Lemma 7.8.4. Let $\mathcal{C}$ be a category. Let $\mathcal{U} = \{ \varphi _ i : U_ i \to U\} _{i\in I}$, and $\mathcal{V} = \{ \psi _ j : V_ j \to U\} _{j\in J}$ be two families of morphisms with the same fixed target. Assume that the fibre products $U_ i \times _ U U_{i'}$ and $V_ j \times _ U V_{j'}$ exist. If $\mathcal{U}$ and $\mathcal{V}$ are tautologically equivalent, then for any presheaf $\mathcal{F}$ on $\mathcal{C}$ the sheaf condition for $\mathcal{F}$ with respect to $\mathcal{U}$ is equivalent to the sheaf condition for $\mathcal{F}$ with respect to $\mathcal{V}$.

**Proof.**
First, note that if $\varphi : A \to B$ is an isomorphism in the category $\mathcal{C}$, then $\varphi ^* : \mathcal{F}(B) \to \mathcal{F}(A)$ is an isomorphism. Let $\beta : J \to I$ be a map and let $\psi _ j : V_ j \to U_{\beta (j)}$ be isomorphisms over $U$ which are assumed to exist by hypothesis. Let us show that the sheaf condition for $\mathcal{V}$ implies the sheaf condition for $\mathcal{U}$. Suppose given sections $s_ i \in \mathcal{F}(U_ i)$ such that

in $\mathcal{F}(U_ i \times _ U U_{i'})$ for all pairs $(i, i') \in I \times I$. Then we can define $s_ j = \psi _ j^*s_{\beta (j)}$. For any pair $(j, j') \in J \times J$ the morphism $\psi _ j \times _{\text{id}_ U} \psi _{j'} : V_ j \times _ U V_{j'} \to U_{\beta (j)} \times _ U U_{\beta (j')}$ is an isomorphism as well. Hence by transport of structure we see that

as well. The sheaf condition w.r.t. $\mathcal{V}$ implies there exists a unique $s$ such that $s|_{V_ j} = s_ j$ for all $j \in J$. By the first remark of the proof this implies that $s|_{U_ i} = s_ i$ for all $i \in \mathop{\mathrm{Im}}(\beta )$ as well. Suppose that $i \in I$, $i \not\in \mathop{\mathrm{Im}}(\beta )$. For such an $i$ we have isomorphisms $U_ i \to V_{\alpha (i)} \to U_{\beta (\alpha (i))}$ over $U$. This gives a morphism $U_ i \to U_ i \times _ U U_{\beta (\alpha (i))}$ which is a section of the projection. Because $s_ i$ and $s_{\beta (\alpha (i))}$ restrict to the same element on the fibre product we conclude that $s_{\beta (\alpha (i))}$ pulls back to $s_ i$ via $U_ i \to U_{\beta (\alpha (i))}$. Thus we see that also $s_ i = s|_{U_ i}$ as desired. $\square$

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