Let $X$ be an irreducible projective variety over $\mathbb{C}$ (note that I do not assume $X$ smooth) and let $ p : X \longrightarrow S$ be a projective surjective morphism. For any open $U \subset S$, I consider the natural map:

$$ \pi : \mathrm{Hilb}^2_{U}(p^{-1}(U)) \longrightarrow S^2(p^{-1}(U)/U)),$$ where $\mathrm{Hilb}^2_U(p^{-1}(U))$ is the punctual Hilbert scheme parametrizing relative subschemes of $p^{-1}(U)$ of length $2$ and $S^2(p^{-1}(U)/U))$ is the relative symmetric square of $p^{-1}(U)$ over $U$. Let me finally denote by $\mathcal{H}_U$ the closure in $\mathrm{Hilb}^2_U(p^{-1}(U))$ of : $$ \pi^{-1}\left(S^{2}(p^{-1}(U)/U) \backslash \Delta_{p^{-1}(U)/U} \right), $$ where $\Delta_{p^{-1}(U)/U)}$ is the relative diagonal.

**Question** : I would like to know if there exists a non-empty $U$ such that $\mathcal{H}_U$ is irreducible?

The comment below this question seems to suggest (by generic smoothness) that it could be true if $X$ is smooth. However I am interested in the general case. I would also be interested in a reference (or a short proof) if the answer to the question happens to be positive.

**Edit** : In a former (naive) version of the question, I asked if the whole relative Hilbert scheme could be irreducible over some non-empty open $U$. The answer is trivially "no", as observed by Jason Starr in the comment below.