Lemma 21.37.8. Assumptions and notation as in Situation 21.37.1. For $\mathcal{F}$ in $\textit{Ab}(\mathcal{C})$ the sheaf $\pi _!\mathcal{F}$ is the sheaf associated to the presheaf

$V \longmapsto \mathop{\mathrm{colim}}\nolimits _{\mathcal{C}_ V^{opp}} \mathcal{F}|_{\mathcal{C}_ V}$

with restriction maps as indicated in the proof.

Proof. Denote $\mathcal{H}$ be the rule of the lemma. For a morphism $h : V' \to V$ of $\mathcal{D}$ there is a pullback functor $h^* : \mathcal{C}_ V \to \mathcal{C}_{V'}$ of fibre categories (Categories, Definition 4.32.6). Moreover for $U \in \mathop{\mathrm{Ob}}\nolimits (\mathcal{C}_ V)$ there is a strongly cartesian morphism $h^*U \to U$ covering $h$. Restriction along these strongly cartesian morphisms defines a transformation of functors

$\mathcal{F}|_{\mathcal{C}_ V} \longrightarrow \mathcal{F}|_{\mathcal{C}_{V'}} \circ h^*.$

Hence a map $\mathcal{H}(V) \to \mathcal{H}(V')$ between colimits, see Categories, Lemma 4.14.7.

To prove the lemma we show that

$\mathop{Mor}\nolimits _{\textit{PSh}(\mathcal{D})}(\mathcal{H}, \mathcal{G}) = \mathop{Mor}\nolimits _{\mathop{\mathit{Sh}}\nolimits (\mathcal{C})}(\mathcal{F}, \pi ^{-1}\mathcal{G})$

for every sheaf $\mathcal{G}$ on $\mathcal{C}$. An element of the left hand side is a compatible system of maps $\mathcal{F}(U) \to \mathcal{G}(p(U))$ for all $U$ in $\mathcal{C}$. Since $\pi ^{-1}\mathcal{G}(U) = \mathcal{G}(p(U))$ by our choice of topology on $\mathcal{C}$ we see the same thing is true for the right hand side and we win. $\square$

## Comments (1)

Comment #837 by on

In the statement the word "to" is missing in "sheaf associated the presheaf".

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