The Stacks project

Proof. It is clear that (3) implies (2).

Let us prove that (2) implies (1). Assume (2). Choose any affine opens $U \subset X$ and $V \subset S$ such that $f(U) \subset V$. We have to show that $\mathcal{O}_ S(V) \to \mathcal{O}_ X(U)$ is of finite presentation. Let $(A_ i, \varphi _{ii'})$ be a directed system of $\mathcal{O}_ S(V)$-algebras. Set $A = \mathop{\mathrm{colim}}\nolimits _ i A_ i$. According to Algebra, Lemma 10.127.3 we have to show that

Consider the schemes $T_ i = \mathop{\mathrm{Spec}}(A_ i)$. They form an inverse system of $V$-schemes over $I$ with transition morphisms $f_{ii'} : T_ i \to T_{i'}$ induced by the $\mathcal{O}_ S(V)$-algebra maps $\varphi _{i'i}$. Set $T := \mathop{\mathrm{Spec}}(A) = \mathop{\mathrm{lim}}\nolimits _ i T_ i$. The formula above becomes in terms of morphism sets of schemes

We first observe that $\mathop{\mathrm{Mor}}\nolimits _ V(T_ i, U) = \mathop{\mathrm{Mor}}\nolimits _ S(T_ i, U)$ and $\mathop{\mathrm{Mor}}\nolimits _ V(T, U) = \mathop{\mathrm{Mor}}\nolimits _ S(T, U)$. Hence we have to show that

and we are given that

Hence it suffices to prove that given a morphism $g_ i : T_ i \to X$ over $S$ such that the composition $T \to T_ i \to X$ ends up in $U$ there exists some $i' \geq i$ such that the composition $g_{i'} : T_{i'} \to T_ i \to X$ ends up in $U$. Denote $Z_{i'} = g_{i'}^{-1}(X \setminus U)$. Assume each $Z_{i'}$ is nonempty to get a contradiction. By Lemma 32.4.8 there exists a point $t$ of $T$ which is mapped into $Z_{i'}$ for all $i' \geq i$. Such a point is not mapped into $U$. A contradiction.

Finally, let us prove that (1) implies (3). Assume (1). Let an inverse directed system $(T_ i, f_{ii'})$ of $S$-schemes be given. Assume the morphisms $f_{ii'}$ are affine and each $T_ i$ is quasi-compact and quasi-separated as a scheme. Let $T = \mathop{\mathrm{lim}}\nolimits _ i T_ i$. Denote $f_ i : T \to T_ i$ the projection morphisms. We have to show:

1. Given morphisms $g_ i, g'_ i : T_ i \to X$ over $S$ such that $g_ i \circ f_ i = g'_ i \circ f_ i$, then there exists an $i' \geq i$ such that $g_ i \circ f_{i'i} = g'_ i \circ f_{i'i}$.

2. Given any morphism $g : T \to X$ over $S$ there exists an $i \in I$ and a morphism $g_ i : T_ i \to X$ such that $g = f_ i \circ g_ i$.

First let us prove the uniqueness part (a). Let $g_ i, g'_ i : T_ i \to X$ be morphisms such that $g_ i \circ f_ i = g'_ i \circ f_ i$. For any $i' \geq i$ we set $g_{i'} = g_ i \circ f_{i'i}$ and $g'_{i'} = g'_ i \circ f_{i'i}$. We also set $g = g_ i \circ f_ i = g'_ i \circ f_ i$. Consider the morphism $(g_ i, g'_ i) : T_ i \to X \times _ S X$. Set

This is an open in $X \times _ S X$, with the property that the morphism $\Delta _{X/S}$ factors through a closed immersion into $W$, see the proof of Schemes, Lemma 26.21.2. Note that the composition $(g_ i, g'_ i) \circ f_ i : T \to X \times _ S X$ is a morphism into $W$ because it factors through the diagonal by assumption. Set $Z_{i'} = (g_{i'}, g'_{i'})^{-1}(X \times _ S X \setminus W)$. If each $Z_{i'}$ is nonempty, then by Lemma 32.4.8 there exists a point $t \in T$ which maps to $Z_{i'}$ for all $i' \geq i$. This is a contradiction with the fact that $T$ maps into $W$. Hence we may increase $i$ and assume that $(g_ i, g'_ i) : T_ i \to X \times _ S X$ is a morphism into $W$. By construction of $W$, and since $T_ i$ is quasi-compact we can find a finite affine open covering $T_ i = T_{1, i} \cup \ldots \cup T_{n, i}$ such that $(g_ i, g'_ i)|_{T_{j, i}}$ is a morphism into $U \times _ V U$ for some pair $(U, V)$ as in the definition of $W$ above. Since it suffices to prove that $g_{i'}$ and $g'_{i'}$ agree on each of the $f_{i'i}^{-1}(T_{j, i})$ this reduces us to the affine case. The affine case follows from Algebra, Lemma 10.127.3 and the fact that the ring map $\mathcal{O}_ S(V) \to \mathcal{O}_ X(U)$ is of finite presentation (see Morphisms, Lemma 29.21.2).

Finally, we prove the existence part (b). Let $g : T \to X$ be a morphism of schemes over $S$. We can find a finite affine open covering $T = W_1 \cup \ldots \cup W_ n$ such that for each $j \in \{ 1, \ldots , n\} $ there exist affine opens $U_ j \subset X$ and $V_ j \subset S$ with $f(U_ j) \subset V_ j$ and $g(W_ j) \subset U_ j$. By Lemmas 32.4.11 and 32.4.13 (after possibly shrinking $I$) we may assume that there exist affine open coverings $T_ i = W_{1, i} \cup \ldots \cup W_{n, i}$ compatible with transition maps such that $W_ j = \mathop{\mathrm{lim}}\nolimits _ i W_{j, i}$. We apply Algebra, Lemma 10.127.3 to the rings corresponding to the affine schemes $U_ j$, $V_ j$, $W_{j, i}$ and $W_ j$ using that $\mathcal{O}_ S(V_ j) \to \mathcal{O}_ X(U_ j)$ is of finite presentation (see Morphisms, Lemma 29.21.2). Thus we can find for each $j$ an index $i_ j \in I$ and a morphism $g_{j, i_ j} : W_{j, i_ j} \to X$ such that $g_{j, i_ j} \circ f_ i|_{W_ j} : W_ j \to W_{j, i} \to X$ equals $g|_{W_ j}$. By part (a) proved above, using the quasi-compactness of $W_{j_1, i} \cap W_{j_2, i}$ which follows as $T_ i$ is quasi-separated, we can find an index $i' \in I$ larger than all $i_ j$ such that

for all $j_1, j_2 \in \{ 1, \ldots , n\} $. Hence the morphisms $g_{j, i_ j} \circ f_{i'i_ j}|_{W_{j, i'}}$ glue to given the desired morphism $T_{i'} \to X$. $\square$