Lemma 7.47.11. Let $\mathcal{C}$ be a category. Let $\{ \mathcal{F}_ i \} _{i\in I}$ be a collection of presheaves of sets on $\mathcal{C}$. For each $U \in \mathop{\mathrm{Ob}}\nolimits (\mathcal{C})$ denote $J(U)$ the set of sieves $S$ with the following property: For every morphism $V \to U$, the maps

$\mathop{\mathrm{Mor}}\nolimits _{\textit{PSh}(\mathcal{C})}(h_ V, \mathcal{F}_ i) \longrightarrow \mathop{\mathrm{Mor}}\nolimits _{\textit{PSh}(\mathcal{C})}(S \times _ U V, \mathcal{F}_ i)$

are bijective for all $i \in I$. Then $J$ defines a topology on $\mathcal{C}$. This topology is the finest topology in which all of the $\mathcal{F}_ i$ are sheaves.

Proof. If we show that $J$ is a topology, then the last statement of the lemma immediately follows. The first and third axioms of a topology are immediately verified. Thus, assume that we have an object $U$, and sieves $S, S'$ of $U$ such that $S \in J(U)$, and for all $V \to U$ in $S(V)$ we have $S' \times _ U V \in J(V)$. We have to show that $S' \in J(U)$. In other words, we have to show that for any $f : W \to U$, the maps

$\mathcal{F}_ i(W) = \mathop{\mathrm{Mor}}\nolimits _{\textit{PSh}(\mathcal{C})}(h_ W, \mathcal{F}_ i) \longrightarrow \mathop{\mathrm{Mor}}\nolimits _{\textit{PSh}(\mathcal{C})}(S' \times _ U W, \mathcal{F}_ i)$

are bijective for all $i \in I$. Pick an element $i \in I$ and pick an element $\varphi \in \mathop{\mathrm{Mor}}\nolimits _{\textit{PSh}(\mathcal{C})}(S' \times _ U W, \mathcal{F}_ i)$. We will construct a section $s \in \mathcal{F}_ i(W)$ mapping to $\varphi$.

Suppose $\alpha : V \to W$ is an element of $S \times _ U W$. According to the definition of pullbacks we see that the composition $f \circ \alpha : V \to W \to U$ is in $S$. Hence $S' \times _ U V$ is in $J(W)$ by assumption on the pair of sieves $S, S'$. Now we have a commutative diagram of presheaves

$\xymatrix{ S' \times _ U V \ar[r] \ar[d] & h_ V \ar[d] \\ S' \times _ U W \ar[r] & h_ W }$

The restriction of $\varphi$ to $S' \times _ U V$ corresponds to an element $s_{V, \alpha } \in \mathcal{F}_ i(V)$. This we see from the definition of $J$, and because $S' \times _ U V$ is in $J(W)$. We leave it to the reader to check that the rule $(V, \alpha ) \mapsto s_{V, \alpha }$ defines an element $\psi \in \mathop{\mathrm{Mor}}\nolimits _{\textit{PSh}(\mathcal{C})}(S \times _ U W, \mathcal{F}_ i)$. Since $S \in J(U)$ we see immediately from the definition of $J$ that $\psi$ corresponds to an element $s$ of $\mathcal{F}_ i(W)$.

We leave it to the reader to verify that the construction $\varphi \mapsto s$ is inverse to the natural map displayed above. $\square$

Comment #3436 by Remy on

"The first and second axioms of a topology are immediately verified." This should be "first and third".

Comment #4290 by Kazuki Masugi on

I can't make sure that $\varphi \mapsto s$ is right inverse to the map in the statement. Would you please describe the precise way to verify?

Comment #4291 by Laurent Moret-Bailly on

I think one can simplify the exposition by reducing (via 00Z7(1)) to the case of just one presheaf.

Comment #4455 by on

Unless somebody tells me that this is wrong I am going to leave it as is. This lemma and all of the material on sieves is never used later in the Stacks project. I would be happy to accept a careful write up of this proof if somebody has the time and interest in doing so. Of course the suggestion of Laurent is a good one: there is an immediate reduction to the case of a single presheaf which should simplify the proof (and the notation used in the proof).

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