The Stacks project

Proof. To calculate $R\mathop{\mathrm{lim}}\nolimits $ on an object $K$ of $D(\mathcal{C} \times \mathbf{N})$ we choose a K-injective representative $\mathcal{I}^\bullet $ whose terms are injective objects of $\textit{Ab}(\mathcal{C} \times \mathbf{N})$, see Injectives, Theorem 19.12.6. We may and do think of $\mathcal{I}^\bullet $ as an inverse system of complexes $(\mathcal{I}_ n^\bullet )$ and then we see that

where the right hand side is the termwise inverse limit.

Let $\mathcal{J} = (\mathcal{J}_ n)$ be an injective object of $\textit{Ab}(\mathcal{C} \times \mathbf{N})$. The morphisms $(U, n) \to (U, n + 1)$ are monomorphisms of $\mathcal{C} \times \mathbf{N}$, hence $\mathcal{J}(U, n + 1) \to \mathcal{J}(U, n)$ is surjective (Lemma 21.12.6). It follows that $\mathcal{J}_{n + 1} \to \mathcal{J}_ n$ is surjective as a map of presheaves.

Note that the functor $i_ n^{-1}$ has an exact left adjoint $i_{n, !}$. Namely, $i_{n, !}\mathcal{F}$ is the inverse system $\ldots 0 \to 0 \to \mathcal{F} \to \ldots \to \mathcal{F}$. Thus the complexes $i_ n^{-1}\mathcal{I}^\bullet = \mathcal{I}_ n^\bullet $ are K-injective by Derived Categories, Lemma 13.31.9.

Because we chose our K-injective complex to have injective terms we conclude that

is a short exact sequence of complexes of abelian sheaves as it is a short exact sequence of complexes of abelian presheaves. Moreover, the products in the middle and the right represent the products in $D(\mathcal{C})$, see Injectives, Lemma 19.13.4 and its proof (this is where we use that $\mathcal{I}_ n^\bullet $ is K-injective). Thus $R\mathop{\mathrm{lim}}\nolimits K$ is a homotopy limit of the inverse system $(K_ n)$ by definition of homotopy limits in triangulated categories. $\square$

Proof. The first statement follows from Injectives, Lemma 19.13.6. Then we may apply More on Algebra, Remark 15.86.10 to $R\mathop{\mathrm{lim}}\nolimits R\Gamma (U, K_ n) = R\Gamma (U, R\mathop{\mathrm{lim}}\nolimits K_ n)$ to get the short exact sequences. $\square$

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 $D(\mathcal{O})$. $\square$

Proof. Set $K_ n = \mathcal{F}_ n$ and $K = R\mathop{\mathrm{lim}}\nolimits \mathcal{F}_ n$. Using the notation of Remark 21.23.4 and assumption (2) we see that for $U \in \mathcal{B}$ we have $\underline{\mathcal{H}}_ n^ m(U) = 0$ when $m \not= 0$ and $\underline{\mathcal{H}}_ n^0(U) = \mathcal{F}_ n(U)$. From Equation (21.23.4.1) and assumption (3) we see that $\underline{\mathcal{H}}^ m(U) = 0$ when $m \not= 0$ and equal to $\mathop{\mathrm{lim}}\nolimits \mathcal{F}_ n(U)$ when $m = 0$. Sheafifying using (1) we find that $\mathcal{H}^ m = 0$ when $m \not= 0$ and equal to $\mathop{\mathrm{lim}}\nolimits \mathcal{F}_ n$ when $m = 0$. Hence $K = \mathop{\mathrm{lim}}\nolimits \mathcal{F}_ n$. Since $H^ m(U, K) = \underline{\mathcal{H}}^ m(U) = 0$ for $m > 0$ (see above) we see that the second assertion holds. $\square$

Proof. Let $\gamma \in H^ m(R\mathop{\mathrm{lim}}\nolimits K_ n)(V)$ map to zero in $H^ m(K_{n(V)})(V)$. Since $H^ m(R\mathop{\mathrm{lim}}\nolimits K_ n)$ is the sheafification of $U \mapsto H^ m(U, R\mathop{\mathrm{lim}}\nolimits K_ n)$ (by Lemma 21.20.3) we can choose $\{ V_ i \to V\} \in \text{Cov}_ V$ and elements $\tilde\gamma _ i \in H^ m(V_ i, R\mathop{\mathrm{lim}}\nolimits K_ n)$ mapping to $\gamma |_{V_ i}$. Then $\tilde\gamma _ i$ maps to $\tilde\gamma _{i, n(V)} \in H^ m(V_ i, K_{n(V)})$. Using that $H^ m(K_{n(V)})$ is the sheafification of $U \mapsto H^ m(U, K_{n(V)})$ (by Lemma 21.20.3 again) we see that after replacing $\{ V_ i \to V\} $ by a refinement we may assume that $\tilde\gamma _{i, n(V)} = 0$ for all $i$. For this covering we consider the short exact sequences

of Lemma 21.23.2. By assumption (1) the group on the left is zero and by assumption (2) the group on the right maps injectively into $H^ m(V_ i, K_{n(V)})$. We conclude $\tilde\gamma _ i = 0$ and hence $\gamma = 0$ as desired. $\square$

Proof. Set $K_ n = \tau _{\geq -n}E$ and $K = R\mathop{\mathrm{lim}}\nolimits K_ n$. The canonical map $E \to K$ comes from the canonical maps $E \to K_ n = \tau _{\geq -n}E$. We have to show that $E \to K$ induces an isomorphism $H^ m(E) \to H^ m(K)$ of cohomology sheaves. In the rest of the proof we fix $m$. If $n \geq -m$, then the map $E \to \tau _{\geq -n}E = K_ n$ induces an isomorphism $H^ m(E) \to H^ m(K_ n)$. To finish the proof it suffices to show that for every $V \in \mathcal{B}$ there exists an integer $n(V) \geq -m$ such that the map $H^ m(K)(V) \to H^ m(K_{n(V)})(V)$ is injective. Namely, then the composition

is a bijection and the second arrow is injective, hence the first arrow is bijective. By property (1) this will imply $H^ m(E) \to H^ m(K)$ is an isomorphism. Set

so that in any case $n(V) \geq -m$. Claim: the maps

are isomorphisms for $n \geq n(V)$ and $\{ V_ i \to V\} \in \text{Cov}_ V$. The claim implies conditions (1) and (2) of Lemma 21.23.6 are satisfied and hence implies the desired injectivity. Recall (Derived Categories, Remark 13.12.4) that we have distinguished triangles

Looking at the associated long exact cohomology sequence the claim follows if

are zero for $n \geq n(V)$ and $\{ V_ i \to V\} \in \text{Cov}_ V$. This follows from our choice of $n(V)$ and the assumption in the lemma. $\square$

Proof. This follows from Lemma 21.23.7 with $p(V, m) = d_ V + \max (0, m)$. $\square$

Proof. Apply Lemma 21.23.7 with $p(V, m) = p(m)$ and $\text{Cov}_ V$ equal to the set of coverings $\{ V_ i \to V\} $ with $V_ i \in \mathcal{B}$ for all $i$. $\square$

Proof. Apply Lemma 21.23.8 with $d_ V = d$ and $\text{Cov}_ V$ equal to the set of coverings $\{ V_ i \to V\} $ with $V_ i \in \mathcal{B}$ for all $i$. $\square$

Proof. Observe that $K = R\mathop{\mathrm{lim}}\nolimits \tau _{\geq -n} K$ by Lemma 21.23.10 with $d = 0$. Let $U \in \mathcal{B}$. By Equation (21.23.4.1) we get a short exact sequence

Condition (2) implies $H^ q(U, \tau _{\geq -n} K) = H^0(U, H^ q(\tau _{\geq -n} K))$ for all $q$ by using the spectral sequence of Derived Categories, Lemma 13.21.3. The spectral sequence converges because $\tau _{\geq -n}K$ is bounded below. If $n > -q$ then we have $H^ q(\tau _{\geq -n}K) = H^ q(K)$. Thus the systems on the left and the right of the displayed short exact sequence are eventually constant with values $H^0(U, H^{q - 1}(K))$ and $H^0(U, H^ q(K))$ and the lemma follows. $\square$

Proof. Set $K = R\mathop{\mathrm{lim}}\nolimits K_ n$. We will use notation as in Remark 21.23.4. Let $U \in \mathcal{B}$. By Lemma 21.23.11 and (2)(a) we have $H^ q(U, K_ n) = H^0(U, H^ q(K_ n))$. Using that the functor $R\Gamma (U, -)$ commutes with derived limits we have

where the final equality follows from More on Algebra, Remark 15.86.10 and assumption (2)(b). Thus $H^ q(U, K)$ is the inverse limit the sections of the sheaves $H^ q(K_ n)$ over $U$. Since $\mathop{\mathrm{lim}}\nolimits H^ q(K_ n)$ is a sheaf we find using assumption (1) that $H^ q(K)$, which is the sheafification of the presheaf $U \mapsto H^ q(U, K)$, is equal to $\mathop{\mathrm{lim}}\nolimits H^ q(K_ n)$. This proves the lemma. $\square$