Lemma 59.94.3. Let $f : X \to S$ be a morphism of schemes. Let $K \in D(X_{\acute{e}tale})$. Assume

1. $K$ is bounded below, i.e., $K \in D^+(X_{\acute{e}tale})$,

2. $f$ is locally acyclic relative to $K$,

3. $f$ is proper, and

4. $K$ has torsion cohomology sheaves.

Then for every geometric point $\overline{s}$ of $S$ and every geometric point $\overline{t}$ of $\mathop{\mathrm{Spec}}(\mathcal{O}^{sh}_{S, \overline{s}})$ both the specialization map $sp : (Rf_*K)_{\overline{s}} \to (Rf_*K)_{\overline{t}}$ and the cospecialization map $cosp : R\Gamma (X_{\overline{t}}, K) \to R\Gamma (X_{\overline{s}}, K)$ are isomorphisms.

Proof. By the proper base change theorem (in the form of Lemma 59.91.13) we have $(Rf_*K)_{\overline{s}} = R\Gamma (X_{\overline{s}}, K)$ and similarly for $\overline{t}$. The “correct” proof would be to show that the argument in Lemma 59.94.2 shows that $sp$ and $cosp$ are inverse isomorphisms in this case. Instead we will show directly that $cosp$ is an isomorphism. From the discussion above we see that $cosp$ is an isomorphism if and only if pullback by $i$

$R\Gamma (X \times _ S \mathop{\mathrm{Spec}}(\mathcal{O}^{sh}_{S, \overline{s}}), Rh_*(K|_{X_{\overline{t}}})) \longrightarrow R\Gamma (X_{\overline{s}}, i^{-1}Rh_*(K|_{X_{\overline{t}}}))$

is an isomorphism in $D^+(\textit{Ab})$. This is true by the proper base change theorem for the proper morphism $f' : X \times _ S \mathop{\mathrm{Spec}}(\mathcal{O}^{sh}_{S, \overline{s}}) \to \mathop{\mathrm{Spec}}(\mathcal{O}^{sh}_{S, \overline{s}})$ by the morphism $\overline{s} \to \mathop{\mathrm{Spec}}(\mathcal{O}^{sh}_{S, \overline{s}})$ and the complex $K' = Rh_*(K|_{X_{\overline{t}}})$. The complex $K'$ is bounded below and has torsion cohomology sheaves by Lemma 59.78.2. Since $\mathop{\mathrm{Spec}}(\mathcal{O}^{sh}_{S, \overline{s}})$ is strictly henselian with $\overline{s}$ lying over the closed point, we see that the source of the displayed arrow equals $(Rf'_*K')_{\overline{s}}$ and the target equals $R\Gamma (X_{\overline{s}}, K')$ and the displayed map is an isomorphism by the already used Lemma 59.91.13. Thus we see that three out of the four arrows in the diagram of Lemma 59.94.2 are isomorphisms and we conclude. $\square$

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