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> How do I make the following sort of argument work in characteristic p?
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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 $Y$. 



What is the replacement for a "sufficiently small disc" in characteristic p?
I think it would be difficult to give a general result that covers everything you want and can do but there is a collection of techniques (maybe better described as a dictionary) that works in many cases.


* 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.


As I said to see how this works in any particular situation one would need to have more details on it, no *a priori* success is guaranteed.


