Lemma 59.94.2. In the situation above, if in addition $f$ is quasi-compact and quasi-separated, then the diagram
is commutative.
Lemma 59.94.2. In the situation above, if in addition $f$ is quasi-compact and quasi-separated, then the diagram
is commutative.
Proof. As in the proof of Lemma 59.94.1 we may replace $S$ by $\mathop{\mathrm{Spec}}(\mathcal{O}^{sh}_{S, \overline{s}})$. Then our maps simplify to $h : X_{\overline{t}} \to X$, $i : X_{\overline{s}} \to X$, and $\beta _{K, \overline{s}, \overline{t}} : K \to Rh_*(K|_{X_{\overline{t}}})$. Using that $(Rf_*K)_{\overline{s}} = R\Gamma (X, K)$ by Theorem 59.53.1 the composition of $sp$ with the base change map $(Rf_*K)_{\overline{t}} \to R\Gamma (X_{\overline{t}}, K)$ is just pullback of cohomology along $h$. This is the same as the map
Now the map $cosp$ first inverts the $=$ sign in this displayed formula, then pulls back along $i$, and finally applies the inverse of $i^{-1}\beta _{K, \overline{s}, \overline{t}}$. Hence we get the desired commutativity. $\square$
Comments (0)