[proofplan] We first verify that both the condition $x_i \neq 0$ and the coordinate ratios defining $\phi_i$ are independent of the representative of a projective point. Then we prove injectivity by normalizing representatives so that their $i$-th coordinate is $1$. Finally, we prove surjectivity by inserting a $1$ in the $i$-th coordinate of an arbitrary element of $K^n$, with the case $n=0$ included by the same empty-tuple convention. [/proofplan]

[step:Check that the chart condition and coordinate ratios are representative independent]Let $x = (x_0,\dots,x_n) \in K^{n+1} \setminus \{0\}$ and let $\lambda \in K^\times$. Define $y = (y_0,\dots,y_n) \in K^{n+1} \setminus \{0\}$ by $y_j = \lambda x_j$ for every $j \in \{0,\dots,n\}$. Since $\lambda \neq 0$, we have \begin{align*} y_i \neq 0 \iff x_i \neq 0. \end{align*} Thus membership in $U_i$ depends only on the projective class. Assume now that $x_i \neq 0$. For every $j \in \{0,\dots,n\}$ with $j \neq i$, the elements $x_i$ and $y_i = \lambda x_i$ are nonzero in $K$, and hence \begin{align*} \frac{y_j}{y_i} = \frac{\lambda x_j}{\lambda x_i} = \frac{x_j}{x_i}. \end{align*} Therefore the tuple of ratios defining $\phi_i([x_0:\cdots:x_n])$ is independent of the representative. Hence $\phi_i: U_i \to K^n$ is well-defined.[/step]

[guided]A point of $\mathbb{P}^n_K$ is not a single vector but an equivalence class of nonzero vectors. Therefore the first thing to verify is that the formula for $\phi_i$ gives the same answer after replacing a representative by a nonzero scalar multiple. Let $x = (x_0,\dots,x_n) \in K^{n+1} \setminus \{0\}$ be a representative, and let $\lambda \in K^\times$. Define another representative $y = (y_0,\dots,y_n) \in K^{n+1} \setminus \{0\}$ by $y_j = \lambda x_j$ for every index $j \in \{0,\dots,n\}$. Since $\lambda$ is nonzero and $K$ is a field, multiplication by $\lambda$ preserves whether an element is zero. Thus \begin{align*} y_i \neq 0 \iff x_i \neq 0. \end{align*} This proves that the subset $U_i$ is well-defined as a subset of [projective space](/page/Projective%20Space): if one representative has nonzero $i$-th coordinate, then every equivalent representative has nonzero $i$-th coordinate. Now suppose $x_i \neq 0$. For each index $j \in \{0,\dots,n\}$ with $j \neq i$, both denominators $x_i$ and $y_i = \lambda x_i$ are nonzero. Since $K$ is a field, division by these elements is defined, and cancellation of the nonzero scalar $\lambda$ gives \begin{align*} \frac{y_j}{y_i} = \frac{\lambda x_j}{\lambda x_i} = \frac{x_j}{x_i}. \end{align*} Thus every coordinate ratio used in the definition of $\phi_i$ is unchanged by replacing the representative $x$ by the equivalent representative $\lambda x$. Consequently the whole tuple \begin{align*} \left(\frac{x_0}{x_i},\ldots,\frac{x_{i-1}}{x_i},\frac{x_{i+1}}{x_i},\ldots,\frac{x_n}{x_i}\right) \end{align*} depends only on the projective point $[x_0:\cdots:x_n]$. This proves that $\phi_i$ is a well-defined map from $U_i$ to $K^n$.[/guided]

[step:Normalize representatives to prove injectivity] Let $p,q \in U_i$ and suppose $\phi_i(p) = \phi_i(q)$. Choose representatives \begin{align*} p = [x_0:\cdots:x_n] \end{align*} and \begin{align*} q = [z_0:\cdots:z_n] \end{align*} with $x_i \neq 0$ and $z_i \neq 0$. Since $\phi_i(p) = \phi_i(q)$, for every $j \in \{0,\dots,n\}$ with $j \neq i$ we have \begin{align*} \frac{x_j}{x_i} = \frac{z_j}{z_i}. \end{align*} Define $a_j := x_j/x_i$ for $j \neq i$. Then \begin{align*} [x_0:\cdots:x_n] = [a_0:\cdots:a_{i-1}:1:a_{i+1}:\cdots:a_n] \end{align*} and \begin{align*} [z_0:\cdots:z_n] = [a_0:\cdots:a_{i-1}:1:a_{i+1}:\cdots:a_n]. \end{align*} Indeed, the first equality follows by multiplying the representative $(x_0,\dots,x_n)$ by $x_i^{-1}$, and the second follows by multiplying $(z_0,\dots,z_n)$ by $z_i^{-1}$. Hence $p=q$. Therefore $\phi_i$ is injective. [/step]

[step:Insert a unit coordinate to prove surjectivity] Let $a \in K^n$. If $n \geq 1$, write \begin{align*} a = (a_0,\dots,a_{i-1},a_{i+1},\dots,a_n), \end{align*} where the displayed tuple is indexed by $\{0,\dots,n\} \setminus \{i\}$. Define $v = (v_0,\dots,v_n) \in K^{n+1}$ by $v_i = 1$ and $v_j = a_j$ for every $j \neq i$. Then $v \neq 0$ because $v_i = 1$, so the projective class $[v_0:\cdots:v_n]$ is defined and lies in $U_i$. By the definition of $\phi_i$, \begin{align*} \phi_i([v_0:\cdots:v_n]) = (a_0,\dots,a_{i-1},a_{i+1},\dots,a_n) = a. \end{align*} If $n=0$, then $i=0$, the space $K^0$ is a singleton, and $U_0 = \mathbb{P}^0_K$ consists of the single class $[1]$. The map $\phi_0: U_0 \to K^0$ sends the unique element of $U_0$ to the unique element of $K^0$, so it is surjective. Combining this with injectivity proves that $\phi_i$ is a bijection. [/step]