The Stacks project

Lemma 83.4.2. Let $(A, I)$ be a henselian pair. Let $X$ be an algebraic space over $A$ such that the structure morphism $f : X \to \mathop{\mathrm{Spec}}(A)$ is proper. Let $i : X_0 \to X$ be the inclusion of $X \times _{\mathop{\mathrm{Spec}}(A)} \mathop{\mathrm{Spec}}(A/I)$. For any sheaf $\mathcal{F}$ on $X_{\acute{e}tale}$ we have $\Gamma (X, \mathcal{F}) = \Gamma (X_0, i^{-1}\mathcal{F})$.

Proof. Choose a surjective proper morphism $Y \to X$ where $Y$ is a scheme, see Cohomology of Spaces, Lemma 68.18.1. Consider the diagram

\[ \xymatrix{ \Gamma (X_0, \mathcal{F}_0) \ar[r] & \Gamma (Y_0, \mathcal{G}_0) \ar@<1ex>[r] \ar@<-1ex>[r] & \Gamma ((Y \times _ X Y)_0, \mathcal{H}_0) \\ \Gamma (X, \mathcal{F}) \ar[r] \ar[u] & \Gamma (Y, \mathcal{G}) \ar@<1ex>[r] \ar@<-1ex>[r] \ar[u] & \Gamma (Y \times _ X Y, \mathcal{H}) \ar[u] } \]

Here $\mathcal{G}$, resp. $\mathcal{H}$ is the pullbackf or $\mathcal{F}$ to $Y$, resp. $Y \times _ X Y$ and the index $0$ indicates base change to $\mathop{\mathrm{Spec}}(A/I)$. By the case of schemes (Étale Cohomology, Lemma 59.91.2) we see that the middle and right vertical arrows are bijective. By Lemma 83.4.1 it follows that the left one is too. $\square$


Comments (2)

Comment #5919 by Harry Gindi on

I'm pretty sure that this diagram is upside-down. The map on global sections goes Γ(X,F)→Γ(X_0,F_0) (and similarly with the higher terms). Otherwise, the proof looks good!


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