Can someone please explain it to me what does the highlighted text mean? I am trying to learn Algebraic Geometry from Fulton and Shafarevich but I am stuck now. This says that to check whether an $m-$tuple of rational functions will define a map or not it is sufficient to check whether it satisfies all the equations of $Y$ or not. Although the converse part is clear to me but I am having difficulty understanding how to interpret this $\varphi_i$'s?
Is it $\varphi_i(y_1,y_2,...,y_m)=\varphi_i(y_i)$?
Let $Y$(need not be irreducible) be a closed set in $\mathbb{A}^m$. Since, $k[X_1,X_2,\dots,X_m]$ is Noetherian, the ideal $\mathfrak{U}_Y$ of the set $Y$ is finitely generated in $k[X_1,X_2,\dots,X_m]$, say $\mathfrak{U}_Y=(P_1,P_2,\dots,P_k)$.
Let us consider the rational functions $\varphi_1,\varphi_2,\dots,\varphi_m$ in $k(X)$ such that they satisfy all the equations of $Y$ i.e., $P_i(\varphi_1,\varphi_2,\dots,\varphi_m)=0$ for each $1\leq i\leq k.$ Then for any $F\in\mathfrak{U}_Y$ we will have $F(\varphi_1,\varphi_2,\dots,\varphi_m)=0.$ In particular for the open set $U$ where each of the $\varphi_i$'s are regular, we will have $F(\varphi_1(x),\varphi_2(x),\dots,\varphi_m(x))=0$ for each $x\in U,F\in\mathfrak{U}_Y$. Hence, for each $x\in U$ we have $(\varphi_1(x),\varphi_2(x),\dots,\varphi_m(x) )\in V(\mathfrak{U}_Y).$ As $Y$ is a closed subset in $\mathbb{A}^m$, $Y=V(S)$ for some $S\subseteq k[X_1,X_2,\dots,X_m]$. Therefore, $V(\mathfrak{U}_Y)=V(\mathfrak{U}_{V_S})=V(S)=Y$. Thus for each $x\in U$ we have $(\varphi_1(x),\varphi_2(x),\dots,\varphi_m(x)) \in Y.$ This implies that $\varphi\equiv \varphi(\varphi_1,\varphi_2,\dots,\varphi_m):X\to Y$ is a rational map.