Lemma 20.34.10. Let $(X, \mathcal{O}_ X)$ be a ringed space. Let $K$ be an object of $D(\mathcal{O}_ X)$. Let $\mathcal{B}$ be a set of opens of $X$. Assume

1. every open of $X$ has a covering whose members are elements of $\mathcal{B}$,

2. $H^ p(U, H^ q(K)) = 0$ for all $p > 0$, $q \in \mathbf{Z}$, and $U \in \mathcal{B}$.

Then $H^ q(U, K) = H^0(U, H^ q(K))$ for $q \in \mathbf{Z}$ and $U \in \mathcal{B}$.

Proof. Observe that $K = R\mathop{\mathrm{lim}}\nolimits \tau _{\geq -n} K$ by Lemma 20.34.9 with $d = 0$. Let $U \in \mathcal{B}$. By Equation (20.34.3.1) we get a short exact sequence

$0 \to R^1\mathop{\mathrm{lim}}\nolimits H^{q - 1}(U, \tau _{\geq -n}K) \to H^ q(U, K) \to \mathop{\mathrm{lim}}\nolimits H^ q(U, \tau _{\geq -n}K) \to 0$

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 Example 20.29.3. 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 $H^0(U, H^ q(K))$. The lemma follows. $\square$

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