The Stacks project

Proof. In this paragraph we reduce to the case where $B$ is an affine scheme. Let $B' \to B$ be an étale morphism of algebraic spaces. Observe that conditions (1) and (2) are preserved if we replace $B$, $X_ i$, $X$ by $B'$, $X_ i \times _ B B'$, $X \times _ B B'$. Let $\{ B_ a \to B\} _{a \in A}$ be an étale covering with $B_ a$ affine, see Properties of Spaces, Lemma 66.6.1. For $a \in A$ denote $X_ a$, $X_{a, i}$ the base changes of $X$ and the diagram to $B_ a$. For $a, b \in A$ denote $X_{a, b}$ and $X_{a, b, i}$ the base changes of $X$ and the diagram to $B_ a \times _ B B_ b$. By Lemma 81.3.2 it suffices to prove that $X_ a = \mathop{\mathrm{colim}}\nolimits X_{a, i}$ and $X_{a, b} = \mathop{\mathrm{colim}}\nolimits X_{a, b, i}$. This reduces us to the case where $B = B_ a$ (an affine scheme) or $B = B_ a \times _ B B_ b$ (a separated scheme). Repeating the argument once more, we conclude that we may assume $B$ is an affine scheme (this uses that the intersection of affine opens in a separated scheme is affine).

Assume $B$ is an affine scheme. Let $Z$ be an algebraic space over $B$. We have to show

is a bijection.

Proof of injectivity. Let $f, g : X \to Z$ be morphisms such that the compositions $f_ i, g_ i : X_ i \to Z$ are the same for all $i$. Choose an affine scheme $Z'$ and an étale morphism $Z' \to Z$. By Properties of Spaces, Lemma 66.6.1 we know we can cover $Z$ by such affines. Set $U = X \times _{f, Z} Z'$ and $U' = X \times _{g, Z} Z'$ and denote $p : U \to X$ and $p' : U' \to X$ the projections. Since $f_ i = g_ i$ for all $i$, we see that

compatible with transition morphisms. By (1) there is a unique isomorphism $\epsilon : U \to U'$ as algebraic spaces over $X$, i.e., with $p = p' \circ \epsilon $ which is compatible with the displayed identifications. Choose an étale covering $\{ h_ a : U_ a \to U\} $ with $U_ a$ affine. By (2) we see that $f \circ p \circ h_ a = g \circ p' \circ \epsilon \circ h_ a = g \circ p \circ h_ a$. Since $\{ h_ a : U_ a \to U\} $ is an étale covering we conclude $f \circ p = g \circ p$. Since the collection of morphisms $p : U \to X$ we obtain in this manner is an étale covering, we conclude that $f = g$.

Proof of surjectivity. Let $f_ i : X_ i \to Z$ be an element of the right hand side of the displayed arrow in the first paragraph of the proof. It suffices to find an étale covering $\{ U_ c \to X\} _{c \in C}$ such that the families $f_{c, i} \in \mathop{\mathrm{lim}}\nolimits _ i \mathop{\mathrm{Mor}}\nolimits _ B(X_ i \times _ X U_ c, Z)$ come from morphisms $f_ c : U_ c \to Z$. Namely, by the uniqueness proved above the morphisms $f_ c$ will agree on $U_ c \times _ X U_ b$ and hence will descend to give the desired morphism $f : X \to Z$. To find our covering, we first choose an étale covering $\{ g_ a : Z_ a \to Z\} _{a \in A}$ where each $Z_ a$ is affine. Then we let $U_{a, i} = X_ i \times _{f_ i, Z} Z_ a$. By (1) we find $U_{a, i} = X_ i \times _ X U_ a$ for some algebraic spaces $U_ a$ étale over $X$. Then we choose étale coverings $\{ U_{a, b} \to U_ a\} _{b \in B_ a}$ with $U_{a, b}$ affine and we consider the morphisms

By (2) we obtain morphisms $f_{a, b} : U_{a, b} \to Z_ a$ compatible with these morphisms. Setting $C = \coprod _{a \in A} B_ a$ and for $c \in C$ corresponding to $b \in B_ a$ setting $U_ c = U_{a, b}$ and $f_ c = g_ a \circ f_{a, b} : U_ c \to Z$ we conclude. $\square$