TITLE: Dominant rational maps
QUESTION [16 upvotes]: The University I go to doesn't have any courses in (classical) Algebraic Geometry so I am trying to learn myself. I am fairly comfortable with the content I have covered so far aside from a so called "easy" results which I fail to understand. So my question:
Why is the concept of a rational map between two varieties being dominant well-defined. Specifically let $X$ and $Y$ be varieties; let $U$ and $V$ be open sets of $X$; and let $(f,U)$ and $(g,V)$ be equivalent rational maps from $X$ to $Y$. Why is the image of $f$ dense in $Y$ if and only if the image of $g$ is dense in $Y$?
Thanks for any help.

REPLY [6 votes]: Thanks for your help Nils and M Turgeon. What I didn't realize was that if $f: X \to Y$ is a continuous map of topological spaces, and $U$ is a subset of $X$, then $f(\overline{U})$ is contained in $\overline{f(U)}$. Is the following a correct proof of this fact?
Since $\overline{f(U)}$ is closed in $Y$, by continuity $f^{-1}(\overline{f(U)})$ is closed in $X$. Since also, $U\subseteq f^{-1}(\overline{f(U)})$, it follows that $\overline{U}\subseteq f^{-1}(\overline{f(U)})$ i.e. $f(\overline{U})\subseteq \overline{f(U)}$.