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Let $f\colon X\to Y$ be a flat morphism of irreducible projective algebraic varieties over $\mathbb{C}$ (or any other algebraically closed field of characteristic 0). Assume that $Y$ is smooth, and the generic fiber of $f$ is smooth (these two assumptions seem to be important).

Let $U\subset Y$ be a Zariski open non-empty subset such that $f\colon f^{-1}(U)\to U$ is smooth. Fix a closed point $y\in Y$.

Let us denote by $S$ the "disk", more precisely $S$ is the spectrum of henselization of $\mathbb{C}[t]$ localized at the ideal generated by $t$. Let $\eta$ be the generic point of $S$.

For any morphism $\nu\colon S\to Y$ such that $\eta$ is mapped to $U$ and the closed point of $S$ to $y$, consider the fibered product $S\times_Y X\to S$. Notice that the generic fiber of this morphism is smooth over $\eta$. Consider the nearby cycle functor of the constant sheaf $\underline{\mathbb{\mathbb{Q}_l}}$. It is a perverse sheaf on $f^{-1}(y)$ which we will denote by $\mathcal{F}_\nu$ to emphasize dependence on the morphism $\nu$.

QUESTION. Is it true that for all choices of $\nu$ as above the perverse sheaves $\mathcal{F}_\nu$ are isomorphic to each other?

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    $\begingroup$ $S$ is the same as $D$? $\endgroup$ – მამუკა ჯიბლაძე Mar 14 '17 at 19:32
  • $\begingroup$ Can't you get a counter-example by taking $f = id : \mathbb{C} \to \mathbb{C}$ and $\nu : \mathbb{C} \to \mathbb{C} : z \mapsto z^m$? (The nearby cycles functor is simply $m$-copies of the constant sheaf.) $\endgroup$ – Geordie Williamson Jun 4 '17 at 3:45
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The answer is no!

Start with any $g:Z \to S$, with smooth generic fiber. Write $Z_0$ for the closed fiber; in order for your conditions to be eventually satisfied this should be assumed smooth as well. Then set $X = Z\times S$, $Y = S\times S$, $f = g\times id.$ Taking the base change of $f$ with the two coordinate-axis embeddings $S \to S\times S$ gives two different schemes over $S$, namely $Z$ and $Z_0\times S$. In the latter case your nearby cycles sheaf will just be the constant perverse sheaf on $Z_0$; hence we will be done if we can find a starting $Z$ whose nearby cycles is not the constant perverse sheaf. For this there are many examples, but the easiest is when $Z_0$ contains an entire component of $Z$.

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  • $\begingroup$ Thank you. On the second thought I am not sure I understand completely your argument. If $Z_0$ contains an entire component of $Z$, that probably means that the map $g$ is not flat. Also if $Z_0$ is smooth and $g$ is flat, then $g$ is smooth. Does it imply that nearby cycles of $Z$ and $Z_0\times S$ are the same? $\endgroup$ – MKO Jun 3 '17 at 8:01

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