Lemma 20.35.1. Let $I$ be an ideal of a ring $A$. Let $X$ be a topological space. Let

\[ \ldots \to \mathcal{F}_3 \to \mathcal{F}_2 \to \mathcal{F}_1 \]

be an inverse system of sheaves of $A$-modules on $X$ such that $\mathcal{F}_ n = \mathcal{F}_{n + 1}/I^ n\mathcal{F}_{n + 1}$. Let $p \geq 0$. Assume

\[ \bigoplus \nolimits _{n \geq 0} H^{p + 1}(X, I^ n\mathcal{F}_{n + 1}) \]

satisfies the ascending chain condition as a graded $\bigoplus _{n \geq 0} I^ n/I^{n + 1}$-module. Then the inverse system $M_ n = H^ p(X, \mathcal{F}_ n)$ satisfies the Mittag-Leffler condition^{1}.

**Proof.**
Set $N_ n = H^{p + 1}(X, I^ n\mathcal{F}_{n + 1})$ and let $\delta _ n : M_ n \to N_ n$ be the boundary map on cohomology coming from the short exact sequence $0 \to I^ n\mathcal{F}_{n + 1} \to \mathcal{F}_{n + 1} \to \mathcal{F}_ n \to 0$. Then $\bigoplus \mathop{\mathrm{Im}}(\delta _ n) \subset \bigoplus N_ n$ is a graded submodule. Namely, if $s \in M_ n$ and $f \in I^ m$, then we have a commutative diagram

\[ \xymatrix{ 0 \ar[r] & I^ n\mathcal{F}_{n + 1} \ar[d]_ f \ar[r] & \mathcal{F}_{n + 1} \ar[d]_ f \ar[r] & \mathcal{F}_ n \ar[d]_ f \ar[r] & 0 \\ 0 \ar[r] & I^{n + m}\mathcal{F}_{n + m + 1} \ar[r] & \mathcal{F}_{n + m + 1} \ar[r] & \mathcal{F}_{n + m} \ar[r] & 0 } \]

The middle vertical map is given by lifting a local section of $\mathcal{F}_{n + 1}$ to a section of $\mathcal{F}_{n + m + 1}$ and then multiplying by $f$; similarly for the other vertical arrows. We conclude that $\delta _{n + m}(fs) = f \delta _ n(s)$. By assumption we can find $s_ j \in M_{n_ j}$, $j = 1, \ldots , N$ such that $\delta _{n_ j}(s_ j)$ generate $\bigoplus \mathop{\mathrm{Im}}(\delta _ n)$ as a graded module. Let $n > c = \max (n_ j)$. Let $s \in M_ n$. Then we can find $f_ j \in I^{n - n_ j}$ such that $\delta _ n(s) = \sum f_ j \delta _{n_ j}(s_ j)$. We conclude that $\delta (s - \sum f_ j s_ j) = 0$, i.e., we can find $s' \in M_{n + 1}$ mapping to $s - \sum f_ js_ j$ in $M_ n$. It follows that

\[ \mathop{\mathrm{Im}}(M_{n + 1} \to M_{n - c}) = \mathop{\mathrm{Im}}(M_ n \to M_{n - c}) \]

Namely, the elements $f_ js_ j$ map to zero in $M_{n - c}$. This proves the lemma.
$\square$

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