Why is the graph of a rational map independent of the choice of representative?

If $\phi:X\rightarrow Y$ is a rational map of varieties, why is the graph of $\phi$ well-defined? That is, if $\langle \rho,U\rangle$ and $\langle \psi,V\rangle$ are two representatives of $\phi$ (so that $U,V\subseteq X$ open, and $\rho\vert_{U\cap V}=\psi\vert_{U\cap V}$), why is the closure of the image of $\rho$ equal to the closure of the image of $\psi$?

-

Because $U\cap V$ is Zariski dense in both $U$ and $V$ (by irreducibility of $X$), and so both graphs are obtained by closing up the graph of $\rho_{| U\cap V}$ (which of course coincides with the graph of $\phi_{| U \cap V}$).