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Consider the picture description given here: Understanding the Atlas of $\mathbb RP^n$. Let $\pi$ be the canonical projection $\pi:\mathbb{R}^{n+1}\setminus\{0\}\to \mathbb{R}P^n$. In the comments, and in Wikipedia, the inverse of the chart $\varphi_i:\pi(U_i)\to \mathbb{R}^n, \varphi_i(x_1,\dots,x_{n+1}) = \left(\frac{x_1}{x_i},\dots,\hat{x}_i,\dots,\frac{x_1}{x_i}\right)$ ,$U_i = \{(x_1,\dots,x_{n+1}) \in \mathbb{R}^{n+1}\setminus\{0\}\mid x_i \neq 0\}$ (where the hat denotes a coordinate that is to be left out), is claimed to be $\varphi_i^{-1}(x_1,\dots,x_n) = (x_1,\dots,x_{i-1},1,x_{i},\dots,x_{n})$. What I don't understand is that how is $\varphi^{-1}_i$ the inverse of $\varphi_i$, when the $i$th coordinate is mapped back to $1$, while it could have been, say, $x_i = 0.5$?

As an additional question, under the assumption that the given mappings are in fact charts, wouldn't they also work for $\overline{U}_i = \{(x_1,\dots,x_{n + 1})\in \mathbb{S}^n \subset \mathbb{R}^{n+1}\mid x_i \neq 0\}$?

Wasradin
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    There's an equivalence relation or class missing. – psl2Z Apr 18 '22 at 17:57
  • @psl2Z What do you mean? – Wasradin Apr 18 '22 at 18:28
  • $x \sim y$ for $x,y \in \mathbb{R}^{n+1}\setminus{0}$ precisely if there is some $\lambda \in \mathbb{R}$ with $y = \lambda x$. The $U_i$ should be $\pi(U_i)$, where $\pi$ is the canonical projection $\mathbb{R}^{n+1}\setminus{0} \to \mathbb{R}P^n$. – psl2Z Apr 18 '22 at 19:09
  • @psl2Z I see! However, even with this modification, I don't really understand why the claimed inverse mapping is in fact an inverse mapping. – Wasradin Apr 18 '22 at 19:39
  • In each equivalence class there is precisely one element with "1" in the $i$-th place. – Арсений Кряжев Apr 18 '22 at 23:22
  • Several mistakes. You need $\pi$ mapping $\Bbb R^{n+1}\setminus {0}$, and you need $\varphi$ applied to the equivalence class $[x_1,\dots,x_{n+1}]$. – Ted Shifrin Apr 19 '22 at 00:07
  • An example for $n=4$ and $i=2$: $\phi_2(x_1, 0.5, x_3, x_4) = (2 x_1, 2 x_3, 2x_4)$, and $\phi^{-1}(2x_1, 2x_3, 2x_4) = (x_1, 1, x_3, x4)$. But as far as as $\mathbb R P^3$ is concerned, $(x_1, 0.5, x_3, x_4)$ and $(2x_1, 1, 2x_3, 2x_4)$ denote the same point. – blamocur Apr 19 '22 at 01:14
  • @blamocur How is $\varphi^{-1}(2x_1, 2x_3, 2x_4) = (x_1, 1, x_3, x_4)$? Shouldn't the image be either $(2x_1, 1, 2x_3, 2x_4)$ or $(x_1, 0.5, x_3, x_4)$ like you wrote? – Wasradin Apr 19 '22 at 06:39
  • @SickSeries You are right, it should be $(2x_1,1,2x_3,2x_4)$ -- thanks for catching the typo. – blamocur Apr 19 '22 at 14:13

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