Dominant rational maps 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.
 A: The reason why it is well-defined is because the definition of a dominant rational map is the following:
Definition: A rational map $(f,U)$ is dominant is there exists a rational map $(g,V)$ equivalent to $(f,U)$ such that the image of $g$ is dense in its codomain.
The key point here is that I could take a smaller open set as the domain for my rational map in such a way that the image is not dense, but still get a dominant rational map. (Added: As MattE mentionned in the comments, the correct definition of rational map in the reducible case makes sure that this never happens.)
Added: For my last paragraph, this all depends on your definition of variety. If you assume (as is often the case in algebraic geometry, but not in the theory of algebraic groups) that a variety is irreducible, then we have the following:
Lemma: Let $(f,U)$ and $(g,V)$ be two equivalent rational maps, where we denote the codomain by $Y$. Then $f(U)$ is dense in $Y$ if and only if $g(V)$ is.
Proof: We have the following general fact from topology: For any subset $O\subseteq U$, we have $f(\overline{O})\subseteq \overline{f(O)}$. But if $O$ is open and non-empty, it is dense in $U$, and so $\overline{O}=U$.
A: 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. $f(\overline{U})\subseteq \overline{f(U)}$.
A: This is a purely topological fact:
First we have $f(U \cap V) = g(U \cap V) \subset g(V)$, as $(f,U)$ and $(g,V)$ are equivalent. If $f(U)$ is dense in $Y$ we get
$Y=\overline{f(U)}=\overline{f(\overline{U \cap V})} = \overline{f(U \cap V)} \subset \overline{g(V)}$
where the third equality follows from continuity of $f$. Note also that I take the closure of $U \cap V$ in $U$, so that $\overline{U \cap V}=U$. Hence $g(V)$ is dense in $Y$ as well.
EDIT: In view of M Turgeons answer I should add that my varieties are always irreducible. I used this fact to conclude that $\overline{U \cap V}=U$, as $U \cap V$ is dense in $X$ (it is a non-empty open subset of an irreducible space).
