Lemma 68.22.4. In Situation 68.22.1. Fix $p \geq 0$.

1. There exists a $c_1 \geq 0$ such that for all $n \geq c_1$ we have

$\mathop{\mathrm{Ker}}( H^ p(X, \mathcal{F}) \to H^ p(X, \mathcal{F}/I^ n\mathcal{F}) ) \subset I^{n - c_1}H^ p(X, \mathcal{F}).$
2. The inverse system

$\left(H^ p(X, \mathcal{F}/I^ n\mathcal{F})\right)_{n \in \mathbf{N}}$

satisfies the Mittag-Leffler condition (see Homology, Definition 12.31.2).

3. In fact for any $p$ and $n$ there exists a $c_2(n) \geq n$ such that

$\mathop{\mathrm{Im}}(H^ p(X, \mathcal{F}/I^ k\mathcal{F}) \to H^ p(X, \mathcal{F}/I^ n\mathcal{F})) = \mathop{\mathrm{Im}}(H^ p(X, \mathcal{F}) \to H^ p(X, \mathcal{F}/I^ n\mathcal{F}))$

for all $k \geq c_2(n)$.

Proof. Let $c_1 = \max \{ c_ p, c_{p + 1}\}$, where $c_ p, c_{p +1}$ are the integers found in Lemma 68.22.3 for $H^ p$ and $H^{p + 1}$. We will use this constant in the proofs of (1), (2) and (3).

Let us prove part (1). Consider the short exact sequence

$0 \to I^ n\mathcal{F} \to \mathcal{F} \to \mathcal{F}/I^ n\mathcal{F} \to 0$

From the long exact cohomology sequence we see that

$\mathop{\mathrm{Ker}}( H^ p(X, \mathcal{F}) \to H^ p(X, \mathcal{F}/I^ n\mathcal{F}) ) = \mathop{\mathrm{Im}}( H^ p(X, I^ n\mathcal{F}) \to H^ p(X, \mathcal{F}) )$

Hence by our choice of $c_1$ we see that this is contained in $I^{n - c_1}H^ p(X, \mathcal{F})$ for $n \geq c_1$.

Note that part (3) implies part (2) by definition of the Mittag-Leffler condition.

Let us prove part (3). Fix an $n$ throughout the rest of the proof. Consider the commutative diagram

$\xymatrix{ 0 \ar[r] & I^ n\mathcal{F} \ar[r] & \mathcal{F} \ar[r] & \mathcal{F}/I^ n\mathcal{F} \ar[r] & 0 \\ 0 \ar[r] & I^{n + m}\mathcal{F} \ar[r] \ar[u] & \mathcal{F} \ar[r] \ar[u] & \mathcal{F}/I^{n + m}\mathcal{F} \ar[r] \ar[u] & 0 }$

This gives rise to the following commutative diagram

$\xymatrix{ H^ p(X, I^ n\mathcal{F}) \ar[r] & H^ p(X, \mathcal{F}) \ar[r] & H^ p(X, \mathcal{F}/I^ n\mathcal{F}) \ar[r]_\delta & H^{p + 1}(X, I^ n\mathcal{F}) \\ H^ p(X, I^{n + m}\mathcal{F}) \ar[r] \ar[u] & H^ p(X, \mathcal{F}) \ar[r] \ar[u]^1 & H^ p(X, \mathcal{F}/I^{n + m}\mathcal{F}) \ar[r] \ar[u] & H^{p + 1}(X, I^{n + m}\mathcal{F}) \ar[u]^ a }$

If $m \geq c_1$ we see that the image of $a$ is contained in $I^{m - c_1} H^{p + 1}(X, I^ n\mathcal{F})$. By the Artin-Rees lemma (see Algebra, Lemma 10.51.3) there exists an integer $c_3(n)$ such that

$I^ N H^{p + 1}(X, I^ n\mathcal{F}) \cap \mathop{\mathrm{Im}}(\delta ) \subset \delta \left(I^{N - c_3(n)}H^ p(X, \mathcal{F}/I^ n\mathcal{F})\right)$

for all $N \geq c_3(n)$. As $H^ p(X, \mathcal{F}/I^ n\mathcal{F})$ is annihilated by $I^ n$, we see that if $m \geq c_3(n) + c_1 + n$, then

$\mathop{\mathrm{Im}}(H^ p(X, \mathcal{F}/I^{n + m}\mathcal{F}) \to H^ p(X, \mathcal{F}/I^ n\mathcal{F})) = \mathop{\mathrm{Im}}(H^ p(X, \mathcal{F}) \to H^ p(X, \mathcal{F}/I^ n\mathcal{F}))$

In other words, part (3) holds with $c_2(n) = c_3(n) + c_1 + n$. $\square$

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