Lemma 7.8.6. Let $\mathcal{C}$ be a category. Let $\mathcal{V} = \{ V_ j \to U\} _{j \in J} \to \mathcal{U} = \{ U_ i \to U\} _{i \in I}$ be a morphism of families of maps with fixed target of $\mathcal{C}$ given by $\text{id} : U \to U$, $\alpha : J \to I$ and $f_ j : V_ j \to U_{\alpha (j)}$. Let $\mathcal{F}$ be a presheaf on $\mathcal{C}$. If

1. the fibre products $U_ i \times _ U U_{i'}$, $U_ i \times _ U V_ j$, $V_ j \times _ U V_{j'}$ exist,

2. $\mathcal{F}$ satisfies the sheaf condition with respect to $\mathcal{V}$, and

3. for every $i \in I$ the map $\mathcal{F}(U_ i) \to \prod _{j \in J} \mathcal{F}(V_ j \times _ U U_ i)$ is injective.

Then $\mathcal{F}$ satisfies the sheaf condition with respect to $\mathcal{U}$.

Proof. By Lemma 7.8.5 the map $\mathcal{F}(U) \to \prod \mathcal{F}(U_ i)$ is injective. Suppose given $s_ i \in \mathcal{F}(U_ i)$ such that $s_ i|_{U_ i \times _ U U_{i'}} = s_{i'}|_{U_ i \times _ U U_{i'}}$ for all $i, i' \in I$. Set $s_ j = f_ j^*(s_{\alpha (j)}) \in \mathcal{F}(V_ j)$. Since the morphisms $f_ j$ are morphisms over $U$ we obtain induced morphisms $f_{jj'} : V_ j \times _ U V_{j'} \to U_{\alpha (i)} \times _ U U_{\alpha (i')}$ compatible with the $f_ j, f_{j'}$ via the projection maps. It follows that

$s_ j|_{V_ j \times _ U V_{j'}} = f_{jj'}^*(s_{\alpha (j)}|_{U_{\alpha (j)} \times _ U U_{\alpha (j')}}) = f_{jj'}^*(s_{\alpha (j')}|_{U_{\alpha (j)} \times _ U U_{\alpha (j')}}) = s_{j'}|_{V_ j \times _ U V_{j'}}$

for all $j, j' \in J$. Hence, by the sheaf condition for $\mathcal{F}$ with respect to $\mathcal{V}$, we get a section $s \in \mathcal{F}(U)$ which restricts to $s_ j$ on each $V_ j$. We are done if we show $s$ restricts to $s_ i$ on $U_ i$ for any $i \in I$. Since $\mathcal{F}$ satisfies (3) it suffices to show that $s$ and $s_ i$ restrict to the same element over $U_ i \times _ U V_ j$ for all $j \in J$. To see this we use

$s|_{U_ i \times _ U V_ j} = s_ j|_{U_ i \times _ U V_ j} = (\text{id} \times f_ j)^*s_{\alpha (j)}|_{U_ i \times _ U U_{\alpha (j)}} = (\text{id} \times f_ j)^*s_ i|_{U_ i \times _ U U_{\alpha (j)}} = s_ i|_{U_ i \times _ U V_ j}$

as desired. $\square$

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