Lemma 59.49.2. Let $X$ be a scheme. Let $\tau \in \{ Zariski, {\acute{e}tale}, smooth, syntomic, fppf\}$. Let $\mathcal{C}_1 \subset \mathcal{C}_2 \subset (\mathit{Sch}/X)_\tau$ be full subcategories with the following properties:

1. For an object $U/X$ of $\mathcal{C}_ t$,

1. if $\{ U_ i \to U\}$ is a covering of $(\mathit{Sch}/X)_\tau$, then $U_ i/X$ is an object of $\mathcal{C}_ t$,

2. $U \times \mathbf{A}^1/X$ is an object of $\mathcal{C}_ t$.

2. $X/X$ is an object of $\mathcal{C}_ t$.

We endow $\mathcal{C}_ t$ with the structure of a site whose coverings are exactly those coverings $\{ U_ i \to U\}$ of $(\mathit{Sch}/X)_\tau$ with $U \in \mathop{\mathrm{Ob}}\nolimits (\mathcal{C}_ t)$. Then

1. The functor $\mathcal{C}_1 \to \mathcal{C}_2$ is fully faithful, continuous, and cocontinuous.

Denote $g : \mathop{\mathit{Sh}}\nolimits (\mathcal{C}_1) \to \mathop{\mathit{Sh}}\nolimits (\mathcal{C}_2)$ the corresponding morphism of topoi. Denote $\mathcal{O}_ t$ the restriction of $\mathcal{O}$ to $\mathcal{C}_ t$. Denote $g_!$ the functor of Modules on Sites, Definition 18.16.1.

1. The canonical map $g_!\mathcal{O}_1 \to \mathcal{O}_2$ is an isomorphism.

Proof. Assertion (a) is immediate from the definitions. In this proof all schemes are schemes over $X$ and all morphisms of schemes are morphisms of schemes over $X$. Note that $g^{-1}$ is given by restriction, so that for an object $U$ of $\mathcal{C}_1$ we have $\mathcal{O}_1(U) = \mathcal{O}_2(U) = \mathcal{O}(U)$. Recall that $g_!\mathcal{O}_1$ is the sheaf associated to the presheaf $g_{p!}\mathcal{O}_1$ which associates to $V$ in $\mathcal{C}_2$ the group

$\mathop{\mathrm{colim}}\nolimits _{V \to U} \mathcal{O}(U)$

where $U$ runs over the objects of $\mathcal{C}_1$ and the colimit is taken in the category of abelian groups. Below we will use frequently that if

$V \to U \to U'$

are morphisms with $U, U' \in \mathop{\mathrm{Ob}}\nolimits (\mathcal{C}_1)$ and if $f' \in \mathcal{O}(U')$ restricts to $f \in \mathcal{O}(U)$, then $(V \to U, f)$ and $(V \to U', f')$ define the same element of the colimit. Also, $g_!\mathcal{O}_1 \to \mathcal{O}_2$ maps the element $(V \to U, f)$ simply to the pullback of $f$ to $V$.

Surjectivity. Let $V$ be a scheme and let $h \in \mathcal{O}(V)$. Then we obtain a morphism $V \to X \times \mathbf{A}^1$ induced by $h$ and the structure morphism $V \to X$. Writing $\mathbf{A}^1 = \mathop{\mathrm{Spec}}(\mathbf{Z}[x])$ we see the element $x \in \mathcal{O}(X \times \mathbf{A}^1)$ pulls back to $h$. Since $X \times \mathbf{A}^1$ is an object of $\mathcal{C}_1$ by assumptions (1)(b) and (2) we obtain the desired surjectivity.

Injectivity. Let $V$ be a scheme. Let $s = \sum _{i = 1, \ldots , n} (V \to U_ i, f_ i)$ be an element of the colimit displayed above. For any $i$ we can use the morphism $f_ i : U_ i \to X \times \mathbf{A}^1$ to see that $(V \to U_ i, f_ i)$ defines the same element of the colimit as $(f_ i : V \to X \times \mathbf{A}^1, x)$. Then we can consider

$f_1 \times \ldots \times f_ n : V \to X \times \mathbf{A}^ n$

and we see that $s$ is equivalent in the colimit to

$\sum \nolimits _{i = 1, \ldots , n} (f_1 \times \ldots \times f_ n : V \to X \times \mathbf{A}^ n, x_ i) = (f_1 \times \ldots \times f_ n : V \to X \times \mathbf{A}^ n, x_1 + \ldots + x_ n)$

Now, if $x_1 + \ldots + x_ n$ restricts to zero on $V$, then we see that $f_1 \times \ldots \times f_ n$ factors through $X \times \mathbf{A}^{n - 1} = V(x_1 + \ldots + x_ n)$. Hence we see that $s$ is equivalent to zero in the colimit. $\square$

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