Equivalence of Definitions of a Regular Map

Definitions.

Definition (Regular Map 1). Let $X\subseteq {\mathbb{P}}^m$ and $Y\subseteq {\mathbb{P}}^n$ be quasiprojective varieties over a field $K$. A map $F:X\to Y$ is a regular map if for every point $p\in X$, there exist homogeneous polynomials $f_0,\dots ,f_n\in K[x_0,\dots ,x_m]$ of the same degree such that $p\notin \mathcal{V}(f_0,\dots ,f_n)$ and $F(a)=[f_0(a):\cdots :f_n(a)]$ for all $a\in X\setminus \mathcal{V}(f_0,\dots ,f_n)$.

Definition (Regular Map 2). Let $X\subseteq {\mathbb{P}}^m$ and $Y\subseteq {\mathbb{P}}^n$ be quasiprojective varieties over a field $K$. A map $F:X\to Y$ is a regular map if for every point $p\in X$, there exists an open neighborhood $U\subseteq X$ of $p$ and $n+1$ homogeneous polynomials $f_0,\dots ,f_n\in K[x_0,\dots ,x_m]$ of the same degree such that for every point $a\in U$, at least one $f_i(a)$ is non-zero, and $F(a)=[f_0(a):\cdots :f_n(a)]$.

Theorem (Equivalence of Definitions).

The two definitions of a regular map are equivalent.

Proof. Let $F:X\to Y$ be a map between quasiprojective varieties.

($1\Rightarrow 2$) Assume $F$ satisfies Definition (1). Let $p\in X$. By assumption, there exist homogeneous polynomials $f_0,\dots ,f_n$ of the same degree such that $p\notin \mathcal{V}(f_0,\dots ,f_n)$. Let $U=X\setminus \mathcal{V}(f_0,\dots ,f_n)$. The set $\mathcal{V}(f_0,\dots ,f_n)$ is closed in the Zariski topology, so $U$ is an open subset of $X$. Since $p\in U$, it is an open neighborhood of $p$. By construction of $U$, for any $a\in U$, at least one $f_i(a)$ is non-zero. Definition (1) states that $F(a)=[f_0(a):\cdots :f_n(a)]$ for all $a\in U$. Thus, $F$ satisfies Definition (2).

($2\Rightarrow 1$) Assume $F$ satisfies Definition (2). Let $p\in X$. By assumption, there exists an open neighborhood $U\subseteq X$ of $p$ and homogeneous polynomials $f_0,\dots ,f_n$ of the same degree such that for any $a\in U$, at least one $f_i(a)\ne 0$ and $F(a)=[f_0(a):\cdots :f_n(a)]$. Since $p\in U$, it follows that at least one $f_i(p)\ne 0$, which implies $p\notin \mathcal{V}(f_0,\dots ,f_n)$. Let $V=X\setminus \mathcal{V}(f_0,\dots ,f_n)$. The set $V$ is an open subset of $X$ containing $U$. Let $G:V\to Y$ be the map defined by $G(a)=[f_0(a):\cdots :f_n(a)]$. Both $F$ and $G$ are regular maps, and they agree on the non-empty open set $U$. Two regular maps that agree on a non-empty open subset of a quasiprojective variety must be identical on the intersection of their domains. Thus, $F(a)=G(a)$ for all $a\in V$. This means $F(a)=[f_0(a):\cdots :f_n(a)]$ for all $a\in X\setminus \mathcal{V}(f_0,\dots ,f_n)$. Therefore, $F$ satisfies Definition (1). $\square$