0708-1300/not homeomorphic - Revision history

Franklin at 17:29, 18 November 2007

2007-11-18T17:29:51Z

← Older revision		Revision as of 13:29, 18 November 2007
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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 is a contraction of <math>S^n</math> which is a contradiction with the fact that no such contraction exists.		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.

Franklin at 17:29, 18 November 2007

2007-11-18T17:29:04Z

← Older revision		Revision as of 13:29, 18 November 2007
Line 2:		Line 2:

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

Franklin at 17:27, 18 November 2007

2007-11-18T17:27:28Z

← Older revision		Revision as of 13:27, 18 November 2007
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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.		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.

Franklin at 17:27, 18 November 2007

2007-11-18T17:27:09Z

← Older revision		Revision as of 13:27, 18 November 2007
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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.		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.

Franklin at 17:26, 18 November 2007

2007-11-18T17:26:06Z

← Older revision		Revision as of 13:26, 18 November 2007
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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}~~oF_0~~</math> is a contraction of <math>S^n</math> which is a contradiction with the fact that no such contraction exists.		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.

Franklin at 17:25, 18 November 2007

2007-11-18T17:25:21Z

← Older revision		Revision as of 13:25, 18 November 2007
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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}~~\circ F_0~~</math> is a contraction of <math>S^n</math> which is a contradiction with the fact that no such contraction exists.		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}oF_0</math> is a contraction of <math>S^n</math> which is a contradiction with the fact that no such contraction exists.

Franklin at 17:24, 18 November 2007

2007-11-18T17:24:40Z

← Older revision		Revision as of 13:24, 18 November 2007
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>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		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}\circ F_0</math> is a contraction of <math>S^n</math> which is a contradiction with the fact that no such contraction exists.
	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_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.

Franklin at 17:21, 18 November 2007

2007-11-18T17:21:39Z

← Older revision		Revision as of 13:21, 18 November 2007
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>~~\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		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>~~\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 : 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.

Franklin at 17:20, 18 November 2007

2007-11-18T17:20:18Z

@@ Line 1: / Line 1: @@
 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
+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
 a homeomorphism. Taking inverse if necessary we may assume <math>n < m</math>. Let

Franklin at 17:18, 18 November 2007

2007-11-18T17:18:33Z

New page

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
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.