Lemma 20.34.6. Let $(X, \mathcal{O}_ X)$ be a ringed space. Let $E \in D(\mathcal{O}_ X)$. Assume that for every $x \in X$ there exist a function $p(x, -) : \mathbf{Z} \to \mathbf{Z}$ and a fundamental system $\mathfrak {U}_ x$ of open neighbourhoods of $x$ such that

$H^ p(U, H^{m - p}(E)) = 0 \text{ for } U \in \mathfrak {U}_ x \text{ and } p > p(x, m)$

Then the canonical map $E \to R\mathop{\mathrm{lim}}\nolimits \tau _{\geq -n} E$ is an isomorphism in $D(\mathcal{O}_ X)$.

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 $x \in X$ there exists an integer $n(x) \geq -m$ such that the map $H^ m(K)_ x \to H^ m(K_{n(x)})_ x$ is injective. Namely, then the composition

$H^ m(E)_ x \to H^ m(K)_ x \to H^ m(K_{n(x)})_ x$

is a bijection and the second arrow is injective, hence the first arrow is bijective. Set

$n(x) = 1 + \max \{ -m, p(x, m - 1) - m, -1 + p(x, m) - m, -2 + p(x, m + 1) - m\} .$

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

$H^{m - 1}(U, K_{n + 1}) \to H^{m - 1}(U, K_ n) \quad \text{and}\quad H^ m(U, K_{n + 1}) \to H^ m(U, K_ n)$

are isomorphisms for $n \geq n(x)$ and $U \in \mathfrak {U}_ x$. The claim implies conditions (1) and (2) of Lemma 20.34.5 are satisfied and hence implies the desired injectivity. Recall (Derived Categories, Remark 13.12.4) that we have distinguished triangles

$H^{-n - 1}(E)[n + 1] \to K_{n + 1} \to K_ n \to H^{-n - 1}(E)[n + 2]$

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

$H^{m + n}(U, H^{-n - 1}(E)),\quad H^{m + n + 1}(U, H^{-n - 1}(E)),\quad H^{m + n + 2}(U, H^{-n - 1}(E))$

are zero for $n \geq n(x)$ and $U \in \mathfrak {U}_ x$. This follows from our choice of $n(x)$ and the assumption in the lemma. $\square$

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