Lemma 77.7.6. In Situation 77.7.1 suppose given an affine morphism $i : Z \to X$ and a quasi-coherent $\mathcal{O}_ Z$-module $\mathcal{H}$ such that $\mathcal{G} = i_*\mathcal{H}$. Let $v : i^*\mathcal{F} \to \mathcal{H}$ be the map adjoint to $u$. Then
$F_{v, zero} = F_{u, zero}$, and
if $i$ is a closed immersion, then $F_{v, surj} = F_{u, surj}$.
Proof. Let $T$ be a scheme over $B$. Denote $i_ T : Z_ T \to X_ T$ the base change of $i$ and $\mathcal{H}_ T$ the pullback of $\mathcal{H}$ to $Z_ T$. Observe that $(i^*\mathcal{F})_ T = i_ T^*\mathcal{F}_ T$ and $i_{T, *}\mathcal{H}_ T = (i_*\mathcal{H})_ T$. The first statement follows from commutativity of pullbacks and the second from Cohomology of Spaces, Lemma 69.11.1. Hence we see that $u_ T$ and $v_ T$ are adjoint maps as well. Thus $u_ T = 0$ if and only if $v_ T = 0$. This proves (1). In case (2) we see that $u_ T$ is surjective if and only if $v_ T$ is surjective because $u_ T$ factors as
and the fact that $i_{T, *}$ is an exact functor fully faithfully embedding the category of quasi-coherent modules on $Z_ T$ into the category of quasi-coherent $\mathcal{O}_{X_ T}$-modules. See Morphisms of Spaces, Lemma 67.14.1. $\square$
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