Lemma 7.25.9. Notation $\mathcal{C}$, $f : V \to U$, $j_ U$, $j_ V$, and $j$ as in Lemma 7.25.8. Via the identifications $\mathop{\mathit{Sh}}\nolimits (\mathcal{C}/V) = \mathop{\mathit{Sh}}\nolimits (\mathcal{C})/h_ V^\# $ and $\mathop{\mathit{Sh}}\nolimits (\mathcal{C}/U) = \mathop{\mathit{Sh}}\nolimits (\mathcal{C})/h_ U^\# $ of Lemma 7.25.4 we have
the functor $j^{-1}$ has the following description
\[ j^{-1}(\mathcal{H} \xrightarrow {\varphi } h_ U^\# ) = (\mathcal{H} \times _{\varphi , h_ U^\# , f} h_ V^\# \to h_ V^\# ). \]
the functor $j_!$ has the following description
\[ j_!(\mathcal{H} \xrightarrow {\varphi } h_ V^\# ) = (\mathcal{H} \xrightarrow {h_ f \circ \varphi } h_ U^\# ) \]
Proof.
Proof of (2). Recall that the identification $\mathop{\mathit{Sh}}\nolimits (\mathcal{C}/V) \to \mathop{\mathit{Sh}}\nolimits (\mathcal{C})/h_ V^\# $ sends $\mathcal{G}$ to $j_{V!}\mathcal{G} \to j_{V!}(*) = h_ V^\# $ and similarly for $\mathop{\mathit{Sh}}\nolimits (\mathcal{C}/U) \to \mathop{\mathit{Sh}}\nolimits (\mathcal{C})/h_ U^\# $. Thus $j_!\mathcal{G}$ is mapped to $j_{U!}(j_!\mathcal{G}) \to j_{U!}(*) = h_ U^\# $ and (2) follows because $j_{U!}j_! = j_{V!}$ by Lemma 7.25.8.
The reader can now prove (1) by using that $j^{-1}$ is the right adjoint to $j_!$ and using that the rule in (1) is the right adjoint to the rule in (2). Here is a direct proof. Suppose that $\varphi : \mathcal{H} \to h_ U^\# $ is an object of $\mathop{\mathit{Sh}}\nolimits (\mathcal{C})/h_ U^\# $. By the proof of Lemma 7.25.4 this corresponds to the sheaf $\mathcal{H}_\varphi $ on $\mathcal{C}/U$ defined by the rule
\[ (a : W \to U) \longmapsto \{ s \in \mathcal{H}(W) \mid \varphi (s) = a\} \]
on $\mathcal{C}/U$. The pullback $j^{-1}\mathcal{H}_\varphi $ to $\mathcal{C}/V$ is given by the rule
\[ (a : W \to V) \longmapsto \{ s \in \mathcal{H}(W) \mid \varphi (s) = f \circ a\} \]
by the description of $j^{-1} = j_{U/V}^{-1}$ as the restriction of $\mathcal{H}_\varphi $ to $\mathcal{C}/V$. On the other hand, applying the rule to the object
\[ \xymatrix{ \mathcal{H}' = \mathcal{H} \times _{\varphi , h_ U^\# , f} h_ V^\# \ar[rr]^-{\varphi '} & & h_ V^\# } \]
of $\mathop{\mathit{Sh}}\nolimits (\mathcal{C})/h_ V^\# $ we get $\mathcal{H}'_{\varphi '}$ given by
\begin{align*} (a : W \to V) \longmapsto & \{ s' \in \mathcal{H}'(W) \mid \varphi '(s') = a\} \\ = & \{ (s, a') \in \mathcal{H}(W) \times h_ V^\# (W) \mid a' = a \text{ and } \varphi (s) = f \circ a'\} \end{align*}
which is exactly the same rule as the one describing $j^{-1}\mathcal{H}_\varphi $ above.
$\square$
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