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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