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Let $\pi:\mathcal{X} \to T$ be a family of smooth projective complex varieties. Assume $T$ is quasi-projective, reduced, irreducible but not smooth and of positive dimension. Let $\mathcal{Z}$ be a codimension $p$ subscheme of $\mathcal{X}$ flat over $T$. Denote by $\pi': \mathcal{Z} \to T$. Under what conditions on $p$ or $\mathcal{Z}$, we can find an element $\xi \in H^{2p}(\mathcal{X},\mathbb{Z})$ such that $\xi_t \in H^{2p}(\mathcal{X}_t,\mathbb{Z})$ (obtained by the pull-back morphism from $H^{2p}(\mathcal{X},\mathbb{Z})$ to $H^{2p}(\mathcal{X}_t,\mathbb{Z})$ induced by $\mathcal{X}_t \hookrightarrow \mathcal{X})$ is the class of $[\mathcal{Z}_t]$ (under the cycle class map) where $\mathcal{Z}_t:=\pi'^{-1}(t)$ for all $t \in T$?

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  • $\begingroup$ Of course, first thing coming to one's mind is the class $[\mathcal{Z}]$. A bullet-proof sufficient condition would be the transversality of $\mathcal{Z}$ to the fibers. $\endgroup$ – Alex Degtyarev Mar 1 '14 at 9:26
  • $\begingroup$ @Degtyarev: The examples I have in mind it is the case. Could you please suggest some reference for this fact. $\endgroup$ – user46578 Mar 1 '14 at 11:36
  • $\begingroup$ This is too obvious to give a reference. The keywords are inverse Hopf homomorphism (and Thom isomorphism) and their geometric interpretation. (Thom isomorphism should be applied to the normal bundle of the fiber.) Any good textbook in algebraic topology would do, but personally I learned it long ago and in Russian, so I cannot recommend any English text. (In fact, I just don't know any satisfactory English textbook :) $\endgroup$ – Alex Degtyarev Mar 1 '14 at 12:01
  • $\begingroup$ I mean, this is a purely topological fact. If $Z, F$ are two submanifolds transversal in $X$ and $i\colon F\to X$ is the inclusion, then $i^![Z]=[Z\cap F]$, where $i^!=D\circ i^*\circ D^{-1}$ is the inverse Hopf homomorphism (composition of the cohomological restriction and Poincare dualities). $\endgroup$ – Alex Degtyarev Mar 1 '14 at 12:05
  • $\begingroup$ Alex, I have difficulties to follow. Are you ware that $\mathcal{X}$ is not smooth? $\endgroup$ – abx Mar 1 '14 at 13:20

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