Cycles that are graph of a morphism. Given projective complex varieties $X$ and $Y$. Let's assume $d=\text{dim}(X)$ and $C_d(X\times Y)$ be the Chow variety of cycles of dimension $d$ on $X\times Y$. Furthermore you can assume $X$ is smooth. Note that the Chow variety is not really a variety since it has infinite number of components (considering different degrees), but each component is a variety. Is the cycles that are irreducible and correspond to the graph of a morphism $X\rightarrow Y$ a Zariski open subset of $C_d(X\times Y)$? If so now assume $U\subset X$ is a Zariski open subset. Is the cycles that are irreducible and its restriction to $U\times Y$ is a graph of a morphism from $U$ to $Y$ a Zariski open subset of $C_d(X\times Y)$?
 A: I think the answer is no.
Here is my example: take $X=Y=\mathbb{P}^1$ with coordinates $[x_0,x_1]$ and $[y_0,y_1]$.
A $1$-cycle $C$ in $X\times Y=\mathbb{P}^1\times \mathbb{P}^1$ is the zero locus of a bi-homogeneous (wrt the two sets of variables) polynomial $p(x_0,x_1,y_0,y_1)$.
Let's first suppose $p$ irreducible.
If $C$ is the graph of a morphism then, for each $x\in X$, we get that $C\cap \{x\}\times Y$ is exactly one point.
This is equivalent to say that for each fixed $[x_0,x_1]\in\mathbb{P}^1$ the eqaution $p(x_0,x_1,y_0,y_1)=0$ has a unique solution in $[y_0,y_1]$, but this happens iff $p$ is of degree $1$ in these latter variables, which is clearly not generic ( if the total degree of $p$ is high enough).
Now $p$, as a standard homogeneous polynomial, is generically irreducible.
I think the same happen for bi-homogeneous polynomials but I have no proof.
Nevertheless, if $p$ has high enough degree and is reducible, its general irreducible factor, by the previous argument, won't give a graph, so I think we are done.
Hope it will help.
