The Stacks project

Proof. Let $Z$ be an algebraic space over $S$. Suppose that $f_ i : X_ i \to Z$ is a family of morphisms such that for each $i \to j$ the composition $X_ i \to X_ j \to Z$ is equal to $f_ i$. We have to construct a morphism of algebraic spaces $f : X \to Z$ such that we can recover $f_ i$ as the composition $X_ i \to X \to Z$. Let $W \to Z$ be a surjective étale morphism of a scheme to $Z$. We may assume that $W$ is a disjoint union of affines and in particular we may assume that $W \to Z$ is separated. For each $i$ set $U_ i = W \times _{Z, f_ i} X_ i$ and denote $h_ i : U_ i \to W$ the projection. Then $U_ i \to X_ i$ forms an object of $\mathop{\mathrm{lim}}\nolimits X_{i, {\acute{e}tale}}$ with $U_ i \to X_ i$ separated. By assumption (4) we can find an étale morphism $U \to X$ and (functorial) isomorphisms $U_ i = X_ i \times _ X U$. By assumption (3) there exists a morphism $h : U \to W$ such that the compositions $U_ i \to U \to W$ are $h_ i$. Let $g : U \to Z$ be the composition of $h$ with the map $W \to Z$. To finish the proof we have to show that $g : U \to Z$ descends to a morphism $X \to Z$. To do this, consider the morphism $(h, h) : U \times _ X U \to W \times _ S W$. Composing with $U_ i \times _{X_ i} U_ i \to U \times _ X U$ we obtain $(h_ i, h_ i)$ which factors through $W \times _ Z W$. Since $U \times _ X U$ is the colimit of the schemes $U_ i \times _{X_ i} U_ i$ by (3) we see that $(h, h)$ factors through $W \times _ Z W$. Hence the two compositions $U \times _ X U \to U \to W \to Z$ are equal. Because each $U_ i \to X_ i$ is surjective and assumption (2) we see that $U \to X$ is surjective. As $Z$ is a sheaf for the étale topology, we conclude that $g : U \to Z$ descends to $f : X \to Z$ as desired. $\square$

Proof. Namely, for an algebraic space $Y$ over $B$ a morphism $X \to Y$ over $B$ is the same thing as a collection of morphism $U_ a \to Y$ which agree on the overlaps $U_ a \times _ X U_ b$ for all $a, b \in A$, see Descent on Spaces, Lemma 74.7.2. $\square$

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. $\square$

Proof. We encourage the reader to look instead at Lemma 81.3.3 and its proof.

Let $Z$ be an algebraic space over $B$. Suppose that $f_ i : X_ i \to Z$ is a family of morphisms such that for each $i \to j$ the composition $X_ i \to X_ j \to Z$ is equal to $f_ i$. We have to construct a morphism of algebraic spaces $f : X \to Z$ over $B$ such that we can recover $f_ i$ as the composition $X_ i \to X \to Z$. Let $W \to Z$ be a surjective étale morphism of a scheme to $Z$. We may assume that $W$ is a disjoint union of affines and in particular we may assume that $W \to Z$ is separated and that $W$ is separated over $B$. For each $i$ set $U_ i = W \times _{Z, f_ i} X_ i$ and denote $h_ i : U_ i \to W$ the projection. Then $U_ i \to X_ i$ forms an object of $\mathop{\mathrm{lim}}\nolimits X_{i, spaces, {\acute{e}tale}}$ with $U_ i \to X_ i$ separated. By assumption (5) we can find a separated étale morphism $U \to X$ of algebraic spaces and (functorial) isomorphisms $U_ i = X_ i \times _ X U$. By assumption (4) there exists a morphism $h : U \to W$ over $B$ such that the compositions $U_ i \to U \to W$ are $h_ i$. Let $g : U \to Z$ be the composition of $h$ with the map $W \to Z$. To finish the proof we have to show that $g : U \to Z$ descends to a morphism $X \to Z$. To do this, consider the morphism $(h, h) : U \times _ X U \to W \times _ S W$. Composing with $U_ i \times _{X_ i} U_ i \to U \times _ X U$ we obtain $(h_ i, h_ i)$ which factors through $W \times _ Z W$. Since $U \times _ X U$ is the colimit of the algebraic spaces $U_ i \times _{X_ i} U_ i$ in the category of algebraic spaces separated over $B$ by (4) we see that $(h, h)$ factors through $W \times _ Z W$. Hence the two compositions $U \times _ X U \to U \to W \to Z$ are equal. Because each $U_ i \to X_ i$ is surjective and assumption (2) we see that $U \to X$ is surjective. As $Z$ is a sheaf for the étale topology, we conclude that $g : U \to Z$ descends to $f : X \to Z$ as desired. $\square$