Proposition 66.17.1 (Gabber). Let $S$ be a scheme. Let $X$ be an algebraic space over $S$. Then $X$ satisfies the sheaf property for the fpqc topology.

## 66.17 Spaces and fpqc coverings

Let $S$ be a scheme. An algebraic space over $S$ is defined as a sheaf in the fppf topology with additional properties. Hence it is not immediately clear that it satisfies the sheaf property for the fpqc topology (see Topologies, Definition 34.9.13). In this section we give Gabber's argument showing this is true. However, when we say that the algebraic space $X$ satisfies the sheaf property for the fpqc topology we really only consider fpqc coverings $\{ f_ i : T_ i \to T\} _{i \in I}$ such that $T, T_ i$ are objects of the big site $(\mathit{Sch}/S)_{fppf}$ (as per our conventions, see Section 66.2).

**Proof.**
Since $X$ is a sheaf for the Zariski topology it suffices to show the following. Given a surjective flat morphism of affines $f : T' \to T$ we have: $X(T)$ is the equalizer of the two maps $X(T') \to X(T' \times _ T T')$. See Topologies, Lemma 34.9.14 (there is a little argument omitted here because the lemma cited is formulated for functors defined on the category of all schemes).

Let $a, b : T \to X$ be two morphisms such that $a \circ f = b \circ f$. We have to show $a = b$. Consider the fibre product

By Spaces, Lemma 65.13.1 the morphism $\Delta _{X/S}$ is a representable monomorphism. Hence $E \to T$ is a monomorphism of schemes. Our assumption that $a \circ f = b \circ f$ implies that $T' \to T$ factors (uniquely) through $E$. Consider the commutative diagram

Since the projection $T' \times _ T E \to T'$ is a monomorphism with a section we conclude it is an isomorphism. Hence we conclude that $E \to T$ is an isomorphism by Descent, Lemma 35.23.17. This means $a = b$ as desired.

Next, let $c : T' \to X$ be a morphism such that the two compositions $T' \times _ T T' \to T' \to X$ are the same. We have to find a morphism $a : T \to X$ whose composition with $T' \to T$ is $c$. Choose an affine scheme $U$ and an étale morphism $U \to X$ such that the image of $|U| \to |X|$ contains the image of $|c| : |T'| \to |X|$. This is possible by Lemmas 66.4.6 and 66.6.1, the fact that a finite disjoint union of affines is affine, and the fact that $|T'|$ is quasi-compact (small argument omitted). Since $U \to X$ is separated (Lemma 66.6.4), we see that

is a surjective, étale, separated morphism of schemes (to see that it is surjective use Lemma 66.4.3 and our choice of $U \to X$). The fact that $c \circ \text{pr}_0 = c \circ \text{pr}_1$ means that we obtain a descent datum on $V/T'/T$ (Descent, Definition 35.34.1) because

The morphism $V \to T'$ is ind-quasi-affine by More on Morphisms, Lemma 37.66.8 (because étale morphisms are locally quasi-finite, see Morphisms, Lemma 29.36.6). By More on Groupoids, Lemma 40.15.3 the descent datum is effective. Say $W \to T$ is a morphism such that there is an isomorphism $\alpha : T' \times _ T W \to V$ compatible with the given descent datum on $V$ and the canonical descent datum on $T' \times _ T W$. Then $W \to T$ is surjective and étale (Descent, Lemmas 35.23.7 and 35.23.29). Consider the composition

The two compositions $b' \circ (\text{pr}_0, 1), b' \circ (\text{pr}_1, 1) : (T' \times _ T T') \times _ T W \to T' \times _ T W \to U$ agree by our choice of $\alpha $ and the corresponding property of $c$ (computation omitted). Hence $b'$ descends to a morphism $b : W \to U$ by Descent, Lemma 35.13.7. The diagram

is commutative. What this means is that we have proved the existence of $a$ étale locally on $T$, i.e., we have an $a' : W \to X$. However, since we have proved uniqueness in the first paragraph, we find that this étale local solution satisfies the glueing condition, i.e., we have $\text{pr}_0^*a' = \text{pr}_1^*a'$ as elements of $X(W \times _ T W)$. Since $X$ is an étale sheaf we find a unique $a \in X(T)$ restricting to $a'$ on $W$. $\square$

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