Lemma 20.34.4. Let $(X, \mathcal{O}_ X)$ be a ringed space. Let $(\mathcal{F}_ n)$ be an inverse system of $\mathcal{O}_ X$-modules. 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. $H^ p(U, \mathcal{F}_ n) = 0$ for $p > 0$ and $U \in \mathcal{B}$,

3. the inverse system $\mathcal{F}_ n(U)$ has vanishing $R^1\mathop{\mathrm{lim}}\nolimits$ for $U \in \mathcal{B}$.

Then $R\mathop{\mathrm{lim}}\nolimits \mathcal{F}_ n = \mathop{\mathrm{lim}}\nolimits \mathcal{F}_ n$ and we have $H^ p(U, \mathop{\mathrm{lim}}\nolimits \mathcal{F}_ n) = 0$ for $p > 0$ and $U \in \mathcal{B}$.

Proof. Set $K_ n = \mathcal{F}_ n$ and $K = R\mathop{\mathrm{lim}}\nolimits \mathcal{F}_ n$. Using the notation of Remark 20.34.3 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 (20.34.3.1) 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. $\square$

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