As rational map $f: X \dashrightarrow Y$ among projective algebraic varieties is not defined everywhere there is also a nontrivial definition of an image of a rational map: it is defined by help of a graph of a rational map.

For this embed $Y$ in a projective space $P^n = P^n_k$ of dimension $n$; now the image of $f$ is the image of the composition still denoted $f: X\dashrightarrow P^n$. It is defined as a regular map on some open dense subset $f_U: U\to P^n$. Let the set theoretical graph be $setgraph f_U \subset U\times P^n$; then define the **graph of the rational map** $f$ as the closure of the $setgraph f_U$ within $X\times P^n$. This does not depend on the choice of the dense open subset $U$ and agrees with the usual graph when $f$ is regular.

Notice that not every $k$-point in the image of a rational map is actually set theoretically in image of the underlying map of sets of $k$-points.

A rational map $f: X\dashrightarrow Y$ is **dominant** if its image as a rational map is the whole of $Y$.

Last revised on March 13, 2013 at 18:25:06. See the history of this page for a list of all contributions to it.