$S^n$ is not homeomorphic to $S^n-1$ [duplicate]

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  • 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!







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marked as duplicate by uniquesolution, José Carlos Santos, Jose Arnaldo Bebita Dris, Tyrone, Adrian Keister Aug 15 at 12:39


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    Remove a point, use this.
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    Aug 15 at 10:53










  • @Wojowu please consider writing an answer.
    – ThatIs
    Aug 15 at 10:54














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!







share|cite|improve this 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.
    – Wojowu
    Aug 15 at 10:53










  • @Wojowu please consider writing an answer.
    – ThatIs
    Aug 15 at 10:54












up vote
0
down vote

favorite









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!







share|cite|improve 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



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









share|cite|improve this question













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edited Aug 15 at 11:03

























asked Aug 15 at 10:51









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.
    – Wojowu
    Aug 15 at 10:53










  • @Wojowu please consider writing an answer.
    – ThatIs
    Aug 15 at 10:54












  • 1




    Remove a point, use this.
    – Wojowu
    Aug 15 at 10:53










  • @Wojowu please consider writing an answer.
    – ThatIs
    Aug 15 at 10:54







1




1




Remove a point, use this.
– Wojowu
Aug 15 at 10:53




Remove a point, use this.
– Wojowu
Aug 15 at 10:53












@Wojowu please consider writing an answer.
– ThatIs
Aug 15 at 10:54




@Wojowu please consider writing an answer.
– ThatIs
Aug 15 at 10:54










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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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    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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      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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        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.






        share|cite|improve this answer












        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.







        share|cite|improve this answer












        share|cite|improve this answer



        share|cite|improve this answer










        answered Aug 15 at 11:00









        Arthur

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