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From Mumford's Algebraic Geometry I, page 32-33.

Let $Z\subset X\times Y$ be a rational map from $X$ to $Y$. Assume $Z$ is regular at a point $x\in X$. Is it trivial that $p_{1}(Z)$ is dense in $X$? Mumford used this fact to prove that $\mathscr{O}_{x,X}\rightarrow \mathscr{O}_{z,Z}$ is injective. I feel unclear about it as I did not see a proof in previous pages. If $Z$ is regular at all points of $X$, then $p_{1}$ should be continuous, bijective and closed. However with only one point I am not sure.

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Is $X$ assumed to be irreducible? From the definition of rational map $p_1(Z)$ is open in $X$, so if $X$ is irreducible then $p_1(Z)$ is dense in it. – Brad Nov 11 '12 at 17:29
I think $X$ is irreducible. – user32240 Nov 11 '12 at 17:49
Something is wrong in your statement. Maybe you mean $Z$ is the graph of a rational map from $X$ to $Y$ and the rational map is defined at $x$ ? – QiL'8 Nov 11 '12 at 21:00
This is verbatim from Mumford. Of course what you said is right. – user32240 Nov 11 '12 at 21:27

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