The Stacks project

Proof. Part (2) is a formal consequence of the existence of the limit $X = \mathop{\mathrm{lim}}\nolimits X_ i$ as an algebraic space over $S$. Choose an element $0 \in I$ (this is possible as a directed set is nonempty). Choose a scheme $U_0$ and a surjective étale morphism $U_0 \to X_0$. Set $R_0 = U_0 \times _{X_0} U_0$ so that $X_0 = U_0/R_0$. For $i \geq 0$ set $U_ i = X_ i \times _{X_0} U_0$ and $R_ i = X_ i \times _{X_0} R_0 = U_ i \times _{X_ i} U_ i$. By Limits, Lemma 32.2.2 we see that $U = \mathop{\mathrm{lim}}\nolimits _{i \geq 0} U_ i$ and $R = \mathop{\mathrm{lim}}\nolimits _{i \geq 0} R_ i$ are schemes. Moreover, the two morphisms $s, t : R \to U$ are the base change of the two projections $R_0 \to U_0$ by the morphism $U \to U_0$, in particular étale. The morphism $R \to U \times _ S U$ defines an equivalence relation as directed a limit of equivalence relations is an equivalence relation. Hence the morphism $R \to U \times _ S U$ is an étale equivalence relation. We claim that the natural map

is an isomorphism of fppf sheaves on the category of schemes over $S$. The claim implies $X = \mathop{\mathrm{lim}}\nolimits X_ i$ is an algebraic space by Spaces, Theorem 65.10.5.

Let $Z$ be a scheme and let $a : Z \to \mathop{\mathrm{lim}}\nolimits X_ i$ be a morphism. Then $a = (a_ i)$ where $a_ i : Z \to X_ i$. Set $W_0 = Z \times _{a_0, X_0} U_0$. Note that $W_0 = Z \times _{a_ i, X_ i} U_ i$ for all $i \geq 0$ by our choice of $U_ i \to X_ i$ above. Hence we obtain a morphism $W_0 \to \mathop{\mathrm{lim}}\nolimits _{i \geq 0} U_ i = U$. Since $W_0 \to Z$ is surjective and étale, we conclude that (70.4.1.1) is a surjective map of sheaves. Finally, suppose that $Z$ is a scheme and that $a, b : Z \to U/R$ are two morphisms which are equalized by (70.4.1.1). We have to show that $a = b$. After replacing $Z$ by the members of an fppf covering we may assume there exist morphisms $a', b' : Z \to U$ which give rise to $a$ and $b$. The condition that $a, b$ are equalized by (70.4.1.1) means that for each $i \geq 0$ the compositions $a_ i', b_ i' : Z \to U \to U_ i$ are equal as morphisms into $U_ i/R_ i = X_ i$. Hence $(a_ i', b_ i') : Z \to U_ i \times _ S U_ i$ factors through $R_ i$, say by some morphism $c_ i : Z \to R_ i$. Since $R = \mathop{\mathrm{lim}}\nolimits _{i \geq 0} R_ i$ we see that $c = \mathop{\mathrm{lim}}\nolimits c_ i : Z \to R$ is a morphism which shows that $a, b$ are equal as morphisms of $Z$ into $U/R$.

Part (1) follows as we have seen above that $U_ i \times _{X_ i} X = U$ and $U \to U_ i$ is affine by construction. $\square$