0708-1300/not homeomorphic: Difference between revisions
No edit summary |
No edit summary |
||
| (8 intermediate revisions by the same user not shown) | |||
| Line 1: | Line 1: | ||
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 \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. |
|||
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 |
|||
<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} : 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}</math> is a contraction of <math>S^n</math> which is a contradiction with the fact that no |
|||
such contraction exists. |
|||
Latest revision as of 12:29, 18 November 2007
Please, read the following carefully. It can contain some mistake.
Assume [math]\displaystyle{ f_0 : R^n \rightarrow 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 \rightarrow 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:S^n\times[0, 1] \rightarrow S^m }[/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:S^n\times[0, 2]\rightarrow S^m }[/math] to be a homotopy of [math]\displaystyle{ f }[/math] to a constant map. But then [math]\displaystyle{ f^{-1}F_{0} }[/math] is a contraction of [math]\displaystyle{ S^n }[/math] which is a contradiction with the fact that no such contraction exists.
