Lemma 99.3.7. Let $S$ be a scheme. Let $B$ be an algebraic space over $S$. Let $i : X' \to X$ be a closed immersion of algebraic spaces over $B$. Let $\mathcal{F}$ be a quasi-coherent $\mathcal{O}_ X$-module and let $\mathcal{G}'$ be a quasi-coherent $\mathcal{O}_{X'}$-module. Then
\[ \mathit{Hom}(\mathcal{F}, i_*\mathcal{G}') = \mathit{Hom}(i^*\mathcal{F}, \mathcal{G}') \]
as functors on $(\mathit{Sch}/B)$.
Proof.
Let $g : T \to B$ be a morphism where $T$ is a scheme. Denote $i_ T : X'_ T \to X_ T$ the base change of $i$. Denote $h : X_ T \to X$ and $h' : X'_ T \to X'$ the projections. Observe that $(h')^*i^*\mathcal{F} = i_ T^*h^*\mathcal{F}$. As a closed immersion is affine (Morphisms of Spaces, Lemma 67.20.6) we have $h^*i_*\mathcal{G} = i_{T, *}(h')^*\mathcal{G}$ by Cohomology of Spaces, Lemma 69.11.1. Thus we have
\begin{align*} \mathit{Hom}(\mathcal{F}, i_*\mathcal{G}')(T) & = \mathop{\mathrm{Hom}}\nolimits _{\mathcal{O}_{X_ T}}(h^*\mathcal{F}, h^*i_*\mathcal{G}') \\ & = \mathop{\mathrm{Hom}}\nolimits _{\mathcal{O}_{X_ T}}(h^*\mathcal{F}, i_{T, *}(h')^*\mathcal{G}) \\ & = \mathop{\mathrm{Hom}}\nolimits _{\mathcal{O}_{X'_ T}}(i_ T^*h^*\mathcal{F}, (h')^*\mathcal{G}) \\ & = \mathop{\mathrm{Hom}}\nolimits _{\mathcal{O}_{X'_ T}}((h')^*i^*\mathcal{F}, (h')^*\mathcal{G}) \\ & = \mathit{Hom}(i^*\mathcal{F}, \mathcal{G}')(T) \end{align*}
as desired. The middle equality follows from the adjointness of the functors $i_{T, *}$ and $i_ T^*$.
$\square$
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