How to prove that $R^nsetminus {0}$ is not contractible












1














If I must prove that $R^{n}setminus{0}$ is not contractible, how may I do so formally. Using the intuitive notion of contractibility at a point as being that any surface homeomorphic to an $n$ sphere, I was thinking that it could be proven by contradiction as follows:
Assume for the sake of contradiction that $R^{n}setminus{0}$ is contractible. Then, as $R^{n}setminus{0}$ can be continuously mapped to $S^{n-1}$, $S^{n-1}$ must also be contractible. However, as the only shape homeomorphic to $S^{n-1}$ that passes through the point under consideration is $S^{n-1}$ itself, there exists no contraction at that point of an $S^{n-1}$ sphere to a point. Thus, we have reached a contradiction, and $R^{n}setminus{0}$ is not contractible.
If indeed the formal definition of contractibility (in terms of null homotopy)is equivalent to the one I have used, could you please tell me a source for the same? (so that I may cite it for an assignment)






general-topology proof-verification geometric-topology







share|cite|improve this question




















  • 1




    you said " $mathbf{R}^n/ {0}$ can be continuously mapped to $S^{n-1}$", Is it $mathbf{R}^n/ {0}$ or $mathbf{R}^n-{0}$?
    – PSG
    Jan 4 at 9:58








  • 1




    When you write "as $R^n setminus {0}$ can be continuously mapped to $S^{n-1}$", you have to write "as $R^n setminus {0}$ is homotopy equivalent to $S^{n-1}$". If you have a continous map from $X$ to $Y$ and $X$ is contractible, $Y$ does not need to be contractible. For example, the inclusion ${N} hookrightarrow S^1$ of the north-pole to the circle is continous and ${N}$ is contractible, but $S^1$ is not.
    – Babelfish
    Jan 4 at 9:59








  • 2




    If it is $mathbf{R}^n -{0}$, homotopy equivalent does the job. If Contractible, $S^{n-1} simmathbf{R}^n -{0}sim * $. Contradiction
    – PSG
    Jan 4 at 10:01








  • 1




    See this question of the user: We are really talking about $R^n setminus {0}$, not $R^n/{0}$, which would be $R^n$ anyway.
    – Babelfish
    Jan 4 at 10:03






  • 1




    I don't understand what you mean with a "shape homeomorphic to $S^{n-1}$ that passes through the point of consideration". Neither do I really understand your "informal" definition of contractability. I think it would also help if you could state the definition of contracibility you know.
    – Babelfish
    Jan 4 at 10:22
















1














If I must prove that $R^{n}setminus{0}$ is not contractible, how may I do so formally. Using the intuitive notion of contractibility at a point as being that any surface homeomorphic to an $n$ sphere, I was thinking that it could be proven by contradiction as follows:
Assume for the sake of contradiction that $R^{n}setminus{0}$ is contractible. Then, as $R^{n}setminus{0}$ can be continuously mapped to $S^{n-1}$, $S^{n-1}$ must also be contractible. However, as the only shape homeomorphic to $S^{n-1}$ that passes through the point under consideration is $S^{n-1}$ itself, there exists no contraction at that point of an $S^{n-1}$ sphere to a point. Thus, we have reached a contradiction, and $R^{n}setminus{0}$ is not contractible.
If indeed the formal definition of contractibility (in terms of null homotopy)is equivalent to the one I have used, could you please tell me a source for the same? (so that I may cite it for an assignment)






general-topology proof-verification geometric-topology







share|cite|improve this question




















  • 1




    you said " $mathbf{R}^n/ {0}$ can be continuously mapped to $S^{n-1}$", Is it $mathbf{R}^n/ {0}$ or $mathbf{R}^n-{0}$?
    – PSG
    Jan 4 at 9:58








  • 1




    When you write "as $R^n setminus {0}$ can be continuously mapped to $S^{n-1}$", you have to write "as $R^n setminus {0}$ is homotopy equivalent to $S^{n-1}$". If you have a continous map from $X$ to $Y$ and $X$ is contractible, $Y$ does not need to be contractible. For example, the inclusion ${N} hookrightarrow S^1$ of the north-pole to the circle is continous and ${N}$ is contractible, but $S^1$ is not.
    – Babelfish
    Jan 4 at 9:59








  • 2




    If it is $mathbf{R}^n -{0}$, homotopy equivalent does the job. If Contractible, $S^{n-1} simmathbf{R}^n -{0}sim * $. Contradiction
    – PSG
    Jan 4 at 10:01








  • 1




    See this question of the user: We are really talking about $R^n setminus {0}$, not $R^n/{0}$, which would be $R^n$ anyway.
    – Babelfish
    Jan 4 at 10:03






  • 1




    I don't understand what you mean with a "shape homeomorphic to $S^{n-1}$ that passes through the point of consideration". Neither do I really understand your "informal" definition of contractability. I think it would also help if you could state the definition of contracibility you know.
    – Babelfish
    Jan 4 at 10:22














1












1








1


2





If I must prove that $R^{n}setminus{0}$ is not contractible, how may I do so formally. Using the intuitive notion of contractibility at a point as being that any surface homeomorphic to an $n$ sphere, I was thinking that it could be proven by contradiction as follows:
Assume for the sake of contradiction that $R^{n}setminus{0}$ is contractible. Then, as $R^{n}setminus{0}$ can be continuously mapped to $S^{n-1}$, $S^{n-1}$ must also be contractible. However, as the only shape homeomorphic to $S^{n-1}$ that passes through the point under consideration is $S^{n-1}$ itself, there exists no contraction at that point of an $S^{n-1}$ sphere to a point. Thus, we have reached a contradiction, and $R^{n}setminus{0}$ is not contractible.
If indeed the formal definition of contractibility (in terms of null homotopy)is equivalent to the one I have used, could you please tell me a source for the same? (so that I may cite it for an assignment)






general-topology proof-verification geometric-topology







share|cite|improve this question















If I must prove that $R^{n}setminus{0}$ is not contractible, how may I do so formally. Using the intuitive notion of contractibility at a point as being that any surface homeomorphic to an $n$ sphere, I was thinking that it could be proven by contradiction as follows:
Assume for the sake of contradiction that $R^{n}setminus{0}$ is contractible. Then, as $R^{n}setminus{0}$ can be continuously mapped to $S^{n-1}$, $S^{n-1}$ must also be contractible. However, as the only shape homeomorphic to $S^{n-1}$ that passes through the point under consideration is $S^{n-1}$ itself, there exists no contraction at that point of an $S^{n-1}$ sphere to a point. Thus, we have reached a contradiction, and $R^{n}setminus{0}$ is not contractible.
If indeed the formal definition of contractibility (in terms of null homotopy)is equivalent to the one I have used, could you please tell me a source for the same? (so that I may cite it for an assignment)






general-topology proof-verification geometric-topology




general-topology proof-verification geometric-topology






share|cite|improve this question















share|cite|improve this question













share|cite|improve this question




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edited Jan 4 at 10:08









Babelfish

1,128520




1,128520










asked Jan 4 at 9:47









Aryaman GuptaAryaman Gupta

336




336








  • 1




    you said " $mathbf{R}^n/ {0}$ can be continuously mapped to $S^{n-1}$", Is it $mathbf{R}^n/ {0}$ or $mathbf{R}^n-{0}$?
    – PSG
    Jan 4 at 9:58








  • 1




    When you write "as $R^n setminus {0}$ can be continuously mapped to $S^{n-1}$", you have to write "as $R^n setminus {0}$ is homotopy equivalent to $S^{n-1}$". If you have a continous map from $X$ to $Y$ and $X$ is contractible, $Y$ does not need to be contractible. For example, the inclusion ${N} hookrightarrow S^1$ of the north-pole to the circle is continous and ${N}$ is contractible, but $S^1$ is not.
    – Babelfish
    Jan 4 at 9:59








  • 2




    If it is $mathbf{R}^n -{0}$, homotopy equivalent does the job. If Contractible, $S^{n-1} simmathbf{R}^n -{0}sim * $. Contradiction
    – PSG
    Jan 4 at 10:01








  • 1




    See this question of the user: We are really talking about $R^n setminus {0}$, not $R^n/{0}$, which would be $R^n$ anyway.
    – Babelfish
    Jan 4 at 10:03






  • 1




    I don't understand what you mean with a "shape homeomorphic to $S^{n-1}$ that passes through the point of consideration". Neither do I really understand your "informal" definition of contractability. I think it would also help if you could state the definition of contracibility you know.
    – Babelfish
    Jan 4 at 10:22














  • 1




    you said " $mathbf{R}^n/ {0}$ can be continuously mapped to $S^{n-1}$", Is it $mathbf{R}^n/ {0}$ or $mathbf{R}^n-{0}$?
    – PSG
    Jan 4 at 9:58








  • 1




    When you write "as $R^n setminus {0}$ can be continuously mapped to $S^{n-1}$", you have to write "as $R^n setminus {0}$ is homotopy equivalent to $S^{n-1}$". If you have a continous map from $X$ to $Y$ and $X$ is contractible, $Y$ does not need to be contractible. For example, the inclusion ${N} hookrightarrow S^1$ of the north-pole to the circle is continous and ${N}$ is contractible, but $S^1$ is not.
    – Babelfish
    Jan 4 at 9:59








  • 2




    If it is $mathbf{R}^n -{0}$, homotopy equivalent does the job. If Contractible, $S^{n-1} simmathbf{R}^n -{0}sim * $. Contradiction
    – PSG
    Jan 4 at 10:01








  • 1




    See this question of the user: We are really talking about $R^n setminus {0}$, not $R^n/{0}$, which would be $R^n$ anyway.
    – Babelfish
    Jan 4 at 10:03






  • 1




    I don't understand what you mean with a "shape homeomorphic to $S^{n-1}$ that passes through the point of consideration". Neither do I really understand your "informal" definition of contractability. I think it would also help if you could state the definition of contracibility you know.
    – Babelfish
    Jan 4 at 10:22








1




1




you said " $mathbf{R}^n/ {0}$ can be continuously mapped to $S^{n-1}$", Is it $mathbf{R}^n/ {0}$ or $mathbf{R}^n-{0}$?
– PSG
Jan 4 at 9:58






you said " $mathbf{R}^n/ {0}$ can be continuously mapped to $S^{n-1}$", Is it $mathbf{R}^n/ {0}$ or $mathbf{R}^n-{0}$?
– PSG
Jan 4 at 9:58






1




1




When you write "as $R^n setminus {0}$ can be continuously mapped to $S^{n-1}$", you have to write "as $R^n setminus {0}$ is homotopy equivalent to $S^{n-1}$". If you have a continous map from $X$ to $Y$ and $X$ is contractible, $Y$ does not need to be contractible. For example, the inclusion ${N} hookrightarrow S^1$ of the north-pole to the circle is continous and ${N}$ is contractible, but $S^1$ is not.
– Babelfish
Jan 4 at 9:59






When you write "as $R^n setminus {0}$ can be continuously mapped to $S^{n-1}$", you have to write "as $R^n setminus {0}$ is homotopy equivalent to $S^{n-1}$". If you have a continous map from $X$ to $Y$ and $X$ is contractible, $Y$ does not need to be contractible. For example, the inclusion ${N} hookrightarrow S^1$ of the north-pole to the circle is continous and ${N}$ is contractible, but $S^1$ is not.
– Babelfish
Jan 4 at 9:59






2




2




If it is $mathbf{R}^n -{0}$, homotopy equivalent does the job. If Contractible, $S^{n-1} simmathbf{R}^n -{0}sim * $. Contradiction
– PSG
Jan 4 at 10:01






If it is $mathbf{R}^n -{0}$, homotopy equivalent does the job. If Contractible, $S^{n-1} simmathbf{R}^n -{0}sim * $. Contradiction
– PSG
Jan 4 at 10:01






1




1




See this question of the user: We are really talking about $R^n setminus {0}$, not $R^n/{0}$, which would be $R^n$ anyway.
– Babelfish
Jan 4 at 10:03




See this question of the user: We are really talking about $R^n setminus {0}$, not $R^n/{0}$, which would be $R^n$ anyway.
– Babelfish
Jan 4 at 10:03




1




1




I don't understand what you mean with a "shape homeomorphic to $S^{n-1}$ that passes through the point of consideration". Neither do I really understand your "informal" definition of contractability. I think it would also help if you could state the definition of contracibility you know.
– Babelfish
Jan 4 at 10:22




I don't understand what you mean with a "shape homeomorphic to $S^{n-1}$ that passes through the point of consideration". Neither do I really understand your "informal" definition of contractability. I think it would also help if you could state the definition of contracibility you know.
– Babelfish
Jan 4 at 10:22










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