Lemma 67.13.7. Let $S$ be a scheme. Let $i : Z \to X$ be a closed immersion of algebraic spaces over $S$. Let $\mathcal{A}$ be a sheaf of rings on $X_{\acute{e}tale}$. Let $\mathcal{B}$ be a sheaf of rings on $Z_{\acute{e}tale}$. Let $\varphi : \mathcal{A} \to i_{small, *}\mathcal{B}$ be a homomorphism of sheaves of rings so that we obtain a morphism of ringed topoi

$f : (\mathop{\mathit{Sh}}\nolimits (Z_{\acute{e}tale}), \mathcal{B}) \longrightarrow (\mathop{\mathit{Sh}}\nolimits (X_{\acute{e}tale}), \mathcal{A}).$

For a sheaf of $\mathcal{A}$-modules $\mathcal{F}$ and a sheaf of $\mathcal{B}$-modules $\mathcal{G}$ the canonical map

$\mathcal{F} \otimes _\mathcal {A} f_*\mathcal{G} \longrightarrow f_*(f^*\mathcal{F} \otimes _\mathcal {B} \mathcal{G}).$

is an isomorphism.

Proof. The map is the map adjoint to the map

$f^*\mathcal{F} \otimes _\mathcal {B} f^* f_*\mathcal{G} = f^*(\mathcal{F} \otimes _\mathcal {A} f_*\mathcal{G}) \longrightarrow f^*\mathcal{F} \otimes _\mathcal {B} \mathcal{G}$

coming from $\text{id} : f^*\mathcal{F} \to f^*\mathcal{F}$ and the adjunction map $f^* f_*\mathcal{G} \to \mathcal{G}$. To see this map is an isomorphism, we may check on stalks (Properties of Spaces, Theorem 66.19.12). Let $\overline{z} : \mathop{\mathrm{Spec}}(k) \to Z$ be a geometric point with image $\overline{x} = i \circ \overline{z} : \mathop{\mathrm{Spec}}(k) \to X$. Working out what our maps does on stalks, we see that we have to show

$\mathcal{F}_{\overline{x}} \otimes _{\mathcal{A}_{\overline{x}}} \mathcal{G}_{\overline{z}} = (\mathcal{F}_{\overline{x}} \otimes _{\mathcal{A}_{\overline{x}}} \mathcal{B}_{\overline{z}}) \otimes _{\mathcal{B}_{\overline{z}}} \mathcal{G}_{\overline{z}}$

which holds true. Here we have used that taking tensor products commutes with taking stalks, the behaviour of stalks under pullback Properties of Spaces, Lemma 66.19.9, and the behaviour of stalks under pushforward along a closed immersion Lemma 67.13.6. $\square$

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