The Stacks project

Proof. Let $\tau = fppf$, resp. $\tau = {\acute{e}tale}$. The lemma follows from Cohomology on Sites, Lemma 21.16.2 applied to the site $\mathcal{X}_\tau$. In order to check the assumptions we use Cohomology on Sites, Remark 21.16.3. Namely, let $\mathcal{B} \subset \mathop{\mathrm{Ob}}\nolimits (\mathcal{X}_\tau )$ be the set of objects lying over affine schemes. In other words, an element of $\mathcal{B}$ is a morphism $x : U \to \mathcal{X}$ with $U$ affine. We check each of the conditions (1) – (4) of the remark in turn:

1. Since $\mathcal{X}$ is quasi-compact, there exists a surjetive and smooth morphism $x : U \to \mathcal{X}$ with $U$ affine (Properties of Stacks, Lemma 100.6.2). Then $h_ x^\# \to *$ is a surjective map of sheaves on $\mathcal{X}_\tau$.

2. Since coverings in $\mathcal{X}_\tau$ are fppf, resp. étale coverings, we see that every covering of $U \in \mathcal{B}$ is refined by a finite affine fppf covering, see Topologies, Lemma 34.7.4, resp. Lemma 34.4.4.

3. Let $x : U \to \mathcal{X}$ and $x' : U' \to \mathcal{X}$ be in $\mathcal{B}$. The product $h_ x^\# \times h_{x'}^\#$ in $\mathop{\mathit{Sh}}\nolimits (\mathcal{X}_\tau )$ is equal to the sheaf on $\mathcal{X}_\tau$ determined by the algebraic space $W = U \times _{x, \mathcal{X}, x'} U'$ over $\mathcal{X}$: for an object $y : V \to \mathcal{X}$ of $\mathcal{X}_\tau$ we have $(h_ x^\# \times h_{x'}^\# )(y) = \{ f : V \to W \mid y = x \circ \text{pr}_1 \circ f = x' \circ \text{pr}_2 \circ f\}$. The algebraic space $W$ is quasi-compact because $\mathcal{X}$ is quasi-separated, see Morphisms of Stacks, Lemma 101.7.8 for example. Hence we can choose an affine scheme $U''$ and a surjective étale morphism $U'' \to W$. Denote $x'' : U'' \to \mathcal{X}$ the composition of $U'' \to W$ and $W \to \mathcal{X}$. Then $h_{x''}^\# \to h_ x^\# \times h_{x'}^\#$ is surjective as desired.

4. Let $x : U \to \mathcal{X}$ and $x' : U' \to \mathcal{X}$ be in $\mathcal{B}$. Let $a, b : U \to U'$ be a morphism over $\mathcal{X}$, i.e., $a, b : x \to x'$ is a morphism in $\mathcal{X}_\tau$. Then the equalizer of $h_ a$ and $h_ b$ is represented by the equalizer of $a, b : U \to U'$ which is affine scheme over $\mathcal{X}$ and hence in $\mathcal{B}$.

This finished the proof. $\square$