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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)

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    $\begingroup$ you said " $\mathbf{R}^n/ \{0\}$ can be continuously mapped to $S^{n-1}$", Is it $\mathbf{R}^n/ \{0\}$ or $\mathbf{R}^n-\{0\}$? $\endgroup$
    – Partha
    Jan 4, 2019 at 9:58
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    $\begingroup$ 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. $\endgroup$
    – Babelfish
    Jan 4, 2019 at 9:59
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    $\begingroup$ If it is $\mathbf{R}^n -\{0\}$, homotopy equivalent does the job. If Contractible, $S^{n-1} \sim\mathbf{R}^n -\{0\}\sim * $. Contradiction $\endgroup$
    – Partha
    Jan 4, 2019 at 10:01
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    $\begingroup$ 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. $\endgroup$
    – Babelfish
    Jan 4, 2019 at 10:03
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    $\begingroup$ 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. $\endgroup$
    – Babelfish
    Jan 4, 2019 at 10:22

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