Here is the function I want to know its inverse:
$\displaystyle\phi_{i}([x_{1},x_{2},...x_{n}])=(\frac{x_{1}}{x_{i}},...\hat{x_{i}},...\frac{x_{n}}{x_{i}})$ where the $\hat{}$ means we omit this variable.
I was told that its inverse is the following function:
$\phi_{i}^{-1}(x_{1},x_{2},...x_{n})=[x_{1},x_{2},...x_{i-1},1,x_{i+1},..x_{n}]$.
But I do not understand how, could anyone explain this to me please?
I suppose $\varphi_i:\mathbb{P}^{n-1}\setminus\{x_i=0\}\rightarrow\mathbb{C}^{n-1}$, let $\psi_i:\mathbb{C}^{n-1}\rightarrow\mathbb{P}^{n-1}\setminus\{x_i=0\}$ defined by $\psi_i(x_1,\ldots,x_{n-1})=\left[x_1,\ldots,x_{i-1},1,x_i,\ldots,x_{n-1}\right]$, let's prove that $\psi_i=\varphi_i^{-1}$. On the one hand, for $(x_1,\ldots,x_{n-1})\in\mathbb{C}^{n-1}$ we have $$ (\varphi_i\circ\psi_i)(x_1,\ldots,x_{n-1})=\varphi_i([x_1,\ldots,x_{i-1},1,x_i,\ldots,x_{n-1}])=(x_1,\ldots,x_{n-1}) $$ On the other hand, if $[x_1,\ldots,x_n]\in\mathbb{P}^{n-1}$ with $x_i\neq 0$, then $$ \begin{aligned} (\psi_i\circ\varphi_i)([x_1,\ldots,x_n]) &= \psi_i\left(\frac{x_1}{x_i},\ldots,\frac{x_{i-1}}{x_i},\frac{x_{i+1}}{x_i},\ldots,\frac{x_n}{x_i}\right) \\ &= \left[\frac{x_1}{x_i},\ldots,\frac{x_{i-1}}{x_i},1,\frac{x_{i+1}}{x_i},\ldots,\frac{x_n}{x_i}\right] \\ &= [x_1,\ldots,x_n] \end{aligned} $$ because by definition of $\mathbb{P}^{n-1}$, you can multiply all the coefficients by $x_i\neq 0$ and this doesn't change the equivalent class. Therefore the inverse of $\varphi_i$ is $\psi_i$.