2023-03-05

227: Stereographic Projection Is Homeomorphism

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A description/proof of that stereographic projection is homeomorphism

Topics


About: topological space

The table of contents of this article


Starting Context



Target Context


  • The reader will have a description and a proof of the proposition that any stereographic projection is a homeomorphism.

Orientation


There is a list of definitions discussed so far in this site.

There is a list of propositions discussed so far in this site.


Main Body


1: Description


For any n-sphere, \(S^n \subseteq \mathbb{R}^{n + 1}\), the stereographic projection, \(f: S^n \setminus \{p_n\} \rightarrow \mathbb{R}^n\) where \(p_n\) is the north pole of \(S^n\), is a homeomorphism.


2: Proof


The line that goes through \(p_n\) and any \(p \in S^n\) is \(\vec{p_n} + (\vec{p} - \vec{p_n}) t\) where \(t \in \mathbb{R}\). The corresponding point, \(q \in \mathbb{R}^n\), is determined by \(\vec{q} = \vec{p_n} + (\vec{p} - \vec{p_n}) t\) where \(q^{n + 1} = 0\). As \(p_n = (0, 0, . . ., 1)\), \(0 = 1 + (p^{n + 1} - 1) t\), \(t = (1 - p^{n + 1})^{-1}\), \(q^i = (1 - p^{n + 1})^{-1} p^i\). \({q^1}^2 + {q^2}^2 + . . . + {q^n}^2 = (1 - p^{n + 1})^{-2} ({p^1}^2 + {p^2}^2 + . . . + {p^n}^2) = (1 - p^{n + 1})^{-2} (1 - {p^{n + 1}}^2) = (1 - p^{n + 1})^{-1} (1 + p^{n + 1})\), \((1 - p^{n + 1}) ({q^1}^2 + {q^2}^2 + . . . + {q^n}^2) = 1 + p^{n + 1}\), \({q^1}^2 + {q^2}^2 + . . . + {q^n}^2 - 1 = p^{n + 1} (1 + {q^1}^2 + {q^2}^2 + . . . + {q^n}^2)\), \(p^{n + 1} = (1 + {q^1}^2 + {q^2}^2 + . . . + {q^n}^2)^{-1} (-1 + {q^1}^2 + {q^2}^2 + . . . + {q^n}^2)\), \(p^i = (1 - p^{n + 1}) q^i = (1 - (1 + {q^1}^2 + {q^2}^2 + . . . + {q^n}^2)^{-1} (-1 + {q^1}^2 + {q^2}^2 + . . . + {q^n}^2)) q^i\).

Let us think of \(T := \{p \in \mathbb{R}^{n + 1}\vert p^{n + 1} \lt 1\}\) as a subspace of \(\mathbb{R}^{n + 1}\). As \(T\) is open on \(\mathbb{R}^{n + 1}\), \((T, id)\) where \(id\) is the identity map, \(id: T \rightarrow T\), is a \(C^\infty\) atlas for \(T\). Let us think of \(f': T \rightarrow \mathbb{R}^n, p \mapsto q, (f' (p))^i = (1 - p^{n + 1})^{-1} p^i\). By the proposition that for any map between \(C^\infty\) manifolds, its continuousness in the topological sense equals its continuousness in the norm sense for the coordinates functions, \(f'\) is continuous. As \(f = f'\vert_{S^n \setminus \{p_n\}}\), by the proposition that any restriction of any continuous map on the domain and the codomain is continuous, \(f\) is continuous.

\(f\) is a bijection, as the reverse map, \(f^{-1}: \mathbb{R}^n \rightarrow S^n \setminus \{p_n\}, q \mapsto p, p^i = (1 - (1 + {q^1}^2 + {q^2}^2 + . . . + {q^n}^2)^{-1} (-1 + {q^1}^2 + {q^2}^2 + . . . + {q^n}^2)) q^i\) for \(i \leq n\), \(p^{n + 1} = (1 + {q^1}^2 + {q^2}^2 + . . . + {q^n}^2)^{-1} (-1 + {q^1}^2 + {q^2}^2 + . . . + {q^n}^2)\), has been already found. Let us think of \({f^{-1}}': \mathbb{R}^n \rightarrow T\). By the proposition that for any map between \(C^\infty\) manifolds, its continuousness in the topological sense equals its continuousness in the norm sense for the coordinates functions, \({f^{-1}}'\) is continuous. By the proposition that any restriction of any continuous map on the domain and the codomain is continuous, \(f^{-1}\) is continuous with the codomain restricted.


3: Note


\(T\) and \(f'\) and \({f^{-1}}'\) have to be introduced in order to evoke the proposition that for any map between \(C^\infty\) manifolds, its continuousness in the topological sense equals its continuousness in the norm sense for the coordinates functions, because \(x^1, x^2, . . ., x^{n + 1}\) is not any coordinates for \(S^n \setminus \{p_n\}\).


References


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