Let $R$ be a complete discrete valuation ring and let $K$ be its field of fractions. Suppose $X$ is a smooth rigid-anaytic space over $K$. Often it is convenient to have a model of $X$ whose reduction has singularities which are as mild as possible--a semistable model. This amounts to having an admissible covering of $X$ by open affinoids $X\_i$, each of which has good reduction, such that the reductions of any pair $X\_i$ and $X\_j$ meet transversally, if at all. (See the paper of Bosch/Lütkebohmert for definitions.) Let us assume such a semi-stable model of $X$ exists. Then the étale cohomology of $X$ can be computed from the combinatorics of the covering $X\_i$, together with the étale cohomology of each $X\_i$, via the weight spectral sequence of Rapaport-Zink. 


Now suppose I have an open affinoid $Z\subset X$ which happens to have good reduction. My question is: Does there admit a semi-stable model of $X$ for which $Z$ belongs to the covering? Failing this, is there some sense one can make of my intuition that the cohomology of the reduction of $Z$ ought to contribute to the cohomology of $X$?


Feel free to edit/criticize my question to smithereens if you like.



Cohomology of rigid-analytic spaces
Here's a first pass at your question; hopefully it will suggest something more definitive.


Let's imagine we were in the simplest case, where $X$ is a disk, with its smooth model
being the formal affine line over $R$, and that $Z$ was the sub-disk of elements of 
absolute value less than or equal the absolute value of the uniformizer. Then we can find a semistable model in which $Z$ is one of the covering opens, by blowing up the formal affine line at the origin. 


So in this test case, the answer seems to be  *yes* .


Now in general, I think that Raynaud (and/or his collaborators or those who followed in his
tradition) will say that the open immersion $Z \rightarrow X$ extends to an open immersion
of formal models. So we can blow up the smooth model of $X$ and the smooth model of $Z$
so that the latter sits inside the former. What I'm not very certain about is how much 
you can control the nature of these blow-ups. (Presumably not at all in general, but you're starting in a fairly nice situation.) 


Have you tried asking Brian Conrad yet?


