Lemma 41.17.2. Let $f : X \to S$ be a morphism of schemes. Let $x_1, \ldots , x_ n \in X$ be points having the same image $s$ in $S$. Assume $f$ is separated and $f$ is unramified at each $x_ i$. Then there exists an étale neighbourhood $(U, u) \to (S, s)$ and a disjoint union decomposition

\[ X_ U = W \amalg \coprod \nolimits _{i, j} V_{i, j} \]

such that

$V_{i, j} \to U$ is a closed immersion passing through $u$,

the fibre $W_ u$ contains no point mapping to any $x_ i$.

In particular, if $f^{-1}(\{ s\} ) = \{ x_1, \ldots , x_ n\} $, then the fibre $W_ u$ is empty.

**Proof.**
Apply Lemma 41.17.1. We may assume $U$ is affine, so $X_ U$ is separated. Then $V_{i, j} \to X_ U$ is a closed map, see Morphisms, Lemma 29.41.7. Suppose $(i, j) \not= (i', j')$. Then $V_{i, j} \cap V_{i', j'}$ is closed in $V_{i, j}$ and its image in $U$ does not contain $u$. Hence after shrinking $U$ we may assume that $V_{i, j} \cap V_{i', j'} = \emptyset $. Moreover, $\bigcup V_{i, j}$ is a closed and open subscheme of $X_ U$ and hence has an open and closed complement $W$. This finishes the proof.
$\square$

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