[proofplan] We construct the correspondence in both directions. A regular map $\varphi:X\to Y$ pulls regular functions on $Y$ back to regular functions on $X$, hence gives a unital $k$-algebra homomorphism after identifying regular functions with coordinate ring elements. Conversely, a homomorphism $\alpha:k[Y]\to k[X]$ determines the images of the coordinate classes $\bar y_1,\dots,\bar y_m$, and these functions define a map $X\to \mathbb A_k^m$ whose image lies in $Y$ because all equations in $I(Y)$ vanish after applying $\alpha$. The two constructions are inverse because the coordinate classes generate $k[Y]$ as a $k$-algebra. [/proofplan]

[step:Pull a regular map back to a coordinate ring homomorphism] Let $\varphi:X\to Y$ be a regular map. Define \begin{align*} \varphi^*:k[Y]\to k[X] \end{align*} by sending a coordinate ring element represented as a regular function $g:Y\to k$ to the regular function $g\circ\varphi:X\to k$. The hypotheses of [citetheorem:9419] apply to both $X$ and $Y$ because they are affine varieties over the [algebraically closed field](/page/Algebraically%20Closed%20Field) $k$. Thus coordinate ring elements are the same as regular functions on these affine varieties. The function $g\circ\varphi:X\to k$ is regular because regularity of a map into an affine variety means that each regular coordinate expression on the target pulls back to a regular function on the source; equivalently, after representing $g$ locally by a polynomial expression in the ambient coordinate functions on $Y$, its pullback is the same polynomial expression in the regular coordinate functions of $\varphi$. Hence $\varphi^*$ is well-defined as a map into $k[X]$. For $g,h\in k[Y]$ and $a\in k$, pointwise evaluation on $X$ gives \begin{align*} \varphi^*(g+h)=\varphi^*(g)+\varphi^*(h). \end{align*} Similarly, \begin{align*} \varphi^*(gh)=\varphi^*(g)\varphi^*(h). \end{align*} For every $a\in k$, viewing $a$ as a constant regular function on $Y$ gives \begin{align*} \varphi^*(a)=a. \end{align*} Also, \begin{align*} \varphi^*(1)=1. \end{align*} Thus $\varphi^*:k[Y]\to k[X]$ is a unital $k$-algebra homomorphism. [/step]

[step:Build a regular map from a coordinate ring homomorphism]Let \begin{align*} \alpha:k[Y]\to k[X] \end{align*} be a unital $k$-algebra homomorphism. For each index $j\in\{1,\dots,m\}$, let $\bar y_j\in k[Y]$ denote the class of the coordinate polynomial $y_j$, and define \begin{align*} f_j:=\alpha(\bar y_j)\in k[X]. \end{align*} Since $X$ is an affine variety over the algebraically closed field $k$, [citetheorem:9419] applies to $X$. Therefore each $f_j$ is a regular function $f_j:X\to k$. Define a map \begin{align*} \varphi_\alpha:X\to \mathbb A_k^m \end{align*} by \begin{align*} \varphi_\alpha(p)=(f_1(p),\dots,f_m(p)). \end{align*} We prove that $\varphi_\alpha(p)\in Y$ for every $p\in X$. Let $H\in I(Y)$ be any defining polynomial. Since $H(\bar y_1,\dots,\bar y_m)=0$ in $k[Y]$, applying $\alpha$ gives \begin{align*} H(f_1,\dots,f_m)=0 \end{align*} in $k[X]$. Interpreting this equality as equality of regular functions on $X$, we obtain, for every $p\in X$, \begin{align*} H(f_1(p),\dots,f_m(p))=0. \end{align*} Since this holds for every $H\in I(Y)$, the point $\varphi_\alpha(p)$ lies in $V(I(Y))=Y$. Therefore $\varphi_\alpha$ is a map $X\to Y$. Its coordinate functions are the regular functions $f_1,\dots,f_m$, so $\varphi_\alpha:X\to Y$ is regular.[/step]

[guided]The homomorphism $\alpha:k[Y]\to k[X]$ tells us where every regular function on $Y$ should pull back to on $X$. To recover a map of varieties, it is enough to recover the coordinate functions of that map. Let $\bar y_j\in k[Y]$ be the class of the $j$th coordinate function on $\mathbb A_k^m$, and define \begin{align*} f_j:=\alpha(\bar y_j)\in k[X] \end{align*} for each $j\in\{1,\dots,m\}$. Since $X$ is an affine variety over the algebraically closed field $k$, [citetheorem:9419] applies to $X$. Hence elements of $k[X]$ are precisely regular functions on $X$, so each $f_j$ is a regular function $f_j:X\to k$. These functions define a candidate map to [affine space](/page/Affine%20Space): \begin{align*} \varphi_\alpha:X\to \mathbb A_k^m \end{align*} by \begin{align*} \varphi_\alpha(p)=(f_1(p),\dots,f_m(p)). \end{align*} The only point that needs checking is that the image actually lands in $Y$, not merely in the ambient affine space $\mathbb A_k^m$. By definition, $Y=V(I(Y))$, so a point $q\in\mathbb A_k^m$ lies in $Y$ exactly when every polynomial in $I(Y)$ vanishes at $q$. Take an arbitrary polynomial $H\in I(Y)$. In the [quotient ring](/page/Quotient%20Ring) $k[Y]=k[y_1,\dots,y_m]/I(Y)$, the class of $H$ is zero, which is the same as saying \begin{align*} H(\bar y_1,\dots,\bar y_m)=0. \end{align*} Applying the $k$-algebra homomorphism $\alpha$ preserves polynomial expressions in the generators, so \begin{align*} H(\alpha(\bar y_1),\dots,\alpha(\bar y_m))=0 \end{align*} in $k[X]$. By the definition of the functions $f_j$, this is \begin{align*} H(f_1,\dots,f_m)=0 \end{align*} in $k[X]$. Evaluating this regular function at an arbitrary point $p\in X$ gives \begin{align*} H(f_1(p),\dots,f_m(p))=0. \end{align*} Thus every polynomial in $I(Y)$ vanishes at $\varphi_\alpha(p)$, so $\varphi_\alpha(p)\in Y$. Since the coordinate functions of $\varphi_\alpha$ are regular, $\varphi_\alpha:X\to Y$ is a regular map.[/guided]

[step:Show the two constructions are inverse] Starting with a regular map $\varphi:X\to Y$, form $\varphi^*:k[Y]\to k[X]$ and then construct $\varphi_{\varphi^*}:X\to Y$ from the images of the coordinate classes. For each $j\in\{1,\dots,m\}$, \begin{align*} \varphi^*(\bar y_j)=\bar y_j\circ\varphi. \end{align*} Thus the $j$th coordinate function of $\varphi_{\varphi^*}$ agrees with the $j$th coordinate function of $\varphi$. Hence \begin{align*} \varphi_{\varphi^*}=\varphi. \end{align*} Conversely, start with a unital $k$-algebra homomorphism $\alpha:k[Y]\to k[X]$ and construct $\varphi_\alpha:X\to Y$. The pullback $\varphi_\alpha^*$ satisfies \begin{align*} \varphi_\alpha^*(\bar y_j)=f_j=\alpha(\bar y_j) \end{align*} for each $j\in\{1,\dots,m\}$. The elements $\bar y_1,\dots,\bar y_m$ generate $k[Y]$ as a $k$-algebra, so two $k$-algebra homomorphisms out of $k[Y]$ that agree on all $\bar y_j$ agree on all of $k[Y]$. Therefore \begin{align*} \varphi_\alpha^*=\alpha. \end{align*} The two assignments are inverse to each other, so the pullback assignment gives the desired bijection. [/step]