Rational map

Remarks

This relation is an equivalence relation. The key fact is that two morphisms $U \to Y$ that agree on a nonempty open subset of $U$ are equal. The reason is that the set where they agree is closed, while a nonempty open subset of the irreducible space $U$ is dense. The union of the domains of all pairs in the class is the largest open set on which $\varphi$ is defined, its domain of definition.

Example: $\varphi \colon \mathbb{P}^2 \dashrightarrow \mathbb{P}^1$, $[x : y : z] \mapsto [x : y]$, is a rational map. It is a morphism on $U = \mathbb{P}^2 \setminus \set{[0 : 0 : 1]}$, and it cannot be extended to a morphism on all of $\mathbb{P}^2$. It is dominant, since it is surjective onto $\mathbb{P}^1$.

Non-example: on $X = k$, the assignment $x \mapsto 1/x$ for $x \ne 0$ and $0 \mapsto 0$ is a function $X \to k$ but not a morphism, so it is not a pair in the definition. Its restriction to $k \setminus \set{0}$ is a morphism and defines a rational map $k \dashrightarrow k$, whose domain of definition is $k \setminus \set{0}$.

Rational maps can only be composed when the first one is dominant; otherwise the image may lie outside the domain of the second. A rational map $X \dashrightarrow \mathbb{A}^1$ is the same as a rational function on $X$.