Definition: A rational map $\pi: X \dashrightarrow Y$ is dominant if for some (and hence every) representative $U \to Y$, the image is dense in $Y$.
I would like to show that if $\pi: X \dashrightarrow Y$ is a dominant rational map, then for every representative $U \to Y$, the image is dense in $Y$.
Let $\alpha: U \to Y$ be a representative such that $\overline{\alpha(U)}=Y$.
I have shown that if $\beta: V \to Y$ is in the same class as $\alpha$, then $\overline{\beta(V)} = Y$.
This is because if $\alpha \sim \beta$, then there exists an open dense subset $Z \subset U \cap V$ such that $\alpha|_Z = \beta|_Z$. Now since the closure of $Z$ in $U$ is just $U$, i.e., $cl(Z,U)=U$, then, $Y=\overline{\alpha(U)}= \overline{\alpha(cl(Z,U))}=\overline{\alpha(Z)}=\overline{\beta(Z)}\subset \overline{\beta(V)}$.
However, I believe we still need to show that if $\beta: V \to Y$ is any other representative, then $\overline{\beta(V)} = Y$.
How do we go about this?
EDIT: Thanks to Daniel Apsley, I realized I was confused about the definition of rational map. For anyone confused about the definition, I think this is may help:
Consider the set $$\{U \to Y \mid U \text{ is an open dense subset of } X, \text{ and } U\to Y \text{ is a morphism of schemes}\}.$$ Now, put an equivalence relation on this set via $(\alpha: U \to Y) \sim (\beta: V \to Y)$ if and only if there exists an open subset $Z \subset U\cap V$ which is dense in $X$ such that $\alpha|_Z = \beta|_Z$.
A rational map is an equivalence class of this equivalence relation.