Proposition 38.37.9. Let $\mathcal{F}$ be a presheaf on one of the sites $(\mathit{Sch}/S)_ h$ constructed in Definition 38.34.12. Then $\mathcal{F}$ is a sheaf if and only if the following conditions are satisfied

1. $\mathcal{F}$ is a sheaf for the Zariski topology,

2. given a morphism $f : X \to Y$ of $(\mathit{Sch}/S)_ h$ with $Y$ affine and $f$ surjective, flat, proper, and of finite presentation, then $\mathcal{F}(Y)$ is the equalizer of the two maps $\mathcal{F}(X) \to \mathcal{F}(X \times _ Y X)$,

3. given an almost blow up square (38.37.0.1) with $X$ affine in the category $(\mathit{Sch}/S)_ h$ the diagram

$\xymatrix{ \mathcal{F}(E) & \mathcal{F}(X') \ar[l] \\ \mathcal{F}(Z) \ar[u] & \mathcal{F}(X) \ar[u] \ar[l] }$

is cartesian in the category of sets.

Proof. Assume $\mathcal{F}$ is a sheaf. Condition (1) holds because a Zariski covering is a h covering, see Lemma 38.34.5. Condition (2) holds because for $f$ as in (2) we have that $\{ X \to Y\}$ is an fppf covering (this is clear) and hence an h covering, see Lemma 38.34.5. Condition (3) holds by Lemma 38.37.7.

Conversely, assume $\mathcal{F}$ satisfies (1), (2), and (3). We will prove $\mathcal{F}$ is a sheaf by applying Lemma 38.34.16. Consider a surjective, finitely presented, proper morphism $f : X \to Y$ in $(\mathit{Sch}/S)_ h$ with $Y$ affine. It suffices to show that $\mathcal{F}(Y)$ is the equalizer of the two maps $\mathcal{F}(X) \to \mathcal{F}(X \times _ Y X)$.

First, assume that $f : X \to Y$ is in addition a closed immersion (in other words, $f$ is a thickening). Then the blow up of $Y$ in $X$ is the empty scheme and this produces an almost blow up square consisting with $\emptyset , \emptyset , X, Y$ at the vertices (compare with the second proof of Lemma 38.37.8). Hence we see that condition (3) tells us that

$\xymatrix{ \mathcal{F}(\emptyset ) & \mathcal{F}(\emptyset ) \ar[l] \\ \mathcal{F}(X) \ar[u] & \mathcal{F}(Y) \ar[u] \ar[l] }$

is cartesian in the category of sets. Since $\mathcal{F}$ is a sheaf for the Zariski topology, we see that $\mathcal{F}(\emptyset )$ is a singleton. Hence we see that $\mathcal{F}(X) = \mathcal{F}(Y)$.

Interlude A: let $T \to T'$ be a morphism of $(\mathit{Sch}/S)_ h$ which is a thickening and of finite presentation. Then $\mathcal{F}(T') \to \mathcal{F}(T)$ is bijective. Namely, choose an affine open covering $T' = \bigcup T'_ i$ and let $T_ i = T \times _{T'} T'_ i$ be the corresponding affine opens of $T$. Then we have $\mathcal{F}(T'_ i) \to \mathcal{F}(T_ i)$ is bijective for all $i$ by the result of the previous paragraph. Using the Zariski sheaf property we see that $\mathcal{F}(T') \to \mathcal{F}(T)$ is injective. Repeating the argument we find that it is bijective. Minor details omitted.

Interlude B: consider an almost blow up square (38.37.0.1) in the category $(\mathit{Sch}/S)_ h$. Then we claim the diagram

$\xymatrix{ \mathcal{F}(E) & \mathcal{F}(X') \ar[l] \\ \mathcal{F}(Z) \ar[u] & \mathcal{F}(X) \ar[u] \ar[l] }$

is cartesian in the category of sets. This is a consequence of condition (3) as follows by choosing an affine open covering of $X$ and arguing as in Interlude A. We omit the details.

Next, let $f : X \to Y$ be a surjective, finitely presented, proper morphism in $(\mathit{Sch}/S)_ h$ with $Y$ affine. Choose a generic flatness stratification

$Y \supset Y_0 \supset Y_1 \supset \ldots \supset Y_ t = \emptyset$

as in Lemma 38.21.4 for $f : X \to Y$. We are going to use all the properties of the stratification without further mention. Set $X_0 = X \times _ Y Y_0$. By the Interlude B we have $\mathcal{F}(Y_0) = \mathcal{F}(Y)$, $\mathcal{F}(X_0) = \mathcal{F}(X)$, and $\mathcal{F}(X_0 \times _{Y_0} X_0) = \mathcal{F}(X \times _ Y X)$.

We are going to prove the result by induction on $t$. If $t = 1$ then $X_0 \to Y_0$ is surjective, proper, flat, and of finite presentation and we see that the result holds by property (2). For $t > 1$ we may replace $Y$ by $Y_0$ and $X$ by $X_0$ (see above) and assume $Y = Y_0$.

Consider the quasi-compact open subscheme $V = Y \setminus Y_1 = Y_0 \setminus Y_1$. Choose a diagram

$\xymatrix{ E \ar[ddd] \ar[rd] & & & D \ar[lll] \ar[ddd] \ar[ld] \\ & Y' \ar[d] & X' \ar[l] \ar[d] \\ & Y & X \ar[l] \\ Z \ar[ru] & & & T \ar[lll] \ar[lu] }$

as in Lemma 38.37.6 for $f : X \to Y \supset V$. Then $f' : X' \to Y'$ is flat and of finite presentation. Also $f'$ is proper (use Morphisms, Lemmas 29.41.4 and 29.41.7 to see this). Thus the image $W = f'(X') \subset Y'$ is an open (Morphisms, Lemma 29.25.10) and closed subscheme of $Y'$. Observe that $Y' \setminus E$ is contained in $W$. By Lemma 38.37.2 this means we may replace $Y'$ by $W$ in the above diagram. In other words, we may and do assume $f'$ is surjective. At this point we know that

$\vcenter { \xymatrix{ \mathcal{F}(E) & \mathcal{F}(Y') \ar[l] \\ \mathcal{F}(Z) \ar[u] & \mathcal{F}(Y) \ar[u] \ar[l] } } \quad \text{and}\quad \vcenter { \xymatrix{ \mathcal{F}(D) & \mathcal{F}(X') \ar[l] \\ \mathcal{F}(T) \ar[u] & \mathcal{F}(X) \ar[u] \ar[l] } }$

are cartesian by Interlude B. Note that $Z \cap Y_1 \to Z$ is a thickening of finite presentation (as $Z$ is set theoretically contained in $Y_1$ as a closed subscheme of $Y$ disjoint from $V$). Thus we obtain a filtration

$Z \supset Z \cap Y_1 \supset Z \cap Y_2 \subset \ldots \subset Z \cap Y_ t = \emptyset$

as above for the restriction $T = Z \times _ Y X \to Z$ of $f$ to $T$. Thus by induction hypothesis we find that $\mathcal{F}(Z) \to \mathcal{F}(T)$ is an injective map of sets whose image is the equalizer of the two maps $\mathcal{F}(T) \to \mathcal{F}(T \times _ Z T)$.

Let $s \in \mathcal{F}(X)$ be in the equalizer of the two maps $\mathcal{F}(X) \to \mathcal{F}(X \times _ Y X)$. By the above we see that the restriction $s|_ T$ comes from a unique element $t \in \mathcal{F}(Z)$ and similarly that the restriction $s|_{X'}$ comes from a unique element $t' \in \mathcal{F}(Y')$. Chasing sections using the restriction maps for $\mathcal{F}$ corresponding to the arrows in the huge commutative diagram above the reader finds that $t$ and $t'$ restrict to the same element of $\mathcal{F}(E)$ because they restrict to the same element of $\mathcal{F}(D)$ and we have (2); here we use that $D \to E$ is surjective, flat, proper, and of finite presentation as the restriction of $X' \to Y'$. Thus by the first of the two cartesian squares displayed above we get a unique section $u \in \mathcal{F}(Y)$ restricting to $t$ and $t'$ on $Z$ and $Y'$. To see that $u$ restrict to $s$ on $X$ use the second diagram. $\square$

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