227: Stereographic Projection Is Homeomorphism

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A description/proof of that stereographic projection is homeomorphism

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Topics

About: topological space

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The table of contents of this article

• Starting Context
• Target Context
• Orientation
• Main Body
• 1: Description
• 2: Proof
• 3: Note

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Starting Context

• The reader knows a definition of topological space.
The reader knows a definition of n-sphere.

The reader knows a definition of stereographic projection.

The reader knows a definition of homeomorphism.

The reader admits the proposition that for any map between $C^{\mathrm{\infty }}$ manifolds, its continuousness in the topological sense equals its continuousness in the norm sense for the coordinates functions.

The reader admits the proposition that any restriction of any continuous map on the domain and the codomain is continuous.

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Target Context

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

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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.

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Main Body

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1: Description

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

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2: Proof

The line that goes through $p_n$ and any $p\in S^n$ is $\overrightarrow{p_n}+(\overrightarrow{p}-\overrightarrow{p_n})t$ where $t\in \mathbb{R}$. The corresponding point, $q\in {\mathbb{R}}^n$, is determined by $\overrightarrow{q}=\overrightarrow{p_n}+(\overrightarrow{p}-\overrightarrow{p_n})t$ 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 $T:=\{p\in {\mathbb{R}}^{n+1}|p^{n+1}<1\}$ 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, $id:T\to T$, is a $C^{\mathrm{\infty }}$ atlas for $T$. Let us think of $f^{\prime }:T\to {\mathbb{R}}^n,p\mapsto q,(f^{\prime }(p){)}^i=(1-p^{n+1}{)}^{-1}{p}^i$. By the proposition that for any map between $C^{\mathrm{\infty }}$ manifolds, its continuousness in the topological sense equals its continuousness in the norm sense for the coordinates functions, $f^{\prime }$ is continuous. As $f=f^{\prime }{|}_{S^n\setminus \{p_n\}}$, 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, $f^{-1}:{\mathbb{R}}^n\to S^n\setminus \{p_n\},q\mapsto p,p^i=(1-(1+{q^1}^2+{q^2}^2+...+{q^n}^2{)}^{-1}(-1+{q^1}^2+{q^2}^2+...+{q^n}^2))q^i$ for $i\le n$, $p^{n+1}=(1+{q^1}^2+{q^2}^2+...+{q^n}^2{)}^{-1}(-1+{q^1}^2+{q^2}^2+...+{q^n}^2)$, has been already found. Let us think of ${f^{-1}}^{\prime }:{\mathbb{R}}^n\to T$. By the proposition that for any map between $C^{\mathrm{\infty }}$ manifolds, its continuousness in the topological sense equals its continuousness in the norm sense for the coordinates functions, ${f^{-1}}^{\prime }$ 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.

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3: Note

$T$ and $f^{\prime }$ and ${f^{-1}}^{\prime }$ have to be introduced in order to evoke the proposition that for any map between $C^{\mathrm{\infty }}$ 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 $S^n\setminus \{p_n\}$.

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References

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