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

From Drorbn

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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> | + | 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 |

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, <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. |

## Revision as of 13:21, 18 November 2007

Please, read the following carefully. It can contain some mistake.

Assume is a homeomorphism. Since is proper we can extend it to a continuous map which in fact will be
a homeomorphism. Taking inverse if necessary we may assume . Let
**Failed to parse (lexing error): F : Sn × [0, 1] --> Sm**

be a homotopy of to a smooth map i.e. is continuous, and is smooth. Since is smooth and 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 not in the image of , but the complement of that point is contractible. This means that we can extend toFailed to parse (lexing error): F_0 : Sn × [0, 2] --> S^mto be a homotopy of to a constant map. But thenFailed to parse (lexing error): f^{－1}\circ F_0is a contraction of which is a contradiction with the fact that no such contraction exists.