Lemma 59.99.2. Let $f : T \to S$ be a morphism of schemes.

For $K$ in $D((\mathit{Sch}/T)_{\acute{e}tale})$ we have $ (Rf_{big, *}K)|_{S_{\acute{e}tale}} = Rf_{small, *}(K|_{T_{\acute{e}tale}}) $ in $D(S_{\acute{e}tale})$.

For $K$ in $D((\mathit{Sch}/T)_{\acute{e}tale}, \mathcal{O})$ we have $ (Rf_{big, *}K)|_{S_{\acute{e}tale}} = Rf_{small, *}(K|_{T_{\acute{e}tale}}) $ in $D(\textit{Mod}(S_{\acute{e}tale}, \mathcal{O}_ S))$.

More generally, let $g : S' \to S$ be an object of $(\mathit{Sch}/S)_{\acute{e}tale}$. Consider the fibre product

Then

For $K$ in $D((\mathit{Sch}/T)_{\acute{e}tale})$ we have $i_ g^{-1}(Rf_{big, *}K) = Rf'_{small, *}(i_{g'}^{-1}K)$ in $D(S'_{\acute{e}tale})$.

For $K$ in $D((\mathit{Sch}/T)_{\acute{e}tale}, \mathcal{O})$ we have $i_ g^*(Rf_{big, *}K) = Rf'_{small, *}(i_{g'}^*K)$ in $D(\textit{Mod}(S'_{\acute{e}tale}, \mathcal{O}_{S'}))$.

For $K$ in $D((\mathit{Sch}/T)_{\acute{e}tale})$ we have $g_{big}^{-1}(Rf_{big, *}K) = Rf'_{big, *}((g'_{big})^{-1}K)$ in $D((\mathit{Sch}/S')_{\acute{e}tale})$.

For $K$ in $D((\mathit{Sch}/T)_{\acute{e}tale}, \mathcal{O})$ we have $g_{big}^*(Rf_{big, *}K) = Rf'_{big, *}((g'_{big})^*K)$ in $D(\textit{Mod}(S'_{\acute{e}tale}, \mathcal{O}_{S'}))$.

## Comments (2)

Comment #3219 by David Hansen on

Comment #3321 by Johan on