The Stacks project

35.13 Fpqc coverings are universal effective epimorphisms

We apply the material above to prove an interesting result, namely Lemma 35.13.7. By Sites, Section 7.12 this lemma implies that the representable presheaves on any of the sites $(\mathit{Sch}/S)_\tau $ are sheaves for $\tau \in \{ Zariski, fppf, {\acute{e}tale}, smooth, syntomic\} $. First we prove a helper lemma.

Lemma 35.13.1. For a scheme $X$ denote $|X|$ the underlying set. Let $f : X \to S$ be a morphism of schemes. Then

\[ |X \times _ S X| \to |X| \times _{|S|} |X| \]

is surjective.

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

Lemma 35.13.2. Let $\{ f_ i : X_ i \to X\} _{i \in I}$ be a family of morphisms of affine schemes. The following are equivalent

  1. for any quasi-coherent $\mathcal{O}_ X$-module $\mathcal{F}$ we have

    \[ \Gamma (X, \mathcal{F}) = \text{Equalizer}\left( \xymatrix{ \prod \nolimits _{i \in I} \Gamma (X_ i, f_ i^*\mathcal{F}) \ar@<1ex>[r] \ar@<-1ex>[r] & \prod \nolimits _{i, j \in I} \Gamma (X_ i \times _ X X_ j, (f_ i \times f_ j)^*\mathcal{F}) } \right) \]
  2. $\{ f_ i : X_ i \to X\} _{i \in I}$ is a universal effective epimorphism (Sites, Definition 7.12.1) in the category of affine schemes.

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

\[ X' = \underline{\mathop{\mathrm{Spec}}}_ X(\mathcal{O}_ X \oplus \mathcal{F}) \]

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

\[ \Gamma (X' \times _ X X_ i, \mathcal{O}_{X' \times _ X X_ i}) = \Gamma (X_ i, \mathcal{O}_{X_ i}) \oplus \Gamma (X_ i, f_ i^*\mathcal{F}) \]

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

\[ \mathop{\mathrm{Mor}}\nolimits (T, \mathbf{A}^1_\mathbf {Z}) = \Gamma (T, \mathcal{O}_ T) \]

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

\[ \prod h_ i^\sharp : \Gamma (T, \mathcal{O}_ T) \to \prod \Gamma (X_ i, \mathcal{O}_{X_ i}) \]

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

\[ \Gamma (Y, \mathcal{G}) = \Gamma (X, p_*\mathcal{G}),\quad \Gamma (Y_ i, g_ i^*\mathcal{G}) = \Gamma (X, f_ i^*p_*\mathcal{G}), \]


\[ \Gamma (Y_ i \times _ Y Y_ j, (g_ i \times g_ j)^*\mathcal{G}) = \Gamma (X, (f_ i \times f_ j)^*p_*\mathcal{G}) \]

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

Lemma 35.13.3. Let $\{ f_ i : X_ i \to X\} _{i \in I}$ be a family of morphisms of schemes.

  1. If the family is universal effective epimorphism in the category of schemes, then $\coprod f_ i$ is surjective.

  2. If $X$ and $X_ i$ are affine and the family is a universal effective epimorphism in the category of affine schemes, then $\coprod f_ i$ is surjective.

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. $\square$

Lemma 35.13.4. Let $\{ f_ i : X_ i \to X\} _{i \in I}$ be a family of morphisms of schemes. If for every morphism $Y \to X$ with $Y$ affine the family of base changes $g_ i : Y_ i \to Y$ forms an effective epimorphism, then the family of $f_ i$ forms a universal effective epimorphism in the category of schemes.

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. $\square$

Lemma 35.13.5. Let $\{ f_ i : X_ i \to X\} _{i \in I}$ be a family of morphisms of affine schemes. Assume the equivalent assumption of Lemma 35.13.2 hold and that moreover for any morphism of affines $Y \to X$ the map

\[ \coprod X_ i \times _ X Y \longrightarrow Y \]

is a submersive map of topological spaces (Topology, Definition 5.6.3). Then our family of morphisms is a universal effective epimorphism in the category of schemes.

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. $\square$

Lemma 35.13.6. Let $\{ f_ i : T_ i \to T\} _{i \in I}$ be a fpqc covering. Suppose that for each $i$ we have an open subset $W_ i \subset T_ i$ such that for all $i, j \in I$ we have $\text{pr}_0^{-1}(W_ i) = \text{pr}_1^{-1}(W_ j)$ as open subsets of $T_ i \times _ T T_ j$. Then there exists a unique open subset $W \subset T$ such that $W_ i = f_ i^{-1}(W)$ for each $i$.

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. $\square$

Lemma 35.13.7. Let $\{ T_ i \to T\} $ be an fpqc covering, see Topologies, Definition 34.9.1. Then $\{ T_ i \to T\} $ is a universal effective epimorphism in the category of schemes, see Sites, Definition 7.12.1. In other words, every representable functor on the category of schemes satisfies the sheaf condition for the fpqc topology, see Topologies, Definition 34.9.12.

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 $\mathop{\mathrm{Spec}}(R) \to S$. $\square$

Lemma 35.13.8. Consider schemes $X, Y, Z$ and morphisms $a, b : X \to Y$ and a morphism $c : Y \to Z$ with $c \circ a = c \circ b$. Set $d = c \circ a = c \circ b$. If there exists an fpqc covering $\{ Z_ i \to Z\} $ such that

  1. for all $i$ the morphism $Y \times _{c, Z} Z_ i \to Z_ i$ is the coequalizer of $(a, 1) : X \times _{d, Z} Z_ i \to Y \times _{c, Z} Z_ i$ and $(b, 1) : X \times _{d, Z} Z_ i \to Y \times _{c, Z} Z_ i$, and

  2. for all $i$ and $i'$ the morphism $Y \times _{c, Z} (Z_ i \times _ Z Z_{i'}) \to (Z_ i \times _ Z Z_{i'})$ is the coequalizer of $(a, 1) : X \times _{d, Z} (Z_ i \times _ Z Z_{i'}) \to Y \times _{c, Z} (Z_ i \times _ Z Z_{i'})$ and $(b, 1) : X \times _{d, Z} (Z_ i \times _ Z Z_{i'}) \to Y \times _{c, Z} (Z_ i \times _ Z Z_{i'})$

then $c$ is the coequalizer of $a$ and $b$.

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. $\square$

Comments (2)

Comment #8277 by anon on

In the proof of 0EUD I think that the target of should be instead of .

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