Lemma 87.32.1. Let $S$ be a scheme. Let $X$ be a formal algebraic space over $S$. Then $X$ satisfies the sheaf property for the fpqc topology.
87.32 Formal algebraic spaces and fpqc coverings
This section is the analogue of Properties of Spaces, Section 66.17. Please read that section first.
Proof. The proof is identical to the proof of Properties of Spaces, Proposition 66.17.1. 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.
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 Lemma 87.11.2 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 a formal affine scheme $U$ and an étale morphism $U \to X$ such that the image of $|U| \to |X_{red}|$ contains the image of $|c| : |T'| \to |X_{red}|$. This is possible by Definition 87.11.1, Properties of Spaces, Lemma 66.4.6, the fact that a finite union of formal affine algebraic spaces is a formal affine algebraic space, and the fact that $|T'|$ is quasi-compact (small argument omitted). The morphism $U \to X$ is representable by schemes (Lemma 87.9.11) and separated (Lemma 87.16.5). Thus
is an étale and separated morphism of schemes. It is also surjective by our choice of $U \to X$ (if you do not want to argue this you can replace $U$ by a disjoint union of formal affine algebraic spaces so that $U \to X$ is surjective everything else still works as well). 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 an unique $a \in X(T)$ restricting to $a'$ on $W$. $\square$
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