Lemma 4.35.10. Let $\mathcal{C}$ be a category. Let $p : \mathcal{S}\to \mathcal{C}$ and $p' : \mathcal{S'}\to \mathcal{C}$ be categories fibred in groupoids. Let $G : \mathcal{S}\to \mathcal{S}'$ be a functor over $\mathcal{C}$. Then $G$ is fully faithful if and only if the diagonal

\[ \Delta _ G : \mathcal{S} \longrightarrow \mathcal{S} \times _{G, \mathcal{S}', G} \mathcal{S} \]

is an equivalence.

**Proof.**
By Lemma 4.35.9 it suffices to look at fibre categories over an object $U$ of $\mathcal{C}$. An object of the right hand side is a triple $(x, x', \alpha )$ where $\alpha : G(x) \to G(x')$ is a morphism in $\mathcal{S}'_ U$. The functor $\Delta _ G$ maps the object $x$ of $\mathcal{S}_ U$ to the triple $(x, x, \text{id}_{G(x)})$. Note that $(x, x', \alpha )$ is in the essential image of $\Delta _ G$ if and only if $\alpha = G(\beta )$ for some morphism $\beta : x \to x'$ in $\mathcal{S}_ U$ (details omitted). Hence in order for $\Delta _ G$ to be an equivalence, every $\alpha $ has to be the image of a morphism $\beta : x \to x'$, and also every two distinct morphisms $\beta , \beta ' : x \to x'$ have to give distinct morphisms $G(\beta ), G(\beta ')$. This proves the lemma.
$\square$

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