Lemma 59.75.4. Let $g : S' \to S$ be a morphism of schemes. Let $\mathcal{F}$ be a sheaf on $S_{\acute{e}tale}$. Let $\overline{s}'$ be a geometric point of $S'$, and let $\overline{t}'$ be a geometric point of $\mathop{\mathrm{Spec}}(\mathcal{O}^{sh}_{S', \overline{s}'})$. Denote $\overline{s} = g(\overline{s}')$ and $\overline{t} = h(\overline{t}')$ where $h : \mathop{\mathrm{Spec}}(\mathcal{O}^{sh}_{S', \overline{s}'}) \to \mathop{\mathrm{Spec}}(\mathcal{O}^{sh}_{S, \overline{s}})$ is the canonical morphism. For any sheaf $\mathcal{F}$ on $S_{\acute{e}tale}$ the specialization map

$sp : (f^{-1}\mathcal{F})_{\overline{s}'} \longrightarrow (f^{-1}\mathcal{F})_{\overline{t}'}$

is equal to the specialization map $sp : \mathcal{F}_{\overline{s}} \to \mathcal{F}_{\overline{t}}$ via the identifications $(f^{-1}\mathcal{F})_{\overline{s}'} = \mathcal{F}_{\overline{s}}$ and $(f^{-1}\mathcal{F})_{\overline{t}'} = \mathcal{F}_{\overline{t}}$ of Lemma 59.36.2.

Proof. Omitted. $\square$

There are also:

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