**Proof.**
Note that by Lemma 8.12.10 we have an equivalence of categories

\[ \mathop{Mor}\nolimits _{\textit{Stacks}/\mathcal{D}}(f^{-1}\mathcal{S}, f^{-1}\mathcal{S}') = \mathop{Mor}\nolimits _{\textit{Stacks}/\mathcal{C}}(\mathcal{S}, f_*f^{-1}\mathcal{S}') \]

Hence (2) follows from (1).

To prove (1) we are going to use Lemma 8.4.8. This lemma tells us that we have to show that $can : \mathcal{S} \to f_*f^{-1}\mathcal{S}$ is fully faithful and that all objects of $f_*f^{-1}\mathcal{S}$ are locally in the essential image.

We quickly describe the functor $can$, see proof of Lemma 8.12.8. To do this we introduce the functor $c'' : \mathcal{S} \to u_{pp}\mathcal{S}$ defined by $c''(x/U) = (U, \text{id} : u(U) \to u(U), x)$, and $c''(\alpha /a) = (a, u(a), \alpha )$. We set $c' : \mathcal{S} \to u_ p\mathcal{S}$ equal to the composition of $c''$ and the canonical functor $u_{pp}\mathcal{S} \to u_ p\mathcal{S}$. We set $c : \mathcal{S} \to f^{-1}\mathcal{S}$ equal to the composition of $c'$ and the canonical functor $u_ p\mathcal{S} \to f^{-1}\mathcal{S}$. Then $can : \mathcal{S} \to f_*f^{-1}\mathcal{S}$ is the functor which to $x/U$ associates the pair $(U, c(x))$ and to $\alpha /a$ the morphism $(a, c(\alpha ))$.

Fully faithfulness. To prove this we are going to use Lemma 8.4.7. Let $U \in \mathop{\mathrm{Ob}}\nolimits (\mathcal{C})$. Let $x, y \in \mathcal{S}_ U$. First off, as $u$ is fully faithful, we have

\[ \mathop{Mor}\nolimits _{(f_*f^{-1}\mathcal{S})_ U}(can(x), can(y)) = \mathop{Mor}\nolimits _{(f^{-1}\mathcal{S})_{u(U)}}(c(x), c(y)) \]

directly from the definition of $f_*$. Similar holds after pulling back to any $U'/U$. Because $f^{-1}\mathcal{S}$ is the stackification of $u_ p\mathcal{S}$, and since $u$ is continuous and cocontinuous the presheaf

\[ U'/U \longmapsto \mathop{Mor}\nolimits _{(f^{-1}\mathcal{S})_{u(U')}}(c(x|_{U'}), c(y|_{U'})) \]

is the sheafification of the presheaf

\[ U'/U \longmapsto \mathop{Mor}\nolimits _{(u_ p\mathcal{S})_{u(U')}}(c'(x|_{U'}), c'(y|_{U'})) \]

Hence to finish the proof of fully faithfulness it suffices to show that for any $U$ and $x, y$ the map

\[ \mathop{Mor}\nolimits _{\mathcal{S}_ U}(x, y) \longrightarrow \mathop{Mor}\nolimits _{(u_ p\mathcal{S})_ U}(c'(x), c'(y)) \]

is bijective. A morphism $f : x \to y$ in $u_ p\mathcal{S}$ over $u(U)$ is given by an equivalence class of diagrams

\[ \xymatrix{ (U', \phi : u(U) \to u(U'), x') \ar[d]_{(c, \text{id}_{u(U)}, \gamma )} \ar[r]_{(a, b, \alpha )} & (U, \text{id} : u(U) \to u(U), y) \\ (U, \text{id} : u(U) \to u(U), x) } \]

with $\gamma $ strongly cartesian and $b = \text{id}_{u(U)}$. But since $u$ is fully faithful we can write $\phi = u(c')$ for some morphism $c' : U \to U'$ and then we see that $a \circ c' = \text{id}_ U$ and $c \circ c' = \text{id}_{U'}$. Because $\gamma $ is strongly cartesian we can find a morphism $\gamma ' : x \to x'$ lifting $c'$ such that $\gamma \circ \gamma ' = \text{id}_ x$. By definition of the equivalence classes defining morphisms in $u_ p\mathcal{S}$ it follows that the morphism

\[ \xymatrix{ (U, \text{id} : u(U) \to u(U), x) \ar[rr]_{(\text{id}, \text{id}, \alpha \circ \gamma ')} & & (U, \text{id} : u(U) \to u(U), y) } \]

of $u_{pp}\mathcal{S}$ induces the morphism $f$ in $u_ p\mathcal{S}$. This proves that the map is surjective. We omit the proof that it is injective.

Finally, we have to show that any object of $f_*f^{-1}\mathcal{S}$ locally comes from an object of $\mathcal{S}$. This is clear from the constructions (details omitted).
$\square$

## Comments (2)

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