The Stacks project

Proof. Follows immediately from the description of points on the fibre product in Schemes, Lemma 26.17.5. $\square$

Proof. Assume (2) holds and let $\mathcal{F}$ be a quasi-coherent $\mathcal{O}_ X$-module. Consider the scheme (Constructions, Section 27.4)

where $\mathcal{O}_ X \oplus \mathcal{F}$ is an $\mathcal{O}_ X$-algebra with multiplication $(f, s)(f', s') = (ff', fs' + f's)$. If $s_ i \in \Gamma (X_ i, f_ i^*\mathcal{F})$ is a section, then $s_ i$ determines a unique element of

Proof of equality omitted. If $(s_ i)_{i \in I}$ is in the equalizer of (1), then, using the equality

which holds for any scheme $T$, we see that these sections define a family of morphisms $h_ i : X' \times _ X X_ i \to \mathbf{A}^1_\mathbf {Z}$ with $h_ i \circ \text{pr}_1 = h_ j \circ \text{pr}_2$ as morphisms $(X' \times _ X X_ i) \times _{X'} (X' \times _ X X_ j) \to \mathbf{A}^1_\mathbf {Z}$. Since we've assume (2) we obtain a morphism $h : X' \to \mathbf{A}^1_\mathbf {Z}$ compatible with the morphisms $h_ i$ which in turn determines an element $s \in \Gamma (X, \mathcal{F})$. We omit the verification that $s$ maps to $s_ i$ in $\Gamma (X_ i, f_ i^*\mathcal{F})$.

Assume (1). Let $T$ be an affine scheme and let $h_ i : X_ i \to T$ be a family of morphisms such that $h_ i \circ \text{pr}_1 = h_ j \circ \text{pr}_2$ on $X_ i \times _ X X_ j$ for all $i, j \in I$. Then

maps into the equalizer and we find that we get a ring map $\Gamma (T, \mathcal{O}_ T) \to \Gamma (X, \mathcal{O}_ X)$ by the assumption of the lemma for $\mathcal{F} = \mathcal{O}_ X$. This ring map corresponds to a morphism $h : X \to T$ such that $h_ i = h \circ f_ i$. Hence our family is an effective epimorphism.

Let $p : Y \to X$ be a morphism of affines. We will show the base changes $g_ i : Y_ i \to Y$ of $f_ i$ form an effective epimorphism by applying the result of the previous paragraph. Namely, if $\mathcal{G}$ is a quasi-coherent $\mathcal{O}_ Y$-module, then

and

by the trivial base change formula (Cohomology of Schemes, Lemma 30.5.1). Thus we see property (1) lemma holds for the family $g_ i$. $\square$

Proof. Omitted. Hint: perform base change by $\mathop{\mathrm{Spec}}(\kappa (x)) \to X$ to see that any $x \in X$ has to be in the image. $\square$

Proof. Let $Y \to X$ be a morphism of schemes. We have to show that the base changes $g_ i : Y_ i \to Y$ form an effective epimorphism. To do this, assume given a scheme $T$ and morphisms $h_ i : Y_ i \to T$ with $h_ i \circ \text{pr}_1 = h_ j \circ \text{pr}_2$ on $Y_ i \times _ Y Y_ j$. Choose an affine open covering $Y = \bigcup V_\alpha $. Set $V_{\alpha , i}$ equal to the inverse image of $V_\alpha $ in $Y_ i$. Then we see that $V_{\alpha , i} \to V_\alpha $ is the base change of $f_ i$ by $V_\alpha \to X$. Thus by assumption the family of restrictions $h_ i|_{V_{\alpha , i}}$ come from a morphism of schemes $h_\alpha : V_\alpha \to T$. We leave it to the reader to show that these agree on overlaps and define the desired morphism $Y \to T$. See discussion in Schemes, Section 26.14. $\square$

Proof. By Lemma 35.13.4 it suffices to base change our family of morphisms by $Y \to X$ with $Y$ affine. Set $Y_ i = X_ i \times _ X Y$. Let $T$ be a scheme and let $h_ i : Y_ i \to Y$ be a family of morphisms such that $h_ i \circ \text{pr}_1 = h_ j \circ \text{pr}_2$ on $Y_ i \times _ Y Y_ j$. Note that $Y$ as a set is the coequalizer of the two maps from $\coprod Y_ i \times _ Y Y_ j$ to $\coprod Y_ i$. Namely, surjectivity by the affine case of Lemma 35.13.3 and injectivity by Lemma 35.13.1. Hence there is a set map of underlying sets $h : Y \to T$ compatible with the maps $h_ i$. By the second condition of the lemma we see that $h$ is continuous! Thus if $y \in Y$ and $U \subset T$ is an affine open neighbourhood of $h(y)$, then we can find an affine open $V \subset Y$ such that $h(V) \subset U$. Setting $V_ i = Y_ i \times _ Y V = X_ i \times _ X V$ we can use the result proved in Lemma 35.13.2 to see that $h|_ V : V \to U \subset T$ comes from a unique morphism of affine schemes $h_ V : V \to U$ agreeing with $h_ i|_{V_ i}$ as morphisms of schemes for all $i$. Glueing these $h_ V$ (see Schemes, Section 26.14) gives a morphism $Y \to T$ as desired. $\square$

Proof. Apply Lemma 35.13.1 to the map $\coprod _{i \in I} T_ i \to T$. It implies there exists a subset $W \subset T$ such that $W_ i = f_ i^{-1}(W)$ for each $i$, namely $W = \bigcup f_ i(W_ i)$. To see that $W$ is open we may work Zariski locally on $T$. Hence we may assume that $T$ is affine. Using Topologies, Definition 34.9.1 we may choose a standard fpqc covering $\{ g_ j : V_ j \to T\} _{j \in J}$ which refines $\{ T_ i \to T\} _{i \in I}$. Let $\alpha : J \to I$ and $h_ j : V_ j \to T_{\alpha (j)}$ be as in Sites, Definition 7.8.1. Then $g_ j^{-1}(W) = h_ j^{-1}(W_{\alpha (j)})$. Thus we may assume $\{ f_ i : T_ i \to T\} $ is a standard fpqc covering. In this case we may apply Morphisms, Lemma 29.25.12 to the morphism $\coprod T_ i \to T$ to conclude that $W$ is open. $\square$

Proof. Let $S$ be a scheme. We have to show the following: Given morphisms $\varphi _ i : T_ i \to S$ such that $\varphi _ i|_{T_ i \times _ T T_ j} = \varphi _ j|_{T_ i \times _ T T_ j}$ there exists a unique morphism $T \to S$ which restricts to $\varphi _ i$ on each $T_ i$. In other words, we have to show that the functor $h_ S = \mathop{\mathrm{Mor}}\nolimits _{\mathit{Sch}}( - , S)$ satisfies the sheaf property for the fpqc topology.

If $\{ T_ i \to T\} $ is a Zariski covering, then this follows from Schemes, Lemma 26.14.1. Thus Topologies, Lemma 34.9.13 reduces us to the case of a covering $\{ X \to Y\} $ given by a single surjective flat morphism of affines.

First proof. By Lemma 35.8.1 we have the sheaf condition for quasi-coherent modules for $\{ X \to Y\} $. By Lemma 35.13.6 the morphism $X \to Y$ is universally submersive. Hence we may apply Lemma 35.13.5 to see that $\{ X \to Y\} $ is a universal effective epimorphism.

Second proof. Let $R \to A$ be the faithfully flat ring map corresponding to our surjective flat morphism $\pi : X \to Y$. Let $f : X \to S$ be a morphism such that $f \circ \text{pr}_1 = f \circ \text{pr}_2$ as morphisms $X \times _ Y X = \mathop{\mathrm{Spec}}(A \otimes _ R A) \to S$. By Lemma 35.13.1 we see that as a map on the underlying sets $f$ is of the form $f = g \circ \pi $ for some (set theoretic) map $g : \mathop{\mathrm{Spec}}(R) \to S$. By Morphisms, Lemma 29.25.12 and the fact that $f$ is continuous we see that $g$ is continuous.

Pick $y \in Y = \mathop{\mathrm{Spec}}(R)$. Choose $U \subset S$ affine open containing $g(y)$. Say $U = \mathop{\mathrm{Spec}}(B)$. By the above we may choose an $r \in R$ such that $y \in D(r) \subset g^{-1}(U)$. The restriction of $f$ to $\pi ^{-1}(D(r))$ into $U$ corresponds to a ring map $B \to A_ r$. The two induced ring maps $B \to A_ r \otimes _{R_ r} A_ r = (A \otimes _ R A)_ r$ are equal by assumption on $f$. Note that $R_ r \to A_ r$ is faithfully flat. By Lemma 35.3.6 the equalizer of the two arrows $A_ r \to A_ r \otimes _{R_ r} A_ r$ is $R_ r$. We conclude that $B \to A_ r$ factors uniquely through a map $B \to R_ r$. This map in turn gives a morphism of schemes $D(r) \to U \to S$, see Schemes, Lemma 26.6.4.

What have we proved so far? We have shown that for any prime $\mathfrak p \subset R$, there exists a standard affine open $D(r) \subset \mathop{\mathrm{Spec}}(R)$ such that the morphism $f|_{\pi ^{-1}(D(r))} : \pi ^{-1}(D(r)) \to S$ factors uniquely through some morphism of schemes $D(r) \to S$. We omit the verification that these morphisms glue to the desired morphism $\mathop{\mathrm{Spec}}(R) \to S$. $\square$

Proof. Namely, for a scheme $T$ a morphism $Z \to T$ is the same thing as a collection of morphism $Z_ i \to T$ which agree on overlaps by Lemma 35.13.7. $\square$