**Proof.**
Assume (1). For $U, x, y$ as in (2) the displayed map $F$ evaluates to the map $F : \mathop{Mor}\nolimits _{\mathcal{S}_{1, V}}(x|_ V, y|_ V) \to \mathop{Mor}\nolimits _{\mathcal{S}_{2, V}}(F(x|_ V), F(y|_ V))$ on an object $V$ of $\mathcal{C}$ lying over $U$. Now, since $F$ is fully faithful, the corresponding map $\mathop{Mor}\nolimits _{\mathcal{S}_1}(x|_ V, y|_ V) \to \mathop{Mor}\nolimits _{\mathcal{S}_2}(F(x|_ V), F(y|_ V))$ is a bijection. Morphisms in the fibre category $\mathcal{S}_{1, V}$ are exactly those morphisms between $x|_ V$ and $y|_ V$ in $\mathcal{S}_1$ lying over $\text{id}_ V$. Similarly, morphisms in the fibre category $\mathcal{S}_{2, V}$ are exactly those morphisms between $F(x|_ V)$ and $F(y|_ V)$ in $\mathcal{S}_2$ lying over $\text{id}_ V$. Thus we find that $F$ induces a bijection between these also. Hence (2) holds.

Assume (2). Suppose given objects $U$, $V$ of $\mathcal{C}$ and $x \in \mathop{\mathrm{Ob}}\nolimits (\mathcal{S}_{1, U})$ and $y \in \mathop{\mathrm{Ob}}\nolimits (\mathcal{S}_{1, V})$. To show that $F$ is fully faithful, it suffices to prove it induces a bijection on morphisms lying over a fixed $f : U \to V$. Choose a strongly Cartesian $f^*y \to y$ in $\mathcal{S}_1$ lying above $f$. This results in a bijection between the set of morphisms $x \to y$ in $\mathcal{S}_1$ lying over $f$ and $\mathop{Mor}\nolimits _{\mathcal{S}_{1, U}}(x, f^*y)$. Since $F$ preserves strongly Cartesian morphisms as a $1$-morphism in the $2$-category of stacks over $\mathcal{C}$, we also get a bijection between the set of morphisms $F(x) \to F(y)$ in $\mathcal{S}_2$ lying over $f$ and $\mathop{Mor}\nolimits _{\mathcal{S}_{2, U}}(F(x), F(f^*y))$. Since $F$ induces a bijection $\mathop{Mor}\nolimits _{\mathcal{S}_{1, U}}(x, f^*y) \to \mathop{Mor}\nolimits _{\mathcal{S}_{2, U}}(F(x), F(f^*y))$ we conclude (1) holds.
$\square$

## Comments (4)

Comment #471 by Kestutis Cesnavicius on

Comment #472 by Kestutis Cesnavicius on

Comment #473 by Pieter Belmans on

Comment #486 by Johan on