[proofplan] We construct the pullback map by substituting the coordinate polynomials $f_1,\ldots,f_m$ for the variables $y_1,\ldots,y_m$. The finite monomial expansion of a polynomial makes the substitution formula well-defined. We then verify directly that substitution preserves addition, multiplication, scalar multiplication, and the unit, and finally identify this algebraic substitution with composition of polynomial functions along $F$. [/proofplan] [step:Define the substitution map on monomial expansions] Let \begin{align*} A:=k[x_1,\ldots,x_n] \end{align*} and \begin{align*} B:=k[y_1,\ldots,y_m]. \end{align*} Let $\mathbb{N}_0:=\{0,1,2,\ldots\}$. For a multi-index $\alpha=(\alpha_1,\ldots,\alpha_m)\in \mathbb{N}_0^m$, define the monomial \begin{align*} y^\alpha:=y_1^{\alpha_1}\cdots y_m^{\alpha_m}\in B \end{align*} and define \begin{align*} f^\alpha:=f_1^{\alpha_1}\cdots f_m^{\alpha_m}\in A. \end{align*} Every polynomial $g\in B$ has a unique finite expansion \begin{align*} g=\sum_{\alpha\in S} c_\alpha y^\alpha \end{align*} where $S\subset \mathbb{N}_0^m$ is finite and $c_\alpha\in k$ for every $\alpha\in S$. Define a map \begin{align*} \Phi:B&\to A \end{align*} by \begin{align*} \Phi(g):=\sum_{\alpha\in S} c_\alpha f^\alpha. \end{align*} The uniqueness of the monomial expansion in $B$ makes $\Phi(g)$ independent of any auxiliary choice of the finite set $S$. [guided] We first turn the intuitive phrase “substitute $f_i$ for $y_i$” into a precise function between rings. Set \begin{align*} A:=k[x_1,\ldots,x_n] \end{align*} and \begin{align*} B:=k[y_1,\ldots,y_m]. \end{align*} The domain is the polynomial algebra whose variables are the target coordinates of $\mathbb{A}^m_k$, and the codomain is the polynomial algebra whose variables are the source coordinates of $\mathbb{A}^n_k$. Let $\mathbb{N}_0:=\{0,1,2,\ldots\}$. For each multi-index $\alpha=(\alpha_1,\ldots,\alpha_m)\in\mathbb{N}_0^m$, define \begin{align*} y^\alpha:=y_1^{\alpha_1}\cdots y_m^{\alpha_m}\in B. \end{align*} Since the coordinate polynomials $f_1,\ldots,f_m$ all lie in the commutative $k$-algebra $A$, we may also define \begin{align*} f^\alpha:=f_1^{\alpha_1}\cdots f_m^{\alpha_m}\in A. \end{align*} Now let $g\in B$. By the definition of a [polynomial ring](/page/Polynomial%20Ring), $g$ has a unique expression as a finite $k$-linear combination of monomials: \begin{align*} g=\sum_{\alpha\in S} c_\alpha y^\alpha \end{align*} for some finite set $S\subset\mathbb{N}_0^m$ and coefficients $c_\alpha\in k$. We define \begin{align*} \Phi(g):=\sum_{\alpha\in S} c_\alpha f^\alpha. \end{align*} This is the formal substitution operation: each monomial $y^\alpha$ is sent to the corresponding product $f^\alpha$, and coefficients from $k$ are left unchanged. The uniqueness of the monomial expansion is the point that guarantees well-definedness. If the same polynomial $g$ is written in its monomial expansion, the coefficients $c_\alpha$ are determined uniquely, so the displayed formula determines a single element of $A$. [/guided] [/step] [step:Verify that substitution is a $k$-algebra homomorphism] We prove that $\Phi$ is a $k$-algebra homomorphism. Let \begin{align*} g=\sum_{\alpha\in S} c_\alpha y^\alpha \end{align*} and \begin{align*} h=\sum_{\beta\in T} d_\beta y^\beta \end{align*} be polynomials in $B$, where $S,T\subset \mathbb{N}_0^m$ are finite and $c_\alpha,d_\beta\in k$. For addition, combining coefficients in the monomial expansion gives \begin{align*} \Phi(g+h)=\sum_{\gamma\in S\cup T}(c_\gamma+d_\gamma)f^\gamma, \end{align*} where missing coefficients are interpreted as $0$. Hence \begin{align*} \Phi(g+h)=\Phi(g)+\Phi(h). \end{align*} For multiplication, the product in $B$ is \begin{align*} gh=\sum_{\alpha\in S}\sum_{\beta\in T} c_\alpha d_\beta y^{\alpha+\beta}. \end{align*} Since $A$ is commutative and $f^{\alpha+\beta}=f^\alpha f^\beta$, we obtain \begin{align*} \Phi(gh)=\sum_{\alpha\in S}\sum_{\beta\in T} c_\alpha d_\beta f^{\alpha+\beta}. \end{align*} Therefore \begin{align*} \Phi(gh)=\left(\sum_{\alpha\in S} c_\alpha f^\alpha\right)\left(\sum_{\beta\in T} d_\beta f^\beta\right)=\Phi(g)\Phi(h). \end{align*} For $\lambda\in k$, scalar multiplication gives \begin{align*} \Phi(\lambda g)=\sum_{\alpha\in S} \lambda c_\alpha f^\alpha=\lambda\Phi(g). \end{align*} Finally, the constant polynomial $1_B$ has expansion $1_B=1\cdot y^0$, and $f^0=1_A$, so \begin{align*} \Phi(1_B)=1_A. \end{align*} Thus $\Phi:B\to A$ is a unital $k$-algebra homomorphism. [/step] [step:Identify the images of the coordinate variables] For each $i\in\{1,\ldots,m\}$, let $e_i\in\mathbb{N}_0^m$ denote the multi-index whose $i$th component is $1$ and whose other components are $0$. Since $y_i=y^{e_i}$, the definition of $\Phi$ gives \begin{align*} \Phi(y_i)=f^{e_i}=f_i. \end{align*} Thus $\Phi$ is a $k$-algebra homomorphism sending each coordinate variable $y_i$ to the corresponding coordinate polynomial $f_i$. [/step] [step:Prove uniqueness from the images of the coordinate variables] Let \begin{align*} \Psi:B\to A \end{align*} be any $k$-algebra homomorphism such that $\Psi(y_i)=f_i$ for every $i\in\{1,\ldots,m\}$. Since $\Psi$ is multiplicative, \begin{align*} \Psi(y^\alpha)=\Psi(y_1)^{\alpha_1}\cdots \Psi(y_m)^{\alpha_m}=f^\alpha \end{align*} for every $\alpha\in\mathbb{N}_0^m$. Since $\Psi$ is $k$-linear, for every polynomial \begin{align*} g=\sum_{\alpha\in S} c_\alpha y^\alpha \end{align*} we have \begin{align*} \Psi(g)=\sum_{\alpha\in S} c_\alpha \Psi(y^\alpha)=\sum_{\alpha\in S} c_\alpha f^\alpha=\Phi(g). \end{align*} Hence $\Psi=\Phi$, so the homomorphism with these prescribed values on $y_1,\ldots,y_m$ is unique. [/step] [step:Interpret the homomorphism as pullback by composition] Define \begin{align*} F^*:=\Phi. \end{align*} By construction, for every $g\in k[y_1,\ldots,y_m]$, \begin{align*} F^*(g)=g(f_1,\ldots,f_m). \end{align*} For a point $a\in\mathbb{A}^n_k$, the polynomial map satisfies \begin{align*} F(a)=(f_1(a),\ldots,f_m(a)). \end{align*} Therefore evaluation at $a$ gives \begin{align*} F^*(g)(a)=g(f_1(a),\ldots,f_m(a))=g(F(a)). \end{align*} Thus $F^*(g)=g\circ F$ as a polynomial function on $\mathbb{A}^n_k$. Under the coordinate-ring identification for [affine space](/page/Affine%20Space) in [citetheorem:9339], this is exactly the pullback homomorphism on coordinate rings. This proves the theorem. [/step]