Lemma 21.23.3. Let $f : (\mathop{\mathit{Sh}}\nolimits (\mathcal{C}), \mathcal{O}) \to (\mathop{\mathit{Sh}}\nolimits (\mathcal{C}'), \mathcal{O}')$ be a morphism of ringed topoi. Then $Rf_*$ commutes with $R\mathop{\mathrm{lim}}\nolimits $, i.e., $Rf_*$ commutes with derived limits.

**Proof.**
Let $(K_ n)$ be an inverse system of objects of $D(\mathcal{O})$. By induction on $n$ we may choose actual complexes $\mathcal{K}_ n^\bullet $ of $\mathcal{O}$-modules and maps of complexes $\mathcal{K}_{n + 1}^\bullet \to \mathcal{K}_ n^\bullet $ representing the maps $K_{n + 1} \to K_ n$ in $D(\mathcal{O})$. In other words, there exists an object $K$ in $D(\mathcal{C} \times \mathbf{N})$ whose associated inverse system is the given one. Next, consider the commutative diagram

of morphisms of topoi. It follows that $R\mathop{\mathrm{lim}}\nolimits R(f \times 1)_*K = Rf_* R\mathop{\mathrm{lim}}\nolimits K$. Working through the definitions and using Lemma 21.23.1 we obtain that $R\mathop{\mathrm{lim}}\nolimits (Rf_*K_ n) = Rf_*(R\mathop{\mathrm{lim}}\nolimits K_ n)$.

Alternate proof in case $\mathcal{C}$ has enough points. Consider the defining distinguished triangle

in $D(\mathcal{O})$. Applying the exact functor $Rf_*$ we obtain the distinguished triangle

in $D(\mathcal{O}')$. Thus we see that it suffices to prove that $Rf_*$ commutes with products in the derived category (which are not just given by products of complexes, see Injectives, Lemma 19.13.4). However, since $Rf_*$ is a right adjoint by Lemma 21.19.1 this follows formally (see Categories, Lemma 4.24.5). Caution: Note that we cannot apply Categories, Lemma 4.24.5 directly as $R\mathop{\mathrm{lim}}\nolimits K_ n$ is not a limit in $D(\mathcal{O})$. $\square$

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## Comments (2)

Comment #2118 by Kestutis Cesnavicius on

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