This is a special case of [Proposition 3.6.1, Tohoku].

Lemma 20.20.4. Let $X$ be a topological space. Let $d \geq 0$ be an integer. Assume

1. $X$ is quasi-compact,

2. the quasi-compact opens form a basis for $X$, and

3. the intersection of two quasi-compact opens is quasi-compact.

4. $H^ p(X, j_!\underline{\mathbf{Z}}_ U) = 0$ for all $p > d$ and any quasi-compact open $j : U \to X$.

Then $H^ p(X, \mathcal{F}) = 0$ for all $p > d$ and any abelian sheaf $\mathcal{F}$ on $X$.

Proof. Let $S = \coprod _{U \subset X} \mathcal{F}(U)$ where $U$ runs over the quasi-compact opens of $X$. For any finite subset $A = \{ s_1, \ldots , s_ n\} \subset S$, let $\mathcal{F}_ A$ be the subsheaf of $\mathcal{F}$ generated by all $s_ i$ (see Modules, Definition 17.4.5). Note that if $A \subset A'$, then $\mathcal{F}_ A \subset \mathcal{F}_{A'}$. Hence $\{ \mathcal{F}_ A\}$ forms a system over the directed partially ordered set of finite subsets of $S$. By Modules, Lemma 17.4.6 it is clear that

$\mathop{\mathrm{colim}}\nolimits _ A \mathcal{F}_ A = \mathcal{F}$

by looking at stalks. By Lemma 20.19.1 we have

$H^ p(X, \mathcal{F}) = \mathop{\mathrm{colim}}\nolimits _ A H^ p(X, \mathcal{F}_ A)$

Hence it suffices to prove the vanishing for the abelian sheaves $\mathcal{F}_ A$. In other words, it suffices to prove the result when $\mathcal{F}$ is generated by finitely many local sections over quasi-compact opens of $X$.

Suppose that $\mathcal{F}$ is generated by the local sections $s_1, \ldots , s_ n$. Let $\mathcal{F}' \subset \mathcal{F}$ be the subsheaf generated by $s_1, \ldots , s_{n - 1}$. Then we have a short exact sequence

$0 \to \mathcal{F}' \to \mathcal{F} \to \mathcal{F}/\mathcal{F}' \to 0$

From the long exact sequence of cohomology we see that it suffices to prove the vanishing for the abelian sheaves $\mathcal{F}'$ and $\mathcal{F}/\mathcal{F}'$ which are generated by fewer than $n$ local sections. Hence it suffices to prove the vanishing for sheaves generated by at most one local section. These sheaves are exactly the quotients of the sheaves $j_!\underline{\mathbf{Z}}_ U$ where $U$ is a quasi-compact open of $X$.

Assume now that we have a short exact sequence

$0 \to \mathcal{K} \to j_!\underline{\mathbf{Z}}_ U \to \mathcal{F} \to 0$

with $U$ quasi-compact open in $X$. It suffices to show that $H^ q(X, \mathcal{K})$ is zero for $q \geq d + 1$. As above we can write $\mathcal{K}$ as the filtered colimit of subsheaves $\mathcal{K}'$ generated by finitely many sections over quasi-compact opens. Then $\mathcal{F}$ is the filtered colimit of the sheaves $j_!\underline{\mathbf{Z}}_ U/\mathcal{K}'$. In this way we reduce to the case that $\mathcal{K}$ is generated by finitely many sections over quasi-compact opens. Note that $\mathcal{K}$ is a subsheaf of $\underline{\mathbf{Z}}_ X$. Thus by Lemma 20.20.3 there exists a finite filtration of $\mathcal{K}$ whose successive quotients $\mathcal{Q}$ fit into a short exact sequence

$0 \to j''_!\underline{\mathbf{Z}}_ W \to j'_!\underline{\mathbf{Z}}_ V \to \mathcal{Q} \to 0$

with $j'' : W \to X$ and $j' : V \to X$ the inclusions of quasi-compact opens. Hence the vanishing of $H^ p(X, \mathcal{Q})$ for $p > d$ follows from our assumption (in the lemma) on the vanishing of the cohomology groups of $j''_!\underline{\mathbf{Z}}_ W$ and $j'_!\underline{\mathbf{Z}}_ V$. Returning to $\mathcal{K}$ this, via an induction argument using the long exact cohomology sequence, implies the desired vanishing for it as well. $\square$

Comment #6303 by Qilin,Yang on

I have been wondering about $Colim_A F_A =F$ .How does it transit from infinity to finiteness? If the stalk of the sheaf ${\ mathscr F}$is a finite or countable rank Abelian group, the proof is understandable. But if the cardinality of the rank of the stalk is bigger than $2^k$, where $k$ is uncountable, I could not understand.

Comment #6415 by on

Since filtered colimits are exact we have $\text{colim} \mathcal{F}_A \subset \mathcal{F}$. The reference shows that the stalks of the colimit are equal to the stalks of $\mathcal{F}$ and hence the inclusion is an equality. OK?

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