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I was reading the solutions to problem 1(iv)(c) in this link: http://math.bu.edu/people/mabeck/Autumn13/sample_exam_f10pc_solutions.pdf

And they seem to claim that $S^2 \setminus \{p\}$ is homeomorphic to $\mathbb R^2$. I am having trouble visualizing this.

I attempted to apply the same reasoning as here Homeomorphism between $\mathbb{R^2}$ and $S^2-N$, the sphere without its north pole but could not construct the homeomorphism in question. Any suggestions?

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    $\begingroup$ Here's a hint: Can you write down a homeomorphism $S^1-\{p\}\to\Bbb R$? You can see this geometrically or, certainly, by recognizing that an open interval in $\Bbb R$ is homeomorphic to $\Bbb R$. $\endgroup$
    – Ted Shifrin
    Apr 27, 2019 at 18:11
  • $\begingroup$ you can find a rotation $R\in SO(3)$ such that $Rp=N$ and then apply the homeomorphism. $\endgroup$
    – Pink Panther
    Apr 27, 2019 at 18:11

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Consider the map $\psi$ from $\mathbb R^2\times\{0\}$ into $S^2\setminus\{(0,0,1)\}$ thus defined: for each $p\in\mathbb R^2\times\{0\}$, consider the line passing through $(1,0,0)$ and $p$ and let $\psi(p)$ be the intersection of that line with the sphere $S^2$ (that is, the point which is distinct from $(0,0,1)$. You will get:$$\psi(x,y,0)=\left(\frac{2x}{1+x^2+y^2},\frac{2y}{1+x^2+y^2},\frac{-1+x^2+y^2}{1+x^2+y^2}\right).$$

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  • $\begingroup$ Thanks for the suggestion! It seems that this idea would generalize for higher dimensions. Can we use the same idea to build homeomorphism between $S^n \setminus \{p\}$ and $\mathbb R^n$? $\endgroup$
    – Firage
    Apr 27, 2019 at 18:19
  • $\begingroup$ Yes, we sure can! $\endgroup$
    – José Carlos Santos
    Apr 27, 2019 at 18:20
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    $\begingroup$ just a comment: check "stereographic projection" on internet for a generalization of the idea of José Carlos $\endgroup$
    – Masacroso
    Apr 27, 2019 at 18:34

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