The Stacks project

Proof. We will prove this by induction on $p$. The base case $p = 0$ follows from Sites, Lemma 7.17.8 part (4). We check the assumptions hold, but we urge the reader to skip this part. Suppose $\mathcal{F} \to *$ is surjective. Choose $K \in S$ and $K \to *$ surjective as in (1). Then $\mathcal{F} \times K \to K$ is surjective. Choose $K' \to \mathcal{F} \times K$ with $K' \in S$ and $K' \to K$ surjective as in (2). Then there is a map $K' \to \mathcal{F}$ and $K' \to *$ is surjective. Hence Sites, Lemma 7.17.8 assumption (4)(a) is satisfied. By Sites, Lemma 7.17.5, assumptions (3) and (5) we see that $K \times K$ is quasi-compact for all $K \in S$. Hence Sites, Lemma 7.17.8 assumption (4)(b) is satisfied. This finishes the proof of the base case.

Induction step. Assume the result holds for $H^ p$ for $p \leq p_0$ and for all topoi $\mathop{\mathit{Sh}}\nolimits (\mathcal{C})$ such that a set $S \subset \mathop{\mathrm{Ob}}\nolimits (\mathop{\mathit{Sh}}\nolimits (\mathcal{C}))$ can be found satisfying (1) – (5). Arguing exactly as in the proof of Lemma 21.16.1 we see that it suffices to show: given a filtered colimit $\mathcal{I} = \mathop{\mathrm{colim}}\nolimits \mathcal{I}_\lambda $ with $\mathcal{I}_\lambda $ injective abelian sheaves, we have $H^{p_0 + 1}(\mathcal{C}, \mathcal{I}) = 0$. Choose $K \to *$ surjective with $K \in S$ as in (1). Denote $K^ n$ the $n$-fold self product of $K$. Consider the spectral sequence

of Lemma 21.13.2. Recall that $H^ q(K^{p + 1}, \mathcal{F}) = H^ q(\mathcal{C}/K^{p + 1}, j^{-1}\mathcal{F})$, for any abelian sheaf on $\mathcal{C}$, see Lemma 21.13.3. We have $j^{-1}\mathcal{I} = \mathop{\mathrm{colim}}\nolimits j^{-1}\mathcal{I}_\lambda $ as $j^{-1}$ commutes with colimits. The restrictions $j^{-1}\mathcal{I}_\lambda $ are injective abelian sheaves on $\mathcal{C}/K^{p + 1}$ by Lemma 21.7.1. Below we will show that the induction hypothesis applies to $\mathcal{C}/K^{p + 1}$ and hence we see that $H^ q(K^{p + 1}, \mathcal{I}) = \mathop{\mathrm{colim}}\nolimits H^ q(K^{p + 1}, \mathcal{I}_\lambda ) = 0$ for $q < p_0 + 1$ (vanishing as $\mathcal{I}_\lambda $ is injective). It follows that

Again using the induction hypothesis, the complex depicted on the right hand side is the colimit over $\Lambda $ of the complexes

These complexes are exact as $\mathcal{I}_\lambda $ is an injective abelian sheaf (follows from the spectral sequence for example). Since filtered colimits are exact in the category of abelian groups we obtain the desired vanishing.

We still have to show that the induction hypothesis applies to the site $\mathcal{C}/K^ n$ for all $n \geq 1$. Recall that $\mathop{\mathit{Sh}}\nolimits (\mathcal{C}/K^ n) = \mathop{\mathit{Sh}}\nolimits (\mathcal{C})/K^ n$, see Sites, Lemma 7.30.3. Thus we may work in $\mathop{\mathit{Sh}}\nolimits (\mathcal{C})/K^ n$. Denote $S_ n \subset \mathop{\mathrm{Ob}}\nolimits (\mathop{\mathit{Sh}}\nolimits (\mathcal{C}/K^ n)$ the set of objects of the form $K' \to K^ n$. We check each property in turn:

1. By (3) and induction there exists a surjection $K' \to K^ n$ with $K' \in S$. Then $(K' \to K^ n) \to (K^ n \to K^ n)$ is a surjection in $\mathop{\mathit{Sh}}\nolimits (\mathcal{C})/K^ n$ and $K^ n \to K^ n$ is the final object of $\mathop{\mathit{Sh}}\nolimits (\mathcal{C})/K^ n$. Hence (1) holds for $S_ n$,

2. Property (2) for $S_ n$ is an immediate consequence of (2) for $S$.

3. Let $a : K_1 \to K^ n$ and $b : K_2 \to K^ n$ be in $S_ n$. Then $(K_1 \to K^ n) \times (K_2 \to K^ n)$ is the object $K_1 \times _{K^ n} K_2 \to K^ n$ of $\mathop{\mathit{Sh}}\nolimits (\mathcal{C})/K^ n$. The subsheaf $K_1 \times _{K^ n} K_2 \subset K_1 \times K_2$ is the equalizer of $a \circ \text{pr}_1$ and $b \circ \text{pr}_2$. Write $a = (a_1, \ldots , a_ n)$ and $b = (b_1, \ldots , b_ n)$. Pick $K_3 \to K_1 \times K_2$ surjective with $K_3 \in S$; this is possibly by assumption (3) for $\mathcal{C}$. Pick

   \[ K_4 \longrightarrow \text{Equalizer}(K_3 \to K_1 \times K_2 \xrightarrow {a_1, b_1} K) \]

   surjective with $K_4 \in S$. This is possible by assumption (4) for $\mathcal{C}$. Pick

   \[ K_5 \longrightarrow \text{Equalizer}(K_4 \to K_1 \times K_2 \xrightarrow {a_2, b_2} K) \]

   surjective with $K_5 \in S$. Again this is possible. Continue in this fashion until we get

   \[ K_{3 + n} \longrightarrow \text{Equalizer}(K_{2 + n} \to K_1 \times K_2 \xrightarrow {a_ n, b_ n} K) \]

   surjective with $K_{3 + n} \in S$. By construction $K_{3 + n} \to K_1 \times _{K^ n} K_2$ is surjective. Hence $(K_{3 + n} \to K^ n)$ is in $S_ n$ and surjects onto the product $(K_1 \to K^ n) \times (K_2 \to K^ n)$. Thus (3) holds for $S_ n$.

4. Property (4) for $S_ n$ is an immediate consequence of property (4) for $S$.

5. Property (5) for $S_ n$ is a consequence of property (5) for $S$. Namely, an object $\mathcal{F} \to K^ n$ of $\mathop{\mathit{Sh}}\nolimits (\mathcal{C})/K^ n$ corresponds to a quasi-compact object of $\mathop{\mathit{Sh}}\nolimits (\mathcal{C}/K^ n)$ if and only if $\mathcal{F}$ is a quasi-compact object of $\mathop{\mathit{Sh}}\nolimits (\mathcal{C})$.

This finishes the proof of the lemma. $\square$