuous map [math]\displaystyle{ f : S^n --\gt S^m }[/math] which in fact will be a homeomorphism. Taking inverse if necessary we may assume [math]\displaystyle{ n \lt m }[/math]. Let [math]\displaystyle{ F : Sn × [0, 1] --\gt Sm }[/math] be a homotopy of [math]\displaystyle{ f }[/math] to a smooth map i.e. [math]\displaystyle{ F }[/math] is continuous, [math]\displaystyle{ F(x, 0) = f(x) }[/math] and [math]\displaystyle{ F(x, 1) }[/math] is smooth. Since [math]\displaystyle{ F(x, 1) }[/math] is smooth and [math]\displaystyle{ n \lt 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]\displaystyle{ S^m }[/math] not in the image of [math]\displaystyle{ F(x, 1) }[/math], but the complement of that point is contractible. This means that we can extend [math]\displaystyle{ F }[/math] to [math]\displaystyle{ F_0 : Sn × [0, 2] --\gt S^m }[/math] to be a homotopy of [math]\displaystyle{ f }[/math] to a constant map. But then [math]\displaystyle{ f^{-1}\circ F_0 }[/math] is a contraction of [math]\displaystyle{ S^n }[/math] which is a contradiction with the fact that no such contraction exists.