Lemma 57.87.1. Let $(A, I)$ be a henselian pair. Let $f : X \to \mathop{\mathrm{Spec}}(A)$ be a proper morphism of schemes. Let $Z = X \times _{\mathop{\mathrm{Spec}}(A)} \mathop{\mathrm{Spec}}(A/I)$. For any sheaf $\mathcal{F}$ on the topological space associated to $X$ we have $\Gamma (X, \mathcal{F}) = \Gamma (Z, \mathcal{F}|_ Z)$.

Proof. We will use Lemma 57.79.4 to prove this. First observe that the underlying topological space of $X$ is spectral by Properties, Lemma 28.2.4. Let $Y \subset X$ be an irreducible closed subscheme. To finish the proof we show that $Y \cap Z = Y \times _{\mathop{\mathrm{Spec}}(A)} \mathop{\mathrm{Spec}}(A/I)$ is connected. Replacing $X$ by $Y$ we may assume that $X$ is irreducible and we have to show that $Z$ is connected. Let $X \to \mathop{\mathrm{Spec}}(B) \to \mathop{\mathrm{Spec}}(A)$ be the Stein factorization of $f$ (More on Morphisms, Theorem 37.48.5). Then $A \to B$ is integral and $(B, IB)$ is a henselian pair (More on Algebra, Lemma 15.11.8). Thus we may assume the fibres of $X \to \mathop{\mathrm{Spec}}(A)$ are geometrically connected. On the other hand, the image $T \subset \mathop{\mathrm{Spec}}(A)$ of $f$ is irreducible and closed as $X$ is proper over $A$. Hence $T \cap V(I)$ is connected by More on Algebra, Lemma 15.11.14. Now $Y \times _{\mathop{\mathrm{Spec}}(A)} \mathop{\mathrm{Spec}}(A/I) \to T \cap V(I)$ is a surjective closed map with connected fibres. The result now follows from Topology, Lemma 5.7.5. $\square$

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