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The Stacks project

Lemma 66.29.5. Let S be a scheme. Let f : X \to Y be a morphism of algebraic spaces over S. Let \mathcal{G} be a quasi-coherent \mathcal{O}_ Y-module. Let \overline{x} be a geometric point of X and let \overline{y} = f \circ \overline{x} be the image in Y. Then there is a canonical isomorphism

(f^*\mathcal{G})_{\overline{x}} = \mathcal{G}_{\overline{y}} \otimes _{\mathcal{O}_{Y, \overline{y}}} \mathcal{O}_{X, \overline{x}}

of the stalk of the pullback with the tensor product of the stalk with the local ring of X at \overline{x}.

Proof. Since f^*\mathcal{G} = f_{small}^{-1}\mathcal{G} \otimes _{f_{small}^{-1}\mathcal{O}_ Y} \mathcal{O}_ X this follows from the description of stalks of pullbacks in Lemma 66.19.9 and the fact that taking stalks commutes with tensor products. A more direct way to see this is as follows. Choose a commutative diagram

\xymatrix{ U \ar[d]_ p \ar[r]_\alpha & V \ar[d]^ q \\ X \ar[r]^ a & Y }

where U and V are schemes, and p and q are surjective étale. By Lemma 66.19.4 we can choose a geometric point \overline{u} of U such that \overline{x} = p \circ \overline{u}. Set \overline{v} = \alpha \circ \overline{u}. Then we see that

\begin{align*} (f^*\mathcal{G})_{\overline{x}} & = (p^*f^*\mathcal{G})_ u \otimes _{\mathcal{O}_{U, u}} \mathcal{O}_{X, \overline{x}} \\ & = (\alpha ^*q^*\mathcal{G})_ u \otimes _{\mathcal{O}_{U, u}} \mathcal{O}_{X, \overline{x}} \\ & = (q^*\mathcal{G})_ v \otimes _{\mathcal{O}_{V, v}} \mathcal{O}_{U, u} \otimes _{\mathcal{O}_{U, u}} \mathcal{O}_{X, \overline{x}} \\ & = (q^*\mathcal{G})_ v \otimes _{\mathcal{O}_{V, v}} \mathcal{O}_{X, \overline{x}} \\ & = (q^*\mathcal{G})_ v \otimes _{\mathcal{O}_{V, v}} \mathcal{O}_{Y, \overline{y}} \otimes _{\mathcal{O}_{Y, \overline{y}}} \mathcal{O}_{X, \overline{x}} \\ & = \mathcal{G}_{\overline{y}} \otimes _{\mathcal{O}_{Y, \overline{y}}} \mathcal{O}_{X, \overline{x}} \end{align*}

Here we have used Lemma 66.29.4 (twice) and the corresponding result for pullbacks of quasi-coherent sheaves on schemes, see Sheaves, Lemma 6.26.4. \square


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