The Stacks project

Lemma 59.94.2. In the situation above, if in addition $f$ is quasi-compact and quasi-separated, then the diagram

\[ \xymatrix{ (Rf_*K)_{\overline{s}} \ar[r] \ar[d]_{sp} & R\Gamma (X_{\overline{s}}, K) \\ (Rf_*K)_{\overline{t}} \ar[r] & R\Gamma (X_{\overline{t}}, K) \ar[u]_{cosp} } \]

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

\[ R\Gamma (X, K) \xrightarrow {\beta _{K, \overline{s}, \overline{t}}} R\Gamma (X, Rh_*(K|_{X_{\overline{t}}})) = R\Gamma (X_{\overline{t}}, K) \]

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. $\square$

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