Lemma 7.18.5. In Situation 7.18.1 assume we have a sheaf $\mathcal{F}$ on $\mathcal{C}$. Then

$\mathcal{F} = \mathop{\mathrm{colim}}\nolimits f_ i^{-1}f_{i, *}\mathcal{F}$

where the transition maps are $f_ j^{-1}\varphi _ a$ for $a : j \to i$ where $\varphi _ a : f_ a^{-1}f_{i, *}\mathcal{F} \to f_{j, *}\mathcal{F}$ is a canonical map satisfying a cocycle condition as in Lemma 7.18.4.

Proof. For the morphism

$\varphi _ a : f_ a^{-1}f_{i, *}\mathcal{F} \to f_{j, *}\mathcal{F}$

we choose the adjoint to the identity map

$f_{i, *}\mathcal{F} \to f_{a, *}f_{j, *}\mathcal{F}$

Hence $\varphi _ a$ is the counit for the adjunction given by $(f_ a^{-1}, f_{a, *})$. We must prove that for all $a : j \to i$ and $b : k \to i$ with composition $c = a \circ b$ we have $\varphi _ c = \varphi _ b \circ f_ b^{-1}\varphi _ a$. This follows from Categories, Lemma 4.24.9. Lastly, we must prove that the map given by adjunction

$\mathop{\mathrm{colim}}\nolimits _{i \in I} f_ i^{-1}f_{i, *}\mathcal{F} \longrightarrow \mathcal{F}$

is an isomorphism. For an object $U$ of $\mathcal{C}$ we need to show the map

$(\mathop{\mathrm{colim}}\nolimits _{i \in I} f_ i^{-1}\mathcal{F}_ i)(U) \to \mathcal{F}(U)$

is bijective. Choose an $i$ and an object $U_ i$ of $\mathcal{C}_ i$ with $u_ i(U_ i) = U$. Then the left hand side is equal to

$(\mathop{\mathrm{colim}}\nolimits _{i \in I} f_ i^{-1}\mathcal{F}_ i)(U) = \mathop{\mathrm{colim}}\nolimits _{a : j \to i} f_{j, *}\mathcal{F}(u_ a(U_ i))$

by Lemma 7.18.4. Since $u_ j(u_ a(U_ i)) = U$ we have $f_{j, *}\mathcal{F}(u_ a(U_ i)) = \mathcal{F}(U)$ for all $a : j \to i$ by definition. Hence the value of the colimit is $\mathcal{F}(U)$ and the proof is complete. $\square$

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