The Stacks project

Proof. The proof is exactly the same as the proof of Lemma 20.35.3. In fact, the result will follow from the arguments given there as soon as we show that $\bigoplus N_ n$ is a graded $\bigoplus _{n \geq 0} I^ n/I^{n + 1}$-submodule of $\bigoplus H^{p + 1}(X, I^ n\mathcal{F}_{n + 1})$ and that $F^ n/F^{n + 1} \subset H^ p(X, \mathcal{F}_{n + 1})$ is contained in the image of $N_ n \to H^ p(X, \mathcal{F}_{n + 1})$. In the proof of Lemma 20.35.2 we have seen the statement on the module structure.

Let $t \in F^ n$. Choose an element $s \in H^ p(X, I^ n\mathcal{F}_{n + 1})$ which maps to the image of $t$ in $H^ p(X, \mathcal{F}_{n + 1})$. We have to show that $s$ is in $N_ n$. Now $F^ n$ is the kernel of the map from $M \to H^ p(X, \mathcal{F}_ n)$ hence for all $m \geq n$ we can map $t$ to an element $t_ m \in H^ p(X, \mathcal{F}_{m + 1})$ which maps to zero in $H^ p(X, \mathcal{F}_ n)$. Consider the cohomology sequence

coming from the short exact sequence $0 \to I^ n\mathcal{F}_{m + 1} \to \mathcal{F}_{m + 1} \to \mathcal{F}_ n \to 0$. We can choose $s_ m \in H^ p(X, I^ n\mathcal{F}_{m + 1})$ mapping to $t_ m$. Comparing the sequence above with the one for $m = n$ we see that $s_ m$ maps to $s$ up to an element in the image of $H^{p - 1}(X, \mathcal{F}_ n) \to H^ p(X, I^ n\mathcal{F}_{n + 1})$. However, this map factors through the map $H^ p(X, I^ n\mathcal{F}_{m + 1}) \to H^ p(X, I^ n\mathcal{F}_{n + 1})$ and we see that $s$ is in the image as desired. $\square$