Let \(\varphi\) be a rational map from a variety \(X\) to a variety \(Y\). Show that there is a largest open set on which \(\varphi\) is represented by a morphism. One says that the rational map is defined at the points of that open set.
Solution
Let \(\varphi:X\dashrightarrow Y\) be the given rational map.
1. Any two representatives of \(\varphi\) glue on the union of their domains.
Let \[ \varphi_U:U\to Y, \qquad \varphi_V:V\to Y \] be two representatives. By the definition of a rational map, \[ \varphi_U|_{U\cap V}=\varphi_V|_{U\cap V}. \] Hence the piecewise rule \[ \psi(x)= \begin{cases} \varphi_U(x),&x\in U,\\ \varphi_V(x),&x\in V \end{cases} \] is well-defined on \(U\cup V\).
It is continuous: for every open \(W\subseteq Y\), \[ \psi^{-1}(W) =\varphi_U^{-1}(W)\cup\varphi_V^{-1}(W), \] which is open in \(U\cup V\).
Let \(h\) be regular on an open subset \(W\subseteq Y\). On \[ \psi^{-1}(W)\cap U \] the pullback \(h\circ\psi\) equals the regular function \(h\circ\varphi_U\), and similarly on the open subset \(\psi^{-1}(W)\cap V\). These opens cover \(\psi^{-1}(W)\), so regularity being local shows that \(h\circ\psi\) is regular. Thus \(\psi\) is a morphism.
2. Let\[ D(\varphi) =\bigcup\{U\subseteq X:U\text{ is open and }\varphi\text{ is represented by a morphism }U\to Y\}. \]Then the representatives glue to a morphism\[ \varphi_D:D(\varphi)\to Y. \]
The set \(D(\varphi)\) is open as a union of opens. On pairwise overlaps all representatives agree by the equivalence relation defining the rational map. Therefore they define a single set map \(\varphi_D\) on the union.
Every point of \(D(\varphi)\) lies in one representative domain \(U\), and on that open neighborhood \(\varphi_D\) equals the morphism \(\varphi_U\). The morphism condition is local on the source, so \(\varphi_D\) is a morphism.
3. The open set \(D(\varphi)\) is the unique largest open subset on which \(\varphi\) is represented by a morphism.
Step 2 shows that \(\varphi\) is represented on \(D(\varphi)\). If \(W\subseteq X\) is any other open subset on which \(\varphi\) is represented by a morphism, then \(W\) is one of the opens in the union defining \(D(\varphi)\). Hence \[ W\subseteq D(\varphi). \] Thus \(D(\varphi)\) is the largest such open set.
4. Q.E.D.
Steps <1>1–<1>3 construct the maximal domain of definition and its representing morphism.
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