The Stacks project

Proof. Let $T$ be an algebraic space over $S$. Let $\{ T_ i \to T\} $ be an fppf covering of $T$ (as in Topologies on Spaces, Section 73.7). Suppose that $(a_ i, b_ i) \in \text{Res}_{Z/B}(X)(T_ i)$ such that $(a_ i, b_ i)|_{T_ i \times _ T T_ j} = (a_ j, b_ j)|_{T_ i \times _ T T_ j}$ for all $i, j$. Then by Descent on Spaces, Lemma 74.7.2 there exists a unique morphism $a : T \to B$ such that $a_ i$ is the composition of $T_ i \to T$ and $a$. Then $\{ T_ i \times _{a_ i, B} Z \to T \times _{a, B} Z\} $ is an fppf covering too and the same lemma implies there exists a unique morphism $b : T \times _{a, B} Z \to X$ such that $b_ i$ is the composition of $T_ i \times _{a_ i, B} Z \to T \times _{a, B} Z$ and $b$. Hence $(a, b) \in \text{Res}_{Z/B}(X)(T)$ restricts to $(a_ i, b_ i)$ over $T_ i$ for all $i$.

Note that the result of the preceding paragraph in particular implies (1).

Let $T$ be an algebraic space over $S$. In order to prove (2) we will construct mutually inverse maps between the displayed sets. In the following when we say “pair” we mean a pair $(a, b)$ fitting into (97.11.0.1).

Let $v : T \to \text{Res}_{Z/B}(X)$ be a natural transformation. Choose a scheme $U$ and a surjective étale morphism $p : U \to T$. Then $v(p) \in \text{Res}_{Z/B}(X)(U)$ corresponds to a pair $(a_ U, b_ U)$ over $U$. Let $R = U \times _ T U$ with projections $t, s : R \to U$. As $v$ is a transformation of functors we see that the pullbacks of $(a_ U, b_ U)$ by $s$ and $t$ agree. Hence, since $\{ U \to T\} $ is an fppf covering, we may apply the result of the first paragraph that deduce that there exists a unique pair $(a, b)$ over $T$.

Conversely, let $(a, b)$ be a pair over $T$. Let $U \to T$, $R = U \times _ T U$, and $t, s : R \to U$ be as above. Then the restriction $(a, b)|_ U$ gives rise to a transformation of functors $v : h_ U \to \text{Res}_{Z/B}(X)$ by the Yoneda lemma (Categories, Lemma 4.3.5). As the two pullbacks $s^*(a, b)|_ U$ and $t^*(a, b)|_ U$ are equal, we see that $v$ coequalizes the two maps $h_ t, h_ s : h_ R \to h_ U$. Since $T = U/R$ is the fppf quotient sheaf by Spaces, Lemma 65.9.1 and since $\text{Res}_{Z/B}(X)$ is an fppf sheaf by (1) we conclude that $v$ factors through a map $T \to \text{Res}_{Z/B}(X)$.

We omit the verification that the two constructions above are mutually inverse. $\square$

Proof. The equality as functors follows immediately from the definitions. The equality as sheaves follows from this because both sides are sheaves according to Lemma 97.11.1 and the fact that a fibre product of sheaves is the same as the corresponding fibre product of pre-sheaves (i.e., functors). $\square$

Proof. Let $U$ be a scheme and let $\xi = (a, b)$ be an element of $\text{Res}_{Z/B}(X)(U)$. We have to prove that the functor

is representable by an algebraic space étale over $U$. Set $Z_ U = U \times _{a, B} Z$ and $W = Z_ U \times _{b, X} X'$. Then $W \to Z_ U \to U$ is as in Lemma 97.9.2 and the sheaf $F$ defined there is identified with the fibre product displayed above. Hence the first assertion of the lemma. The second assertion follows from this and Lemma 97.9.1 which guarantees that $F \to U$ is surjective in the situation above. $\square$

Proof. Omitted. Hint: Exercise in the functorial point of view in algebraic geometry. $\square$

Proof. By Proposition 97.10.4 the functors $\mathit{Mor}_ B(Z, X)$ and $\mathit{Mor}_ B(Z, Z)$ are algebraic spaces. Hence this follows from the cartesian diagram of Lemma 97.11.4 and the fact that fibre products of algebraic spaces exist and are given by the fibre product in the underlying category of sheaves of sets (see Spaces, Lemma 65.7.2). $\square$