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. $\square$

Proof. The limit $X$ is an algebraic space by Lemma 70.4.1. The equality is formal, see Categories, Lemma 4.14.10. $\square$

Proof. Observe that $\mathop{\mathrm{lim}}\nolimits X_ i$ and $\mathop{\mathrm{lim}}\nolimits Y_ i$ exist by Lemma 70.4.1. Pick $0 \in I$ and choose an affine scheme $V_0$ and an étale morphism $V_0 \to Y_0$. Then the morphisms $V_ i = Y_ i \times _{Y_0} V_0 \to U_ i = X_ i \times _{Y_0} V_0$ are closed immersions of affine schemes. Hence the morphism $V = Y \times _{Y_0} V_0 \to U = X \times _{Y_0} V_0$ is a closed immersion because $V = \mathop{\mathrm{lim}}\nolimits V_ i$, $U = \mathop{\mathrm{lim}}\nolimits U_ i$ and because a limit of closed immersions of affine schemes is a closed immersion: a filtered colimit of surjective ring maps is surjective. Since the étale morphisms $V \to Y$ form an étale covering of $Y$ as we vary our choice of $V_0 \to Y_0$ we see that the lemma is true. $\square$

Proof. Observe that $\mathop{\mathrm{lim}}\nolimits X_ i$ exists by Lemma 70.4.1. Pick $0 \in I$ and choose an affine scheme $V_0$ and an étale morphism $U_0 \to X_0$. Then the affine schemes $U_ i = X_ i \times _{X_0} U_0$ are reduced. Hence $U = X \times _{X_0} U_0$ is a reduced affine scheme as a limit of reduced affine schemes: a filtered colimit of reduced rings is reduced. Since the étale morphisms $U \to X$ form an étale covering of $X$ as we vary our choice of $U_0 \to X_0$ we see that the lemma is true. $\square$

Proof. It is clear that (3) implies (2). We will assume (2) and prove (3). The proof is rather formal and we encourage the reader to find their own proof.

Let us first prove that (3) holds when $T_ i$ is in addition assumed separated for all $i$. Choose $i \in I$ and choose a surjective étale morphism $U_ i \to T_ i$ where $U_ i$ is affine. Using Lemma 70.4.2 we see that with $U = U_ i \times _{T_ i} T$ and $U_{i'} = U_ i \times _{T_ i} T_{i'}$ we have $U = \mathop{\mathrm{lim}}\nolimits _{i' \geq i} U_{i'}$. Of course $U$ and $U_{i'}$ are affine (see Lemma 70.4.1). Since $T_ i$ is separated, the fibre product $V_ i = U_ i \times _{T_ i} U_ i$ is an affine scheme as well and we obtain affine schemes $V = V_ i \times _{T_ i} T$ and $V_{i'} = V_ i \times _{T_ i} T_{i'}$ with $V = \mathop{\mathrm{lim}}\nolimits _{i' \geq i} V_{i'}$. Observe that $U \to T$ and $U_ i \to T_ i$ are surjective étale and that $V = U \times _ T U$ and $V_{i'} = U_{i'} \times _{T_{i'}} U_{i'}$. Note that $\mathop{\mathrm{Mor}}\nolimits (T, X)$ is the equalizer of the two maps $\mathop{\mathrm{Mor}}\nolimits (U, X) \to \mathop{\mathrm{Mor}}\nolimits (V, X)$; this is true for example because $X$ as a sheaf on $(\mathit{Sch}/S)_{fppf}$ is the coequalizer of the two maps $h_ V \to h_ u$. Similarly $\mathop{\mathrm{Mor}}\nolimits (T_{i'}, X)$ is the equalizer of the two maps $\mathop{\mathrm{Mor}}\nolimits (U_{i'}, X) \to \mathop{\mathrm{Mor}}\nolimits (V_{i'}, X)$. And of course the same thing is true with $X$ replaced with $Y$. Condition (2) says that the diagrams of in (3) are fibre products in the case of $U = \mathop{\mathrm{lim}}\nolimits U_ i$ and $V = \mathop{\mathrm{lim}}\nolimits V_ i$. It follows formally that the same thing is true for $T = \mathop{\mathrm{lim}}\nolimits T_ i$.

In the general case, choose an affine scheme $U$, an $i \in I$, and a surjective étale morphism $U \to T_ i$. Repeating the argument of the previous paragraph we still achieve the proof: the schemes $V_{i'}$, $V$ are no longer affine, but they are still quasi-compact and separated and the result of the preceding paragraph applies. $\square$