Question

Show that there is a homeomorphism from $S^n-(p)$ to $S^n-\{(0, \ldots, 0,1)\}$ for any $p ; S^n$. (You might want to show first that there is a homeomorphism from $S^n-(p)$ to $S^n-(q)$ for some suitable $q$ which has one more coordinate zero than $p$ has. Alternatively, use the fact that any unit vector is part of an orthonormal basis.)

   Show that there is a homeomorphism from $S^n-(p)$ to $S^n-\{(0, \ldots, 0,1)\}$ for any $p ; S^n$. (You might want to show first that there is a homeomorphism from $S^n-(p)$ to $S^n-(q)$ for some suitable $q$ which has one more coordinate zero than $p$ has. Alternatively, use the fact that any unit vector is part of an orthonormal basis.)

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Combinatorial Group Theory: A Topological Approach

Combinatorial Group Theory: A Topological Approach

Daniel E. Cohen 1st Edition

Chapter 4, Problem 3

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Step 1

It is defined as the set of points in \( \mathbb{R}^{n+1} \) that are at a unit distance from the origin, i.e., \[ S^n = \{ (x_1, x_2, \ldots, x_{n+1}) \in \mathbb{R}^{n+1} : x_1^2 + x_2^2 + \ldots + x_{n+1}^2 = 1 \}. \] Show more…

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Show that there is a homeomorphism from $S^n-(p)$ to $S^n-\{(0, \ldots, 0,1)\}$ for any $p ; S^n$. (You might want to show first that there is a homeomorphism from $S^n-(p)$ to $S^n-(q)$ for some suitable $q$ which has one more coordinate zero than $p$ has. Alternatively, use the fact that any unit vector is part of an orthonormal basis.)

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