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The Stacks project

Lemma 7.34.1. Let u : \mathcal{C} \to \mathcal{D} be a functor. Let v : \mathcal{D} \to \textit{Sets} be a functor and set w = v \circ u. Denote q, resp., p the stalk functor (7.32.1.1) associated to v, resp. w. Then (u_ p\mathcal{F})_ q = \mathcal{F}_ p functorially in the presheaf \mathcal{F} on \mathcal{C}.

Proof. This is a simple categorical fact. We have

\begin{align*} (u_ p\mathcal{F})_ q & = \mathop{\mathrm{colim}}\nolimits _{(V, y)} \mathop{\mathrm{colim}}\nolimits _{U, \phi : V \to u(U)} \mathcal{F}(U) \\ & = \mathop{\mathrm{colim}}\nolimits _{(V, y, U, \phi : V \to u(U))} \mathcal{F}(U) \\ & = \mathop{\mathrm{colim}}\nolimits _{(U, x)} \mathcal{F}(U) \\ & = \mathcal{F}_ p \end{align*}

The first equality holds by the definition of u_ p and the definition of the stalk functor. Observe that y \in v(V). In the second equality we simply combine colimits. To see the third equality we apply Categories, Lemma 4.17.5 to the functor F of diagram categories defined by the rule

F((V, y, U, \phi : V \to u(U))) = (U, v(\phi )(y)).

This makes sense because w(U) = v(u(U)). Let us check the hypotheses of Categories, Lemma 4.17.5. Observe that F has a right inverse, namely (U, x) \mapsto (u(U), x, U, \text{id} : u(U) \to u(U)). Again this makes sense because x \in w(U) = v(u(U)). On the other hand, there is always a morphism

(V, y, U, \phi : V \to u(U)) \longrightarrow (u(U), v(\phi )(y), U, \text{id} : u(U) \to u(U))

in the fibre category over (U, x) which shows the fibre categories are connected. The fourth and final equality is clear. \square


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