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Let $f\colon X\to Y$ be a faithfully flat morphism of smooth projective varieties over $\mathbb{C}$. Assume that the generic fiber of $f$ is smooth. Then there exists a non-empty Zariski open subset $U\subset Y$ such that the morphism $f\colon f^{-1}(U)\to U$ is smooth.

For a closed point $x\in X$ we denote by $B(x,\varepsilon)$ the open ball in $X$ of radius $\varepsilon$ centered at $x$ with respect to a fixed Kahler metric.

Question. Fix $x\in X$. Does there exist $\varepsilon_0>0$ such that for any $\varepsilon \in (0,\varepsilon_0)$ the following is satisfied:

for any $z\in B(x,\varepsilon)\cap f^{-1}(U)$ the intersection $f^{-1}(f(z))\cap B(x,\varepsilon)$ has Betti numbers independent of $z$ and $\varepsilon$?

Remarks. (1) If anything like that is true, the above Betti numbers should not depend on the Kahler metric for small $\varepsilon_0$.

(2) If $Y$ is a curve, then the question is known to have positive answer as it is a part of a topological approach to the construction of the vanishing cycle.

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  • $\begingroup$ The Betti numbers are constant over $U$. Doesn't that imply what you want? $\endgroup$
    – Ariyan Javanpeykar
    Commented Mar 22, 2017 at 13:04
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    $\begingroup$ @Ariyan: the OP's question is (analytic) local on $X$ (not $Y$!). Without properness of the map, in general the Betti numbers are certainly not constant over $U$. $\endgroup$
    – R. van Dobben de Bruyn
    Commented Mar 22, 2017 at 15:27
  • $\begingroup$ By the rank theorem (of real manifolds), on the smooth locus the map is locally given by some coordinate projection. Then the question becomes how the $\varepsilon$-balls behave with respect to projection. There are certainly open sets which do not have this property w.r.t. a coordinate projection, but the question is about (small!) metric balls. I think for $x \in U$ it might not be too bad, but when $x \not \in U$ this approach doesn't really help. $\endgroup$
    – R. van Dobben de Bruyn
    Commented Mar 22, 2017 at 15:34
  • $\begingroup$ @R.vanDobbendeBruyn Whoops. You're right, I misread the question. $\endgroup$
    – Ariyan Javanpeykar
    Commented Mar 23, 2017 at 10:06

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