## 20.35 Inverse systems and cohomology

We prove some results on inverse systems of sheaves of modules.

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 condition1.

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. $\square$

Lemma 20.35.2. 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 $A$-modules on $X$ such that $\mathcal{F}_ n = \mathcal{F}_{n + 1}/I^ n\mathcal{F}_{n + 1}$. Let $p \geq 0$. Given $n$ define

$N_ n = \bigcap \nolimits _{m \geq n} \mathop{\mathrm{Im}}\left( H^{p + 1}(X, I^ n\mathcal{F}_{m + 1}) \to H^{p + 1}(X, I^ n\mathcal{F}_{n + 1}) \right)$

If $\bigoplus N_ n$ 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 condition2.

Proof. The proof is exactly the same as the proof of Lemma 20.35.1. 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 the boundary maps $\delta _ n : M_ n \to H^{p + 1}(X, I^ n\mathcal{F}_{n + 1})$ have image contained in $N_ n$.

Suppose that $\xi \in N_ n$ and $f \in I^ k$. Choose $m \gg n + k$. Choose $\xi ' \in H^{p + 1}(X, I^ n\mathcal{F}_{m + 1})$ lifting $\xi$. We consider the diagram

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

constructed as in the proof of Lemma 20.35.1. We get an induced map on cohomology and we see that $f \xi ' \in H^{p + 1}(X, I^{n + k}\mathcal{F}_{m + 1})$ maps to $f \xi$. Since this is true for all $m \gg n + k$ we see that $f\xi$ is in $N_{n + k}$ as desired.

To see the boundary maps $\delta _ n$ have image contained in $N_ n$ we consider the diagrams

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

for $m \geq n$. Looking at the induced maps on cohomology we conclude. $\square$

Lemma 20.35.3. 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

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

satisfies the ascending chain condition as a graded $\bigoplus _{n \geq 0} I^ n/I^{n + 1}$-module. Then the limit topology on $M = \mathop{\mathrm{lim}}\nolimits H^ p(X, \mathcal{F}_ n)$ is the $I$-adic topology.

Proof. Set $F^ n = \mathop{\mathrm{Ker}}(M \to H^ p(X, \mathcal{F}_ n))$ for $n \geq 1$ and $F^0 = M$. Observe that $I F^ n \subset F^{n + 1}$. In particular $I^ n M \subset F^ n$. Hence the $I$-adic topology is finer than the limit topology. For the converse, we will show that given $n$ there exists an $m \geq n$ such that $F^ m \subset I^ nM$3. We have injective maps

$F^ n/F^{n + 1} \longrightarrow H^ p(X, \mathcal{F}_{n + 1})$

whose image is contained in the image of $H^ p(X, I^ n\mathcal{F}_{n + 1}) \to H^ p(X, \mathcal{F}_{n + 1})$. Denote

$E_ n \subset H^ p(X, I^ n\mathcal{F}_{n + 1})$

the inverse image of $F^ n/F^{n + 1}$. Then $\bigoplus E_ n$ is a graded $\bigoplus I^ n/I^{n + 1}$-submodule of $\bigoplus H^ p(X, I^ n\mathcal{F}_{n + 1})$ and $\bigoplus E_ n \to \bigoplus F^ n/F^{n + 1}$ is a homomorphism of graded modules; details omitted. By assumption $\bigoplus E_ n$ is generated by finitely many homogeneous elements over $\bigoplus I^ n/I^{n + 1}$. Since $E_ n \to F^ n/F^{n + 1}$ is surjective, we see that the same thing is true of $\bigoplus F^ n/F^{n + 1}$. Hence we can find $r$ and $c_1, \ldots , c_ r \geq 0$ and $a_ i \in F^{c_ i}$ whose images in $\bigoplus F^ n/F^{n + 1}$ generate. Set $c = \max (c_ i)$.

For $n \geq c$ we claim that $I F^ n = F^{n + 1}$. The claim shows that $F^{n + c} = I^ nF^ c \subset I^ nM$ as desired. To prove the claim suppose $a \in F^{n + 1}$. The image of $a$ in $F^{n + 1}/F^{n + 2}$ is a linear combination of our $a_ i$. Therefore $a - \sum f_ i a_ i \in F^{n + 2}$ for some $f_ i \in I^{n + 1 - c_ i}$. Since $I^{n + 1 - c_ i} = I \cdot I^{n - c_ i}$ as $n \geq c_ i$ we can write $f_ i = \sum g_{i, j} h_{i, j}$ with $g_{i, j} \in I$ and $h_{i, j}a_ i \in F^ n$. Thus we see that $F^{n + 1} = F^{n + 2} + IF^ n$. A simple induction argument gives $F^{n + 1} = F^{n + e} + IF^ n$ for all $e > 0$. It follows that $IF^ n$ is dense in $F^{n + 1}$. Choose generators $k_1, \ldots , k_ r$ of $I$ and consider the continuous map

$u : (F^ n)^{\oplus r} \longrightarrow F^{n + 1},\quad (x_1, \ldots , x_ r) \mapsto \sum k_ i x_ i$

(in the limit topology). By the above the image of $(F^ m)^{\oplus r}$ under $u$ is dense in $F^{m + 1}$ for all $m \geq n$. By the open mapping lemma (More on Algebra, Lemma 15.36.5) we find that $u$ is open. Hence $u$ is surjective. Hence $IF^ n = F^{n + 1}$ for $n \geq c$. This concludes the proof. $\square$

Lemma 20.35.4. 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 $p \geq 0$. Given $n$ define

$N_ n = \bigcap \nolimits _{m \geq n} \mathop{\mathrm{Im}}\left( H^ p(X, I^ n\mathcal{F}_{m + 1}) \to H^ p(X, I^ n\mathcal{F}_{n + 1}) \right)$

If $\bigoplus N_ n$ satisfies the ascending chain condition as a graded $\bigoplus _{n \geq 0} I^ n/I^{n + 1}$-module, then the limit topology on $M = \mathop{\mathrm{lim}}\nolimits H^ p(X, \mathcal{F}_ n)$ is the $I$-adic topology.

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

$H^{p - 1}(X, \mathcal{F}_ n) \to H^ p(X, I^ n\mathcal{F}_{m + 1}) \to H^ p(X, \mathcal{F}_{m + 1}) \to H^ p(X, \mathcal{F}_ n)$

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. $\square$

[1] In fact, there exists a $c \geq 0$ such that $\mathop{\mathrm{Im}}(M_ n \to M_{n - c})$ is the stable image for all $n \geq c$.
[2] In fact, there exists a $c \geq 0$ such that $\mathop{\mathrm{Im}}(M_ n \to M_{n - c})$ is the stable image for all $n \geq c$.
[3] In fact, there exist a $c \geq 0$ such that $F^{n + c} \subset I^ nM$ for all $n$.

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