Theorem 41.15.1. Let $X$ and $Y$ be two schemes over a base scheme $S$. Let $S_0$ be a closed subscheme of $S$ with the same underlying topological space (for example if the ideal sheaf of $S_0$ in $S$ has square zero). Denote $X_0$ (resp. $Y_0$) the base change $S_0 \times _ S X$ (resp. $S_0 \times _ S Y$). If $X$ is étale over $S$, then the map

\[ \mathop{\mathrm{Mor}}\nolimits _ S(Y, X) \longrightarrow \mathop{\mathrm{Mor}}\nolimits _{S_0}(Y_0, X_0) \]

is bijective.

**Proof.**
After base changing via $Y \to S$, we may assume that $Y = S$. In this case the theorem states that any $S$-morphism $\sigma _0 : S_0 \to X$ actually factors uniquely through a section $S \to X$ of the étale structure morphism $f : X \to S$.

Uniqueness. Suppose we have two sections $\sigma , \sigma '$ through which $\sigma _0$ factors. Because $X \to S$ is étale we see that $\Delta : X \to X \times _ S X$ is an open immersion (Morphisms, Lemma 29.35.13). The morphism $(\sigma , \sigma ') : S \to X \times _ S X$ factors through this open because for any $s \in S$ we have $(\sigma , \sigma ')(s) = (\sigma _0(s), \sigma _0(s))$. Thus $\sigma = \sigma '$.

To prove existence we first reduce to the affine case (we suggest the reader skip this step). Let $X = \bigcup X_ i$ be an affine open covering such that each $X_ i$ maps into an affine open $S_ i$ of $S$. For every $s \in S$ we can choose an $i$ such that $\sigma _0(s) \in X_ i$. Choose an affine open neighbourhood $U \subset S_ i$ of $s$ such that $\sigma _0(U_0) \subset X_{i, 0}$. Note that $X' = X_ i \times _ S U = X_ i \times _{S_ i} U$ is affine. If we can lift $\sigma _0|_{U_0} : U_0 \to X'_0$ to $U \to X'$, then by uniqueness these local lifts will glue to a global morphism $S \to X$. Thus we may assume $S$ and $X$ are affine.

Existence when $S$ and $X$ are affine. Write $S = \mathop{\mathrm{Spec}}(A)$ and $X = \mathop{\mathrm{Spec}}(B)$. Then $A \to B$ is étale and in particular smooth (of relative dimension $0$). As $|S_0| = |S|$ we see that $S_0 = \mathop{\mathrm{Spec}}(A/I)$ with $I \subset A$ locally nilpotent. Thus existence follows from Algebra, Lemma 10.138.17.
$\square$

## Comments (1)

Comment #1777 by OS on