**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 68.11.1. 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

\[ \mathcal{F}_ T \to i_{T, *}i_ T^*\mathcal{F}_ T \xrightarrow {i_{T, *}v_ T} i_{T, *}\mathcal{H}_ T \]

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 66.14.1.
$\square$

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