$S^n$ is not homeomorphic to $S^n-1$ [duplicate]
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This question already has an answer here:
How do I prove that $S^n$ is homeomorphic to $S^m Rightarrow m=n$?
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My geometric intuition is clear about the fact that $S^n ncong S^n-1$ $forall n geq 2$ . It's very easy to do it for lower dimensions using simple Analysis arguments. ($S^n$ is the n-sphere in $Bbb R^n+1$)
But I really want to learn an Elementary proof for the general case (i.e. $forall n geq 2$) without using tools from Algebraic Topology, using basic General Topology arguments.
How to come up with an Elemenatry proof for the fact?
The question linked with this one, I have already visited. It accepts an answer that gives a wiki-link .As I mentioned in the question I was looking for an elementary proof using minimal machineries and hence I've posted this question!
general-topology
marked as duplicate by uniquesolution, José Carlos Santos, Jose Arnaldo Bebita Dris, Tyrone, Adrian Keister Aug 15 at 12:39
This question has been asked before and already has an answer. If those answers do not fully address your question, please ask a new question.
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up vote
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favorite
This question already has an answer here:
How do I prove that $S^n$ is homeomorphic to $S^m Rightarrow m=n$?
2 answers
My geometric intuition is clear about the fact that $S^n ncong S^n-1$ $forall n geq 2$ . It's very easy to do it for lower dimensions using simple Analysis arguments. ($S^n$ is the n-sphere in $Bbb R^n+1$)
But I really want to learn an Elementary proof for the general case (i.e. $forall n geq 2$) without using tools from Algebraic Topology, using basic General Topology arguments.
How to come up with an Elemenatry proof for the fact?
The question linked with this one, I have already visited. It accepts an answer that gives a wiki-link .As I mentioned in the question I was looking for an elementary proof using minimal machineries and hence I've posted this question!
general-topology
marked as duplicate by uniquesolution, José Carlos Santos, Jose Arnaldo Bebita Dris, Tyrone, Adrian Keister Aug 15 at 12:39
This question has been asked before and already has an answer. If those answers do not fully address your question, please ask a new question.
1
Remove a point, use this.
â Wojowu
Aug 15 at 10:53
@Wojowu please consider writing an answer.
â ThatIs
Aug 15 at 10:54
add a comment |Â
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up vote
0
down vote
favorite
This question already has an answer here:
How do I prove that $S^n$ is homeomorphic to $S^m Rightarrow m=n$?
2 answers
My geometric intuition is clear about the fact that $S^n ncong S^n-1$ $forall n geq 2$ . It's very easy to do it for lower dimensions using simple Analysis arguments. ($S^n$ is the n-sphere in $Bbb R^n+1$)
But I really want to learn an Elementary proof for the general case (i.e. $forall n geq 2$) without using tools from Algebraic Topology, using basic General Topology arguments.
How to come up with an Elemenatry proof for the fact?
The question linked with this one, I have already visited. It accepts an answer that gives a wiki-link .As I mentioned in the question I was looking for an elementary proof using minimal machineries and hence I've posted this question!
general-topology
This question already has an answer here:
How do I prove that $S^n$ is homeomorphic to $S^m Rightarrow m=n$?
2 answers
My geometric intuition is clear about the fact that $S^n ncong S^n-1$ $forall n geq 2$ . It's very easy to do it for lower dimensions using simple Analysis arguments. ($S^n$ is the n-sphere in $Bbb R^n+1$)
But I really want to learn an Elementary proof for the general case (i.e. $forall n geq 2$) without using tools from Algebraic Topology, using basic General Topology arguments.
How to come up with an Elemenatry proof for the fact?
The question linked with this one, I have already visited. It accepts an answer that gives a wiki-link .As I mentioned in the question I was looking for an elementary proof using minimal machineries and hence I've posted this question!
This question already has an answer here:
How do I prove that $S^n$ is homeomorphic to $S^m Rightarrow m=n$?
2 answers
general-topology
edited Aug 15 at 11:03
asked Aug 15 at 10:51
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ThatIs
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marked as duplicate by uniquesolution, José Carlos Santos, Jose Arnaldo Bebita Dris, Tyrone, Adrian Keister Aug 15 at 12:39
This question has been asked before and already has an answer. If those answers do not fully address your question, please ask a new question.
marked as duplicate by uniquesolution, José Carlos Santos, Jose Arnaldo Bebita Dris, Tyrone, Adrian Keister Aug 15 at 12:39
This question has been asked before and already has an answer. If those answers do not fully address your question, please ask a new question.
1
Remove a point, use this.
â Wojowu
Aug 15 at 10:53
@Wojowu please consider writing an answer.
â ThatIs
Aug 15 at 10:54
add a comment |Â
1
Remove a point, use this.
â Wojowu
Aug 15 at 10:53
@Wojowu please consider writing an answer.
â ThatIs
Aug 15 at 10:54
1
1
Remove a point, use this.
â Wojowu
Aug 15 at 10:53
Remove a point, use this.
â Wojowu
Aug 15 at 10:53
@Wojowu please consider writing an answer.
â ThatIs
Aug 15 at 10:54
@Wojowu please consider writing an answer.
â ThatIs
Aug 15 at 10:54
add a comment |Â
1 Answer
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One could use the cellular approximation theorem: any map $f:S^n-1to S^n$ is nullhomotopic, but no homeomorphism $h:S^nto S^n$ is nullhomotopic. Thus the two domains $S^n-1$ and $S^n$ of $f$ and $h$ cannot be homeomorphic.
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1 Answer
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1 Answer
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active
oldest
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active
oldest
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active
oldest
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up vote
0
down vote
One could use the cellular approximation theorem: any map $f:S^n-1to S^n$ is nullhomotopic, but no homeomorphism $h:S^nto S^n$ is nullhomotopic. Thus the two domains $S^n-1$ and $S^n$ of $f$ and $h$ cannot be homeomorphic.
add a comment |Â
up vote
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One could use the cellular approximation theorem: any map $f:S^n-1to S^n$ is nullhomotopic, but no homeomorphism $h:S^nto S^n$ is nullhomotopic. Thus the two domains $S^n-1$ and $S^n$ of $f$ and $h$ cannot be homeomorphic.
add a comment |Â
up vote
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up vote
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down vote
One could use the cellular approximation theorem: any map $f:S^n-1to S^n$ is nullhomotopic, but no homeomorphism $h:S^nto S^n$ is nullhomotopic. Thus the two domains $S^n-1$ and $S^n$ of $f$ and $h$ cannot be homeomorphic.
One could use the cellular approximation theorem: any map $f:S^n-1to S^n$ is nullhomotopic, but no homeomorphism $h:S^nto S^n$ is nullhomotopic. Thus the two domains $S^n-1$ and $S^n$ of $f$ and $h$ cannot be homeomorphic.
answered Aug 15 at 11:00
Arthur
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Remove a point, use this.
â Wojowu
Aug 15 at 10:53
@Wojowu please consider writing an answer.
â ThatIs
Aug 15 at 10:54