Lemma 8.4.8. Let $\mathcal{C}$ be a site. Let $\mathcal{S}_1$, $\mathcal{S}_2$ be stacks over $\mathcal{C}$. Let $F : \mathcal{S}_1 \to \mathcal{S}_2$ be a $1$-morphism which is fully faithful. Then the following are equivalent

1. $F$ is an equivalence,

2. for every $U \in \mathop{\mathrm{Ob}}\nolimits (\mathcal{C})$ and for every $x \in \mathop{\mathrm{Ob}}\nolimits (\mathcal{S}_{2, U})$ there exists a covering $\{ f_ i : U_ i \to U\}$ such that $f_ i^*x$ is in the essential image of the functor $F : \mathcal{S}_{1, U_ i} \to \mathcal{S}_{2, U_ i}$.

Proof. The implication (1) $\Rightarrow$ (2) is immediate. To see that (2) implies (1) we have to show that every $x$ as in (2) is in the essential image of the functor $F$. To do this choose a covering as in (2), $x_ i \in \mathop{\mathrm{Ob}}\nolimits (\mathcal{S}_{1, U_ i})$, and isomorphisms $\varphi _ i : F(x_ i) \to f_ i^*x$. Then we get a descent datum for $\mathcal{S}_1$ relative to $\{ f_ i : U_ i \to U\}$ by taking

$\varphi _{ij} : x_ i|_{U_ i \times _ U U_ j} \longrightarrow x_ j|_{U_ i \times _ U U_ j}$

the arrow such that $F(\varphi _{ij}) = \varphi _ j^{-1} \circ \varphi _ i$. This descent datum is effective by the axioms of a stack, and hence we obtain an object $x_1$ of $\mathcal{S}_1$ over $U$. We omit the verification that $F(x_1)$ is isomorphic to $x$ over $U$. $\square$

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).