The closure of graph of dense open sets

Question

$U$ is a quasi-projective variety in $\mathbb{P}^n$, and V is an open dense set in $U$. There is a morphism $\phi:U\to Y\subset\mathbb{P^n}$.$U, Y$ are irreducible. Then how can we get $$ \overline{\{(x,\phi(x))|x\in U\}}=\overline{(x,\phi(x))|x\in V\}}?$$

P.S.

(1) $\{(x,\phi(x))\}$ is defined on $U \times Y$, but the topo on $U\times Y$ is not the product topo of $U$ and $Y$.

(2) I already know $\phi(V)$ is dense in $\phi(U)$.

Answer 1

Here is a more general proof which works without any reference to the Euclidean topology:

Lemma 1: For any continuous map of topological spaces $f:X\to Y$ and any dense subset $U\subset X$, $f(U)$ is dense in $f(X)$.

Lemma 2: Let $X$ be a topological space and $Y$ a subset of $X$. Any closed subset of $X$ which contains $Y$ also contains $\overline{Y}$.

Proof: The definition of the closure of a subset is as the smallest closed subset containing it. If there were a closed subset which contained $Y$ but not $\overline{Y}$, this would contradict the definition of $\overline{Y}$.

Lemma 3: For any topological space $X$, any subset $Y$ with the induced topology, and any $Z\subset Y$ which is dense in $Y$, we have that the closure of $Z$ in $X$ is the same as the closure of $Y$ in $X$.

Proof: On the one hand, we have that $\overline{Z}\subset \overline{Y}$ by lemma 2 because $\overline{Y}$ is a closed subset of $X$ containing $Z$. On the other hand, as $Z$ is dense in $Y$, $\overline{Z}$ contains $Y$, so $\overline{Z}$ is a closed set containing $Y$ and by lemma 2 we have that $\overline{Z}\supset\overline{Y}$.

In your situation, we apply lemma 1 to the map $f:U\to U\times Y$ given by $u\mapsto (u,\phi(u))$ and the subset $V$. To verify that $f$ is a continuous map, we write it as the composite of the diagonal map $U\to U\times U$ by $u\mapsto (u,u)$ and $id\times\phi:U\times U\to U\times Y$, both of which are easily seen to be continuous. We conclude that $f(V)\subset f(U)$ is dense. By an application of lemma 3, we see that the closures of $f(V)$ and $f(U)$ are the same inside $X\times Y$. $\blacksquare$

Answer 2

Okay, I think I solved it..

1) Any open set in $U$ is Eulcidean dense.(This is because only several surfaces are removed).

2) The closure of any open set in U is U. (Because Zariski closure is Euclidean closed).

3) Any open set in U is Zariski dense. ( Because its closure is $U$).

Let's get back to the problem.

$\forall x \in U, \exists \{x_n\}$ converges to $x$ in Euclidean space. Because $\phi$ is a polynomial map in a neighborhood of $x$, so $\phi(x_i) \to \phi(x)$. And the closure of graph of $V$ is Euclidean closed, so $(x, \phi(x))$ is in the closure.