By Samuel Zaidman

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For an irreducible component X of Xs , let X denote the maximal subset of X which does not intersect any of the other irreducible components of Xs . Notice that X is open in Xs and X is smooth at all points of X . Lemma Let X be an irreducible component of Xs , and let σ be the generic point of X in Xη . Then every open neighborhood of σ in Xη contains a closed subset of the form π −1 (Y), where Y is a nonempty open subset of X . Proof. Let Y = Spf(A) be a nonempty open affine subscheme of X with Ys ⊂ X .

There is an étale morphism from an open neighborhood of , the zero point of the projective line, to the affine scheme Spec(k ◦ [T , u]/(T u − a)). The claim now follows from the fact that the preimage an of ϕ −1 ( )\{ } in Xη under the reduction map is identified with the closed disc E(0; |a|) ⊗ k , and the preimage of the infinity point of ϕ −1 ( ) is identified with the open disc D(0; |a|) ⊗ k . Since X = π −1 (Y) coincides with (X/Y )η , we get the required fact. 36 CHAPTER 2 an (b)=⇒(a) Let X be the formal completion X/Y of a nondegenerate strictly poly◦ stable curve X over k along an irreducible component Y of Xs proper over k, and let 1 , .

T). If denotes the homeomorphic embedding [0, 1] → Xη : t → 0t , it follows that Ur ∩ ([0, 1]) coincides with either (]β, 1[) for some 0 ≤ β < 1, if |f (0)| ≤ r, or ([0, 1[), if |f (0)| > r. Thus, we may assume that the formal scheme X is not smooth at . Step 4. 4], there exist an étale morphism S → S(m) and an open subscheme X ⊂ X ×S(m) S such that Xs contains a point over the point and the induced morphism X → S is geometrically elementary. By Step 2, we may assume that the point is k-rational, and we can replace the triple (X, X , ) by the triple (X , X , ), where X is the irreducible component of Xs that contains , and so we may assume that there is an étale morphism X → Z = T( , ) × S with a smooth formal scheme S such that the induced morphism ψ : X → S is geometrically elementary.