Lemma 34.13.3. Let $S$, $\mathcal{C}$, $F$ satisfy conditions (1), (2), (a), and (b) of Lemma 34.13.1 and denote $F' : (\mathit{Sch}/S)^{opp} \to \textit{Sets}$ the unique extension constructed in the lemma. Let $\tau \in \{ Zariski, {\acute{e}tale}, smooth, syntomic, fppf\} $. Assume

for any standard $\tau $-covering $\{ V_ i \to V\} _{i = 1, \ldots , n}$ of affines in $\mathit{Sch}/S$ such that $V \to S$ factors through an affine open $U \subset S$ and $V \to U$ is of finite presentation, the sheaf condition hold for $F$ and $\{ V_ i \to V\} _{i = 1, \ldots , n}$^{1}.

Then $F'$ satisfies the sheaf condition for all $\tau $-coverings.

**Proof.**
Let $X$ be a scheme over $S$ and let $\{ X_ i \to X\} _{i \in I}$ be a $\tau $-covering. Let $s_ i \in F'(X_ i)$ be elements such that $s_ i$ and $s_ j$ map to the same element of $F'(X_ i \times _ X X_ j)$ for all $i, j \in I$. We have to show that there is a unique element $s \in F'(X)$ restricting to $s_ i \in F'(X_ i)$ for all $i \in I$.

Special case: $X$ is an affine such that the structure morphism maps into an affine open $U$ of $S$ and the covering $\{ X_ i \to X\} _{i \in I}$ is a standard $\tau $-covering. In this case we can write

\[ X = \mathop{\mathrm{lim}}\nolimits V_ k \]

as a cofiltered limit with $V_ k \to U$ of finite presentation and $V_ k$ affine. See Algebra, Lemma 10.127.2. By Lemma 34.13.2 there exists a $k$ and a standard $\tau $-covering $\{ V_{k, i} \to V_ k\} _{i \in I}$ whose base change to $X$ is the given covering. For $k' \geq k$ denote $\{ V_{k', i} \to V_{k'}\} _{i \in I}$ the base change to $V_{k'}$ of our covering. Then we see that

\begin{align*} F'(X) & = \mathop{\mathrm{colim}}\nolimits _{k' \geq k} F(V_ k) \\ & = \mathop{\mathrm{colim}}\nolimits _{k' \geq k} \text{Equalizer}( \xymatrix{ \prod F(V_{k', i}) \ar@<1ex>[r] \ar@<-1ex>[r] & \prod F(V_{k', i} \times _{V_{k'}} V_{k', j}) } \\ & = \text{Equalizer}( \xymatrix{ \mathop{\mathrm{colim}}\nolimits _{k' \geq k} \prod F(V_{k', i}) \ar@<1ex>[r] \ar@<-1ex>[r] & \mathop{\mathrm{colim}}\nolimits _{k' \geq k} \prod F(V_{k', i} \times _{V_{k'}} V_{k', j}) } \\ & = \text{Equalizer}( \xymatrix{ \prod F'(X_ i) \ar@<1ex>[r] \ar@<-1ex>[r] & \prod F'(X_ i \times _ X X_ j) } \end{align*}

The first equality holds by construction of $F'$. The second holds by assumption (c). The third holds because filtered colimits are exact. The fourth again holds by construction of $F'$. In this way we find that the sheaf property holds for $F'$ with respect to $\{ X_ i \to X\} _{i \in I}$.

General case. Choose an affine open covering $X = \bigcup U_ k$ such that each $U_ k$ maps into an affine open of $S$. For every $k$ we may choose a standard $\tau $-covering $\{ V_{k, j} \to U_ k\} _{j = 1, \ldots , m_ k}$ which refines $\{ X_ i \times _ X U_ k \to U_ k\} _{i \in I}$. For each $j \in \{ 1, \ldots , m_ k\} $ choose an index $i_{k, j} \in I$ and a morphism $g_{k, j} : V_{k, j} \to X_{i_{k, j}}$ over $X$. Let $s_{k, j}$ be the element of $F'(V_{k, j})$ we get by restricting $s_{i_{k, j}}$ via $g_{k, j}$. Observe that $s_{k, j}$ and $s_{k', j'}$ restrict to the same element of $F'(V_{k, j} \times _ X V_{k', j'})$ for all $k$ and $k'$ and all $j \in \{ 1, \ldots , m_ k\} $ and $j' \in \{ 1, \ldots , m_{k'}\} $; verification omitted. In particular, by the result of the previous paragraph there is a unique element $s_ k \in F'(U_ k)$ restricting to $s_{k, j}$ for all $j$. With this notation we are ready to finish the proof.

Proof of uniqueness of $s$: this is true because $F'$ satisfies the sheaf property for Zariski coverings and $s|_{U_ k}$ must be equal to $s_ k$ because both restrict to $s_{k, j}$ for all $j$. This uniqueness then shows that $s_ k$ and $s_{k'}$ must restrict to the same section of $F'$ over (the non-affine scheme) $U_ k \cap U_{k'}$ because these sections restrict to the same section over the $\tau $-covering $\{ V_{k, j} \times _ X V_{k', j'} \to U_ k \cap U_{k'}\} $. Thus by the sheaf property for Zariski coverings, there is a unique section $s$ of $F'$ over $X$ whose restriction to $U_ k$ is $s_ k$. We omit the verification (similar to the above) that $s$ restricts to $s_ i$ over $X_ i$.
$\square$

## Comments (0)