Formal algebraic spaces are fpqc sheaves

Lemma 86.32.1. Let $S$ be a scheme. Let $X$ be a formal algebraic space over $S$. Then $X$ satisfies the sheaf property for the fpqc topology.

Proof. The proof is identical to the proof of Properties of Spaces, Proposition 65.17.1. 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 34.9.13.

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

$E = X \times _{\Delta _{X/S}, X \times _ S X, (a, b)} T.$

By Lemma 86.11.2 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

$\xymatrix{ T' \times _ T E \ar[r] \ar[d] & E \ar[d] \\ T' \ar[r] \ar@/^5ex/[u] \ar[ru] & T }$

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 86.11.1, Properties of Spaces, Lemma 65.4.6, 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 86.9.11) and separated (Lemma 86.16.5). Thus

$V = U \times _{X, c} T' \longrightarrow T'$

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

\begin{align*} V \times _{T'} (T' \times _ T T') & = U \times _{X, c \circ \text{pr}_0} (T' \times _ T T') \\ & = (T' \times _ T T') \times _{c \circ \text{pr}_1, X} U \\ & = (T' \times _ T T') \times _{T'} V \end{align*}

The morphism $V \to T'$ is ind-quasi-affine by More on Morphisms, Lemma 37.64.8 (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

$b' : T' \times _ T W \longrightarrow V = U \times _{X, c} T' \longrightarrow U$

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

$\xymatrix{ T' \times _ T W \ar[r] \ar[d] & W \ar[r]_ b & U \ar[d] \\ T' \ar[rr]^ c & & X }$

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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