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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 $\psi$?

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up vote 3 down vote accepted

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 $\phi_{| U \cap V}$).

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Ah, I was forgetting that the image of a dense set is dense in the image. –  Zev Chonoles Oct 3 '10 at 19:31
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