[proofplan] The proof is the standard contravariant construction. A morphism gives a pullback map by composing regular functions, and this immediately respects the $k$-algebra operations. Conversely, a $k$-algebra homomorphism is determined by the images of the coordinate classes on $Y$; choosing polynomial representatives for those images defines coordinate functions on $X$. The defining equations of $Y$ are respected because relations in $I(Y)$ map to zero in $k[X]$, so the resulting polynomial map lands in $Y$, and uniqueness follows from the coordinate functions. [/proofplan] [step:Pull back regular functions along a morphism] Let \begin{align*} \phi: X \to Y \end{align*} be a morphism of affine varieties. By the definition of morphism between affine varieties, if $g \in k[Y]$ is a regular function on $Y$, then the composite function \begin{align*} g \circ \phi: X \to k \end{align*} is a regular function on $X$, hence represents an element of $k[X]$. Therefore the assignment \begin{align*} \phi^*: k[Y] \to k[X], \quad g \mapsto g \circ \phi \end{align*} is well-defined. For $g_1,g_2 \in k[Y]$ and $a \in k$, the pointwise identities on $X$ give \begin{align*} \phi^*(g_1+g_2)=(g_1+g_2)\circ\phi=(g_1\circ\phi)+(g_2\circ\phi)=\phi^*(g_1)+\phi^*(g_2). \end{align*} Similarly, \begin{align*} \phi^*(g_1g_2)=(g_1g_2)\circ\phi=(g_1\circ\phi)(g_2\circ\phi)=\phi^*(g_1)\phi^*(g_2). \end{align*} Also, \begin{align*} \phi^*(a)=a \end{align*} because the constant function with value $a$ pulls back to the constant function with value $a$, and \begin{align*} \phi^*(1)=1. \end{align*} Thus $\phi^*$ is a unital $k$-algebra homomorphism. [/step] [step:Recover coordinate functions from a $k$-algebra homomorphism] Let \begin{align*} \alpha: k[Y] \to k[X] \end{align*} be a unital $k$-algebra homomorphism. For each $j \in \{1,\ldots,m\}$, let $\overline{y}_j \in k[Y]$ denote the residue class of the coordinate polynomial $y_j \in k[y_1,\ldots,y_m]$. Define \begin{align*} f_j := \alpha(\overline{y}_j) \in k[X]. \end{align*} Since $k[X]=k[x_1,\ldots,x_n]/I(X)$, choose a polynomial representative \begin{align*} F_j \in k[x_1,\ldots,x_n] \end{align*} whose residue class in $k[X]$ is $f_j$. Define a map \begin{align*} \phi: X \to \mathbb{A}^m_k, \quad x \mapsto (F_1(x),\ldots,F_m(x)). \end{align*} This map is independent of the values of the representatives on $X$: if $F_j'$ is another representative of $f_j$, then $F_j-F_j' \in I(X)$, so $F_j(x)=F_j'(x)$ for every $x \in X$. [/step] [step:Show the recovered map lands in $Y$] We prove that $\phi(X) \subset Y$. Let \begin{align*} h \in I(Y) \end{align*} be a polynomial relation vanishing on $Y$. Its residue class in $k[Y]$ is zero, so \begin{align*} \alpha(\overline{h})=0 \in k[X], \end{align*} where $\overline{h}$ denotes the residue class of $h$ in $k[Y]$. Since $\alpha$ is a $k$-algebra homomorphism and $h$ is a polynomial in the coordinate classes $\overline{y}_1,\ldots,\overline{y}_m$, we have \begin{align*} \alpha(\overline{h})=h(\alpha(\overline{y}_1),\ldots,\alpha(\overline{y}_m))=h(f_1,\ldots,f_m). \end{align*} Thus $h(f_1,\ldots,f_m)=0$ in $k[X]$. In terms of the chosen representatives, this means \begin{align*} h(F_1,\ldots,F_m)(x)=0 \end{align*} for every $x \in X$. Therefore, for each $x \in X$, every polynomial in $I(Y)$ vanishes at $\phi(x)$. Hence \begin{align*} \phi(x) \in V(I(Y)). \end{align*} Because $Y$ is an [affine variety](/page/Affine%20Variety) over the [algebraically closed field](/page/Algebraically%20Closed%20Field) $k$, the affine Nullstellensatz gives $V(I(Y))=Y$. Thus $\phi(x)\in Y$ for every $x\in X$, so $\phi$ is a map \begin{align*} \phi: X \to Y. \end{align*} [guided] We must check that the polynomial formula just constructed actually has values in $Y$, not merely in the ambient [affine space](/page/Affine%20Space) $\mathbb{A}^m_k$. The set $Y$ is cut out by its vanishing ideal $I(Y)$: a point belongs to $Y$ precisely when all polynomials in $I(Y)$ vanish at that point, using the affine Nullstellensatz over the algebraically closed field $k$. Take an arbitrary relation \begin{align*} h \in I(Y). \end{align*} The residue class $\overline{h}$ of $h$ in the coordinate ring $k[Y]=k[y_1,\ldots,y_m]/I(Y)$ is zero. Applying the homomorphism $\alpha$ gives \begin{align*} \alpha(\overline{h})=0 \in k[X]. \end{align*} Now use the fact that $\alpha$ is a $k$-algebra homomorphism. Since $h$ is obtained from the coordinate polynomials $y_1,\ldots,y_m$ by addition, multiplication, and scalar multiplication over $k$, applying $\alpha$ to the class of $h$ is the same as evaluating the same polynomial expression at the images of the coordinate classes: \begin{align*} \alpha(\overline{h})=h(\alpha(\overline{y}_1),\ldots,\alpha(\overline{y}_m)). \end{align*} By definition, $\alpha(\overline{y}_j)=f_j$, so \begin{align*} h(f_1,\ldots,f_m)=0 \in k[X]. \end{align*} Each $f_j$ is represented by the polynomial $F_j \in k[x_1,\ldots,x_n]$. Therefore the equality $h(f_1,\ldots,f_m)=0$ in $k[X]$ says exactly that the polynomial $h(F_1,\ldots,F_m)$ lies in $I(X)$, meaning \begin{align*} h(F_1,\ldots,F_m)(x)=0 \end{align*} for every $x \in X$. But $\phi(x)=(F_1(x),\ldots,F_m(x))$, so this is the same as \begin{align*} h(\phi(x))=0 \end{align*} for every $x \in X$. Since this holds for every $h \in I(Y)$, every defining relation of $Y$ vanishes at $\phi(x)$. Hence $\phi(x)\in V(I(Y))$. Finally, because $k$ is algebraically closed and $Y$ is an affine variety, the affine Nullstellensatz identifies $V(I(Y))$ with $Y$. Thus $\phi(x)\in Y$ for every $x\in X$, and the constructed map is genuinely a map $\phi:X\to Y$. [/guided] [/step] [step:Verify that the recovered map is a morphism] The map \begin{align*} \phi: X \to Y, \quad x \mapsto (F_1(x),\ldots,F_m(x)) \end{align*} has coordinate functions obtained by restricting polynomials $F_1,\ldots,F_m \in k[x_1,\ldots,x_n]$ to $X$. By the definition of a morphism of affine varieties, this makes $\phi$ a morphism. [/step] [step:Compute the induced pullback on coordinate generators] We show that $\phi^*=\alpha$. For each $j \in \{1,\ldots,m\}$ and each $x \in X$, \begin{align*} \phi^*(\overline{y}_j)(x)=\overline{y}_j(\phi(x))=F_j(x). \end{align*} The residue class of $F_j$ in $k[X]$ is $f_j=\alpha(\overline{y}_j)$, so \begin{align*} \phi^*(\overline{y}_j)=\alpha(\overline{y}_j) \end{align*} for every $j \in \{1,\ldots,m\}$. The coordinate ring $k[Y]$ is generated as a $k$-algebra by $\overline{y}_1,\ldots,\overline{y}_m$. Since $\phi^*$ and $\alpha$ are both unital $k$-algebra homomorphisms and agree on these generators, they agree on all of $k[Y]$. Hence \begin{align*} \phi^*=\alpha. \end{align*} [/step] [step:Prove uniqueness from the coordinate pullbacks] Let \begin{align*} \psi: X \to Y \end{align*} be a morphism such that $\psi^*=\alpha$. For each $j \in \{1,\ldots,m\}$, equality of pullbacks gives \begin{align*} \psi^*(\overline{y}_j)=\alpha(\overline{y}_j)=f_j. \end{align*} Thus, for every $x \in X$, \begin{align*} \overline{y}_j(\psi(x))=F_j(x). \end{align*} Therefore the $j$-th coordinate of $\psi(x)$ equals the $j$-th coordinate of $\phi(x)$ for every $j \in \{1,\ldots,m\}$. Hence \begin{align*} \psi(x)=\phi(x) \end{align*} for every $x \in X$, so $\psi=\phi$. We have shown that every morphism induces a unital $k$-algebra homomorphism by pullback, and every unital $k$-algebra homomorphism arises from a unique morphism. This proves the theorem. [/step]