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Mar 4, 2014 at 20:45 comment added Cantlog It would suffice to contract vertical divisors (in the boundary), but I think this is impossible in general in higher dimension.
Mar 4, 2014 at 2:18 comment added Piotr Achinger Thank you for the useful reference! Do you believe this should be true for $S$ henselian and $X$ of higher dimension?
Mar 2, 2014 at 17:26 comment added Cantlog There is an example in the book "Néron models", §6.7, Lemma 6 (remove the non-contractibe irreducible component in the closed fiber and let $X$ be an affine open subscheme with non empty closed fiber). The DVR is not henselian. On the other hand, if S is henselian, and $X$ is normal of relative dimension $1$, then the answer to your question is yes by contracting superfluous irreducible components (i.e. those of positive dimension) in the boundary, see op. cit.
Mar 2, 2014 at 5:31 vote accept Piotr Achinger
Mar 2, 2014 at 4:50 history edited Piotr Achinger CC BY-SA 3.0
missing apostrophe
Mar 2, 2014 at 3:42 answer added Jason Starr timeline score: 6
Mar 1, 2014 at 8:19 history asked Piotr Achinger CC BY-SA 3.0