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. $\square$

In your comment you can use Markdown and LaTeX style mathematics (enclose it like $\pi$). A preview option is available if you wish to see how it works out (just click on the eye in the toolbar).