# What is the replacement for a “sufficiently small disc” in characteristic p?

How do I make the following sort of argument work in characteristic p?

Let $f:X \to Y$ be a proper equidimensional map of smooth algebraic varieties, assume all fibres are reduced. Say at some point $y \in Y$, I have computed the differential and know that $df(T_x X) \supset V$ for all $x \in X_y$ and some vector subspace $V \subset T_y Y$. Then over a sufficiently small polydisc $D$ such that $T_y D + V = T_y Y$, the total space of $X_D$ is smooth. Thus the same is true of a (say) one-dimensional thickening $\tilde{D}$ of $D$, and so moving $D$ a little bit in $\tilde{D}$ produces many deformation equivalent smooth manifolds (with boundary) $X_{D'}$.

In particular if I want to know something about $H^*(X_y)$, I can first thicken to the smooth $X_D$ which is homotopy equivalent to it, then move $X_D$ to the diffeomorphic $X_{D'}$, which I prefer because say $D'$ avoids some bad points in $Y$.

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Are you looking for a nearby cycles functor in étale cohomology? – S. Carnahan Jun 23 '11 at 9:31
I don't know, you tell me... – Vivek Shende Jun 23 '11 at 12:03
Vivek -- if you're still in London then you should find Wansu Kim and ask him. He organised a study group on nearby cycles last term. – Kevin Buzzard Jun 24 '11 at 18:04

• A polydisc should be replaced by the strict Henselisation of some (smooth) subvariety $T$ passing through $y$. The fact that $X_y$ is homotopy equivalent to $X_D$ is then replaced by the proper base change theorem.
• The transversality property has a direct analogue in that one may arrange that $T$ is transversal to $f$ and hence that $X_T$ is smooth.
• Deformation of $D$ is replaced by choosing, locally on $Y$, a smooth map $Y\to Z$ such that $T$ is one of the fibres. To get closer to the topological situation one should probably also assume that a section has been chosen. Deformations correspond to fibres over other other points of $Z$ (or rather the Henselisation at the corresponding point of the section).
• To compare (the analogues of) $X_D$ and $X_{D'}$ one probably also would need the smooth base change theorem together with the proper base change theorem.