The Stacks project

Lemma 7.30.2. In the situation of Lemma 7.30.1, the functor $j_{\mathcal{F}, *}$ is the one associates to $\varphi : \mathcal{G} \to \mathcal{F}$ the sheaf

\[ U \longmapsto \{ \alpha : \mathcal{F}|_ U \to \mathcal{G}|_ U \text{ such that } \alpha \text{ is a right inverse to }\varphi |_ U \} . \]

Proof. For any $\varphi : \mathcal{G} \to \mathcal{F}$, let us use the notation $\mathcal{G}/\mathcal{F}$ to denote the corresponding object of $\mathop{\mathit{Sh}}\nolimits (\mathcal{C})/\mathcal{F}$. We have

\[ (j_{\mathcal{F}, *}(\mathcal{G}/\mathcal{F}))(U) = \mathop{\mathrm{Mor}}\nolimits _{\mathop{\mathit{Sh}}\nolimits (\mathcal{C})}(h_ U^\# , j_{\mathcal{F}, *}(\mathcal{G}/\mathcal{F})) = \mathop{\mathrm{Mor}}\nolimits _{\mathop{\mathit{Sh}}\nolimits (\mathcal{C})/\mathcal{F}}(j_{\mathcal{F}}^{-1}h_ U^{\# }, (\mathcal{G}/\mathcal{F})). \]

By Lemma 7.30.1 this set is the fiber over the element $h_ U^\# \times \mathcal{F} \to \mathcal{F}$ under the map of sets

\[ \mathop{\mathrm{Mor}}\nolimits _{\mathop{\mathit{Sh}}\nolimits (\mathcal{C})}(h_ U^\# \times \mathcal{F}, \mathcal{G}) \xrightarrow {\varphi \circ } \mathop{\mathrm{Mor}}\nolimits _{\mathop{\mathit{Sh}}\nolimits (\mathcal{C})}(h_ U^\# \times \mathcal{F}, \mathcal{F}). \]

By the adjunction in Lemma 7.26.2, we have

\begin{align*} \mathop{\mathrm{Mor}}\nolimits _{\mathop{\mathit{Sh}}\nolimits (\mathcal{C})}(h_ U^{\# }\times \mathcal{F}, \mathcal{G}) & = \mathop{\mathrm{Mor}}\nolimits _{\mathop{\mathit{Sh}}\nolimits (\mathcal{C})}(h_ U^{\# },\mathop{\mathcal{H}\! \mathit{om}}\nolimits (\mathcal{F}, \mathcal{G})) \\ & = \mathop{\mathrm{Mor}}\nolimits _{\mathop{\mathit{Sh}}\nolimits (\mathcal{C}/U)}(\mathcal{F}|_{\mathcal{C}/U}, \mathcal{G}|_{\mathcal{C}/U}), \\ \mathop{\mathrm{Mor}}\nolimits _{\mathop{\mathit{Sh}}\nolimits (\mathcal{C})}(h_ U^{\# } \times \mathcal{F}, \mathcal{F}) & = \mathop{\mathrm{Mor}}\nolimits _{\mathop{\mathit{Sh}}\nolimits (\mathcal{C})}(h_ U^{\# },\mathop{\mathcal{H}\! \mathit{om}}\nolimits (\mathcal{F},\mathcal{F})) \\ & = \mathop{\mathrm{Mor}}\nolimits _{\mathop{\mathit{Sh}}\nolimits (\mathcal{C}/U)}(\mathcal{F}|_{\mathcal{C}/U}, \mathcal{F}|_{\mathcal{C}/U}), \end{align*}

and under the adjunction, the map $\varphi \circ $ becomes the map

\[ \mathop{\mathrm{Mor}}\nolimits _{\mathop{\mathit{Sh}}\nolimits (\mathcal{C}/U)}(\mathcal{F}|_{\mathcal{C}/U}, \mathcal{G}|_{\mathcal{C}/U}) \longrightarrow \mathop{\mathrm{Mor}}\nolimits _{\mathop{\mathit{Sh}}\nolimits (\mathcal{C}/U)}(\mathcal{F}|_{\mathcal{C}/U}, \mathcal{F}|_{\mathcal{C}/U}),\quad \psi \longmapsto \varphi |_{\mathcal{C}/U} \circ \psi , \]

the element $h_ U^\# \times \mathcal{F} \to \mathcal{F}$ becomes $\text{id}_{\mathcal{F}|_{\mathcal{C}/U}}$. Therefore $(j_{\mathcal{F}, *}\mathcal{G}/\mathcal{F})(U)$ is isomorphic to the fiber of $\text{id}_{\mathcal{F}|_{\mathcal{C}/U}}$ under the map

\[ \mathop{\mathrm{Mor}}\nolimits _{\mathop{\mathit{Sh}}\nolimits (\mathcal{C}/U)}(\mathcal{F}|_{\mathcal{C}/U}, \mathcal{G}|_{\mathcal{C}/U}) \xrightarrow {\varphi |_{\mathcal{C}/U}\circ } \mathop{\mathrm{Mor}}\nolimits _{\mathop{\mathit{Sh}}\nolimits (\mathcal{C}/U)}(\mathcal{F}|_{\mathcal{C}/U}, \mathcal{F}|_{\mathcal{C}/U}), \]

which is $\{ \alpha : \mathcal{F}|_ U \to \mathcal{G}|_ U \text{ such that } \alpha \text{ is a right inverse to }\varphi |_ U \} $ as desired. $\square$

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