[proofplan] A projective point is represented by a nonzero vector in $k^{n+1}$, so at least one homogeneous coordinate is nonzero; this proves the covering statement. For a fixed index $i$, the affine chart map is given by ratios $x_j/x_i$, which are unchanged under rescaling of homogeneous coordinates. Its inverse inserts $1$ in the $i$-th homogeneous coordinate, and the two compositions are identities. Regularity follows from the standard description of the induced variety structure on $U_i$: the ratios $x_j/x_i$ are regular coordinates on $U_i$, and inserting affine coordinates into homogeneous coordinates defines a regular projective morphism. [/proofplan] [step:Use nonzero homogeneous representatives to prove the covering] Let $p\in \mathbb P_k^n$. By definition of projective space, $p$ is represented by a nonzero vector \begin{align*} a=(a_0,\dots,a_n)\in k^{n+1}\setminus\{0\}. \end{align*} Since $a\neq 0$, there exists an index $i\in\{0,\dots,n\}$ such that $a_i\neq 0$. Hence \begin{align*} p=[a_0:\cdots:a_n]\in U_i. \end{align*} Because $p$ was arbitrary, $\mathbb P_k^n\subseteq \bigcup_{i=0}^n U_i$. The reverse inclusion follows from $U_i\subseteq \mathbb P_k^n$ for each $i$, so \begin{align*} \mathbb P_k^n=\bigcup_{i=0}^n U_i. \end{align*} [/step] [step:Define the affine-coordinate map and verify it is well-defined] Fix an index $i\in\{0,\dots,n\}$. Let $A_i:=\{0,\dots,n\}\setminus\{i\}$, ordered increasingly, and identify $\mathbb A_k^n$ with [affine space](/page/Affine%20Space) having coordinate functions $(t_j)_{j\in A_i}$ in that order. Define \begin{align*} \phi_i:U_i\to \mathbb A_k^n,\quad [x_0:\cdots:x_n]\mapsto (x_j/x_i)_{j\in A_i}. \end{align*} This formula is well-defined on projective equivalence classes. Indeed, if $\lambda\in k^\times$ and \begin{align*} [y_0:\cdots:y_n]=[\lambda x_0:\cdots:\lambda x_n], \end{align*} then for every $j\in A_i$, \begin{align*} \frac{y_j}{y_i}=\frac{\lambda x_j}{\lambda x_i}=\frac{x_j}{x_i}. \end{align*} Since $x_i\neq 0$ on $U_i$, all displayed ratios are defined. [guided] Fix an index $i\in\{0,\dots,n\}$. The point of restricting to $U_i$ is that the $i$-th homogeneous coordinate is never zero, so it can be used as a denominator. Let \begin{align*} A_i:=\{0,\dots,n\}\setminus\{i\}. \end{align*} We order $A_i$ increasingly and view $\mathbb A_k^n$ as the affine space whose coordinate functions are denoted $(t_j)_{j\in A_i}$. Define \begin{align*} \phi_i:U_i\to \mathbb A_k^n,\quad [x_0:\cdots:x_n]\mapsto (x_j/x_i)_{j\in A_i}. \end{align*} The only possible ambiguity is the choice of homogeneous representative. Suppose a point of $U_i$ is also represented by \begin{align*} (y_0,\dots,y_n)=(\lambda x_0,\dots,\lambda x_n) \end{align*} for some scalar $\lambda\in k^\times$. Then $y_i=\lambda x_i\neq 0$, and for each $j\in A_i$ we have \begin{align*} \frac{y_j}{y_i}=\frac{\lambda x_j}{\lambda x_i}=\frac{x_j}{x_i}. \end{align*} Thus the affine coordinates produced by the formula do not depend on the representative. This proves that $\phi_i$ is a well-defined map from $U_i$ to $\mathbb A_k^n$. [/guided] [/step] [step:Construct the inverse by inserting one homogeneous coordinate] Define \begin{align*} \psi_i:\mathbb A_k^n\to U_i,\quad (u_j)_{j\in A_i}\mapsto [u_0:\cdots:u_{i-1}:1:u_{i+1}:\cdots:u_n]. \end{align*} The displayed homogeneous coordinate vector is nonzero because its $i$-th coordinate is $1$. Its image lies in $U_i$ because the $i$-th homogeneous coordinate is nonzero. [/step] [step:Check that the two maps are inverse to each other] Let $p=[x_0:\cdots:x_n]\in U_i$. Since $x_i\neq 0$, \begin{align*} (\psi_i\circ \phi_i)(p)=\left[\frac{x_0}{x_i}:\cdots:\frac{x_{i-1}}{x_i}:1:\frac{x_{i+1}}{x_i}:\cdots:\frac{x_n}{x_i}\right]. \end{align*} Multiplying all homogeneous coordinates by $x_i\in k^\times$ gives \begin{align*} (\psi_i\circ \phi_i)(p)=[x_0:\cdots:x_n]=p. \end{align*} Thus $\psi_i\circ \phi_i=\operatorname{id}_{U_i}$. Conversely, let $u=(u_j)_{j\in A_i}\in \mathbb A_k^n$. Then $\psi_i(u)$ has $i$-th homogeneous coordinate equal to $1$, so applying $\phi_i$ gives \begin{align*} (\phi_i\circ \psi_i)(u)=(u_j/1)_{j\in A_i}=(u_j)_{j\in A_i}=u. \end{align*} Hence $\phi_i\circ \psi_i=\operatorname{id}_{\mathbb A_k^n}$. [/step] [step:Verify regularity using the induced affine chart structure] It remains to check that $\phi_i$ and $\psi_i$ are morphisms of varieties. On the open subvariety $U_i$, the induced affine chart structure is described by the regular coordinate functions \begin{align*} x_j/x_i:U_i\to k \end{align*} for $j\in A_i$. Therefore the coordinate pullback of $\phi_i$ sends each affine coordinate function $t_j$ on $\mathbb A_k^n$ to the regular function $x_j/x_i$ on $U_i$, so $\phi_i$ is regular. For $\psi_i$, the homogeneous coordinate functions are given by the affine coordinate functions $u_j$ for $j\in A_i$ and by the constant function $1$ in the $i$-th slot. These functions do not vanish simultaneously because the $i$-th one is $1$, and they therefore define a regular morphism \begin{align*} \psi_i:\mathbb A_k^n\to \mathbb P_k^n. \end{align*} Its image lies in $U_i$, so it is a regular morphism $\mathbb A_k^n\to U_i$. Since $\phi_i$ and $\psi_i$ are inverse regular morphisms, $\phi_i$ is an isomorphism of varieties. When $n=0$, the set $A_0$ is empty, $U_0=\mathbb P_k^0$, and $\mathbb A_k^0$ is a one-point variety. The same formulas give the unique maps between these one-point varieties, so the argument includes this case. [guided] We now verify that the inverse bijections are actually inverse morphisms of varieties. The induced variety structure on the standard [open set](/page/Open%20Set) $U_i$ is the one for which the ratios \begin{align*} x_j/x_i:U_i\to k \end{align*} are regular coordinate functions for all $j\in A_i$. This is the dehomogenized coordinate description of the chart $x_i\neq 0$: dividing all homogeneous coordinates by $x_i$ normalizes the $i$-th coordinate to $1$. To prove that $\phi_i$ is regular, it is enough to check its affine coordinate functions. The coordinate function $t_j:\mathbb A_k^n\to k$ pulls back along $\phi_i$ to \begin{align*} t_j\circ \phi_i=x_j/x_i. \end{align*} This is regular on $U_i$ by the chart structure just described. Since this holds for every affine coordinate $t_j$ with $j\in A_i$, the map \begin{align*} \phi_i:U_i\to \mathbb A_k^n \end{align*} is regular. For the inverse map, the defining homogeneous coordinate functions on $\mathbb A_k^n$ are the affine coordinate functions $u_j$ for $j\in A_i$ and the constant function $1$ in the $i$-th coordinate. Thus \begin{align*} \psi_i(u)=[u_0:\cdots:u_{i-1}:1:u_{i+1}:\cdots:u_n]. \end{align*} These functions are regular on affine space. They also have no common zero, because the $i$-th function is identically $1$. Therefore they define a regular morphism from $\mathbb A_k^n$ to projective space. Its image lies in $U_i$ because the $i$-th homogeneous coordinate is always nonzero, so the same map is a regular morphism \begin{align*} \psi_i:\mathbb A_k^n\to U_i. \end{align*} We have already checked that $\psi_i\circ\phi_i=\operatorname{id}_{U_i}$ and $\phi_i\circ\psi_i=\operatorname{id}_{\mathbb A_k^n}$. Thus $\phi_i$ is a regular morphism with regular inverse $\psi_i$, which is precisely an isomorphism of varieties. If $n=0$, then there are no coordinates $u_j$, the tuple is empty, and both $\mathbb P_k^0$ and $\mathbb A_k^0$ are one-point varieties; the same construction gives the unique isomorphism between them. [/guided] [/step]