Lemma 81.10.1 (0AEQ)—The Stacks project

Lemma 81.10.1. Let $S$ be a scheme. Let $f : Y \to X$ be a morphism of algebraic spaces over $S$. Let $Z \subset X$ closed subspace such that $f^{-1}Z \to Z$ is integral and universally injective. Let $\overline{y}$ be a geometric point of $Y$ and $\overline{x} = f(\overline{y})$. We have

\[ (Rf_*Q)_{\overline{x}} = Q_{\overline{y}} \]

in $D(\textit{Ab})$ for any object $Q$ of $D(Y_{\acute{e}tale})$ supported on $|f^{-1}Z|$.

Proof. Consider the commutative diagram of algebraic spaces

\[ \xymatrix{ f^{-1}Z \ar[r]_{i'} \ar[d]_{f'} & Y \ar[d]_ f \\ Z \ar[r]^ i & X } \]

By Cohomology of Spaces, Lemma 69.9.4 we can write $Q = Ri'_*K'$ for some object $K'$ of $D(f^{-1}Z_{\acute{e}tale})$. By Morphisms of Spaces, Lemma 67.53.7 we have $K' = (f')^{-1}K$ with $K = Rf'_*K'$. Then we have $Rf_*Q = Rf_*Ri'_*K' = Ri_*Rf'_*K' = Ri_*K$. Let $\overline{z}$ be the geometric point of $Z$ corresponding to $\overline{x}$ and let $\overline{z}'$ be the geometric point of $f^{-1}Z$ corresponding to $\overline{y}$. We obtain the result of the lemma as follows

\[ Q_{\overline{y}} = (Ri'_*K')_{\overline{y}} = K'_{\overline{z}'} = (f')^{-1}K_{\overline{z}'} = K_{\overline{z}} = Ri_*K_{\overline{x}} = Rf_*Q_{\overline{x}} \]

The middle equality holds because of the description of the stalk of a pullback given in Properties of Spaces, Lemma 66.19.9. $\square$

The Stacks project

Comments (0)

Post a comment