Lemma 68.22.3. In Situation 68.22.1. For every $p \geq 0$ there exists an integer $c \geq 0$ such that

1. the multiplication map $I^{n - c} \otimes H^ p(X, I^ c\mathcal{F}) \to H^ p(X, I^ n\mathcal{F})$ is surjective for all $n \geq c$, and

2. the image of $H^ p(X, I^{n + m}\mathcal{F}) \to H^ p(X, I^ n\mathcal{F})$ is contained in the submodule $I^{m - c} H^ p(X, I^ n\mathcal{F})$ for all $n \geq 0$, $m \geq c$.

Proof. By Lemma 68.22.2 we can find $d_1, \ldots , d_ t \geq 0$, and $x_ i \in H^ p(X, I^{d_ i}\mathcal{F})$ such that $\bigoplus _{n \geq 0} H^ p(X, I^ n\mathcal{F})$ is generated by $x_1, \ldots , x_ t$ over $B = \bigoplus _{n \geq 0} I^ n$. Take $c = \max \{ d_ i\}$. It is clear that (1) holds. For (2) let $b = \max (0, n - c)$. Consider the commutative diagram of $A$-modules

$\xymatrix{ I^{n + m - c - b} \otimes I^ b \otimes H^ p(X, I^ c\mathcal{F}) \ar[r] \ar[d] & I^{n + m - c} \otimes H^ p(X, I^ c\mathcal{F}) \ar[r] & H^ p(X, I^{n + m}\mathcal{F}) \ar[d] \\ I^{n + m - c - b} \otimes H^ p(X, I^ n\mathcal{F}) \ar[rr] & & H^ p(X, I^ n\mathcal{F}) }$

By part (1) of the lemma the composition of the horizontal arrows is surjective if $n + m \geq c$. On the other hand, it is clear that $n + m - c - b \geq m - c$. Hence part (2). $\square$

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