I am slightly confused by the definition of a dominant rational map in Hartshorne, specifically because of a comment he makes about the equivalence relation. In Chapter 1.4, he defines a rational map in the usual way. That is, given a variety $X$ with open sets $U$ and $V$, a rational map is an equivalence class of pairs $\left(U, \phi_{U} \right)$ where two pairs $\left(U, \phi_{U} \right)$ and $\left(V, \phi_{V} \right)$ are equivalent if $\phi_{U}$ and $\phi_{V}$ agree on the intersection $U \cap V$. He then defines a dominant rational map as a rational map where for some (and hence every) such pair, the image of $\phi_{U}$ is dense in $Y$.
My issue here is with the "and hence every" part. Why is it that becase the image of $\phi_{U}$ is dense in $Y$ and $\phi_{U}$ and $\phi_{V}$ agree on $U \cap V$, then the image of $\phi_{V}$ must also be dense in $Y$? My intuition tells me that this has something to do with the fact that the variety $X$ is irreducible, but I still can't see it.
Any help would be appreciated.
Thanks
It suffices to show that if $\phi_U$ is dominant, then $\phi_{U\cap V}$ will also be dominant for $U\cap V \neq \emptyset$. Note that $$ \phi_{U\cap V}(U\cap V) = \phi_U(U\cap V). $$ Now, suppose for a contradiction that there is some nonempty open set $W\subset Y$ such that $W\cap \phi_{U\cap V}(U\cap V) = \emptyset$. By the above equality, $W\cap \phi_U(U\cap V) = \emptyset$. But, we have that $$ \phi_U(\phi_U^{-1}(W) \cap U\cap V) = W \cap \phi_U(U\cap V) = \emptyset, $$ so $\phi_U^{-1}(W)\cap U\cap V = \emptyset$. But because $\phi_U$ is dominant, $W\cap \phi_U(U)\neq \emptyset$, so the preimage $\phi_U^{-1}(W)$ is nonempty and open in $X$. Thus, it must intersect $U\cap V$ because $X$ is irreducible. A contradiction.