Assume we have an irreducible algebraic cycle $Z$ on $X\times Y$ where $X$ and $Y$ are projective varieties ($X$ is smooth) such that restriction of $Z$ to $U\times Y$ where $U\subset X$ is a Zariski open is a graph of a regular map from $U$ to $Y$. Is any cycle close enough to $Z$ on the Chow variety also graph of a morphism when restricted to $U\times Y$? In other words is the family of irreducible cycles with the aforementioned property a Zariski open subset of the Chow variety?


It seems that one can modify the proof of the fact $\text{Hom}$ scheme is an open sub-scheme of the Hilbert scheme to this setting, I just wanted to make sure this is correct and whether there is a simpler proof.



Cycles that are graphs of morphisms
Not as stated. Let $X$ be $\mathbb P^2$, $Y$ be the blowup of $\mathbb P^2$ on one point $P$, $Z$ the graph of the blowup map $Y \to X$, which is the graph of a regular map on the open set $U$ defined as the complement of $P$.


We can deform $Z$ in a family that moves the point $P$. The cycles in that family will each be regular on some open set, but it won't be the same open set $U$.


Allowing $U$ to vary, the answer is positive, since this occurs if and only if the intersection number of $Z$ with a general fiber of the projection to $X$ is $1$, and that's an open (and closed) condition in the Chow variety.


