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Let $S^n$ be the $n$-sphere, $N=(0,0,...,0,1)$ and be the north pole of $S^n$. I am trying to show that stereographic projection gives a homeomorphism $\sigma: S^n \setminus \{N\} \to \mathbb{R}^n$.

I know that a map can be constructed by considering a line from the north pole to a point $x=(x_1,...,x_{n+1}) \in S^n$ and taking $x$ to the point that this line intersects in the plane formed by setting $x_{n+1}=0$. Ultimately we end up with a map $\sigma: S^n\setminus\{N\} \to \mathbb{R}^n$ given by $\sigma(x)=(tx_1,...,tx_n)$, where $t=\frac{1}{1-x_{n+1}}$.

It's not hard to see that $\sigma$ is injective: If $\sigma(x)=(tx_1,...,tx_n)=(ty_1,...,ty_n)=\sigma(y)$, then $tx_i=ty_i$ for all $i$, so $x_i=y_i$ and $x=y$.

For surjectivity, let $y=(y_1,...,y_n) \in \mathbb{R}^n$. Then $\sigma(x)$ where $x=(t^{-1}y_1,...,t^{-1}y_n,y_{n+1})$ and $y_{n+1}=t\cdot \sqrt{1-(t^{-1}y_1)^2-\cdots - (t^{-1}y_n)^2}$ equals $y$, so $\sigma$ is surjective.

It follows that $\sigma$ is a bijection, so now we just need to show that it is continuous. Unfortunately my point-set topology is a bit rusty, and am having a hard time doing this part. Let $U \subseteq \mathbb{R}^n$ and let $x=(x_1,...,x_n) \in U$. I am unsure of how to find an open set around $\sigma^{-1}(x)$ that is contained in $\sigma^{-1}(U)$. How do we go about doing this?

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    $\begingroup$ You can derive explicit formulas for $\sigma $ and $\sigma^{-1}$, which are in terms of functions known to be continuous. $\endgroup$ – user127096 Mar 25 '14 at 1:26
  • $\begingroup$ @cheapeffectivedietpills: Can we define $\sigma(x)=t \cdot x$? I'm less sure about $\sigma^{-1}(y)$... can we use something like $\sigma^{-1}(y)=t^{-1} \cdot y + (0,...,0,y_{n+1})$? But $y_{n+1}$ varies, so I'm not sure how that works. $\endgroup$ – Alex Petzke Mar 25 '14 at 3:12
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    $\begingroup$ For the two dimensional sphere, the formulas are here, for both directions. The generalization to higher dimensions should be straightforward. $\endgroup$ – user127096 Mar 25 '14 at 3:36
  • $\begingroup$ Why does your surjective proof work? It has $y_i$s on both sides? $\endgroup$ – mavavilj Sep 24 '18 at 22:19
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You may not need to do that. Remember that if f=(f1,...fn) and each component is continous, it will folow that your function is obvious continous. Hope it helps

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