Lemma 4.35.8. Let $p : \mathcal{S}\to \mathcal{C}$ and $p' : \mathcal{S'}\to \mathcal{C}$ be categories fibred in groupoids, and suppose that $G : \mathcal{S}\to \mathcal{S}'$ is a functor over $\mathcal{C}$.

1. Then $G$ is faithful (resp. fully faithful, resp. an equivalence) if and only if for each $U\in \mathop{\mathrm{Ob}}\nolimits (\mathcal{C})$ the induced functor $G_ U : \mathcal{S}_ U\to \mathcal{S}'_ U$ is faithful (resp. fully faithful, resp. an equivalence).

2. If $G$ is an equivalence, then $G$ is an equivalence in the $2$-category of categories fibred in groupoids over $\mathcal{C}$.

Proof. Let $x, y$ be objects of $\mathcal{S}$ lying over the same object $U$. Consider the commutative diagram

$\xymatrix{ \mathop{\mathrm{Mor}}\nolimits _\mathcal {S}(x, y) \ar[rd]_ p \ar[rr]_ G & & \mathop{\mathrm{Mor}}\nolimits _{\mathcal{S}'}(G(x), G(y)) \ar[ld]^{p'} \\ & \mathop{\mathrm{Mor}}\nolimits _\mathcal {C}(U, U) & }$

From this diagram it is clear that if $G$ is faithful (resp. fully faithful) then so is each $G_ U$.

Suppose $G$ is an equivalence. For every object $x'$ of $\mathcal{S}'$ there exists an object $x$ of $\mathcal{S}$ such that $G(x)$ is isomorphic to $x'$. Suppose that $x'$ lies over $U'$ and $x$ lies over $U$. Then there is an isomorphism $f : U' \to U$ in $\mathcal{C}$, namely, $p'$ applied to the isomorphism $x' \to G(x)$. By the axioms of a category fibred in groupoids there exists an arrow $f^*x \to x$ of $\mathcal{S}$ lying over $f$. Hence there exists an isomorphism $\alpha : x' \to G(f^*x)$ such that $p'(\alpha ) = \text{id}_{U'}$ (this time by the axioms for $\mathcal{S}'$). All in all we conclude that for every object $x'$ of $\mathcal{S}'$ we can choose a pair $(o_{x'}, \alpha _{x'})$ consisting of an object $o_{x'}$ of $\mathcal{S}$ and an isomorphism $\alpha _{x'} : x' \to G(o_{x'})$ with $p'(\alpha _{x'}) = \text{id}_{p'(x')}$. From this point on we proceed as usual (see proof of Lemma 4.2.19) to produce an inverse functor $F : \mathcal{S}' \to \mathcal{S}$, by taking $x' \mapsto o_{x'}$ and $\varphi ' : x' \to y'$ to the unique arrow $\varphi _{\varphi '} : o_{x'} \to o_{y'}$ with $\alpha _{y'}^{-1} \circ G(\varphi _{\varphi '}) \circ \alpha _{x'} = \varphi '$. With these choices $F$ is a functor over $\mathcal{C}$. We omit the verification that $G \circ F$ and $F \circ G$ are $2$-isomorphic to the respective identity functors (in the $2$-category of categories fibred in groupoids over $\mathcal{C}$).

Suppose that $G_ U$ is faithful (resp. fully faithful) for all $U\in \mathop{\mathrm{Ob}}\nolimits (\mathcal C)$. To show that $G$ is faithful (resp. fully faithful) we have to show for any objects $x, y\in \mathop{\mathrm{Ob}}\nolimits (\mathcal{S})$ that $G$ induces an injection (resp. bijection) between $\mathop{\mathrm{Mor}}\nolimits _\mathcal {S}(x, y)$ and $\mathop{\mathrm{Mor}}\nolimits _{\mathcal{S}'}(G(x), G(y))$. Set $U = p(x)$ and $V = p(y)$. It suffices to prove that $G$ induces an injection (resp. bijection) between morphism $x \to y$ lying over $f$ to morphisms $G(x) \to G(y)$ lying over $f$ for any morphism $f : U \to V$. Now fix $f : U \to V$. Denote $f^*y \to y$ a pullback. Then also $G(f^*y) \to G(y)$ is a pullback. The set of morphisms from $x$ to $y$ lying over $f$ is bijective to the set of morphisms between $x$ and $f^*y$ lying over $\text{id}_ U$. (By the second axiom of a category fibred in groupoids.) Similarly the set of morphisms from $G(x)$ to $G(y)$ lying over $f$ is bijective to the set of morphisms between $G(x)$ and $G(f^*y)$ lying over $\text{id}_ U$. Hence the fact that $G_ U$ is faithful (resp. fully faithful) gives the desired result.

Finally suppose for all $G_ U$ is an equivalence for all $U$, so it is fully faithful and essentially surjective. We have seen this implies $G$ is fully faithful, and thus to prove it is an equivalence we have to prove that it is essentially surjective. This is clear, for if $z'\in \mathop{\mathrm{Ob}}\nolimits (\mathcal{S}')$ then $z'\in \mathop{\mathrm{Ob}}\nolimits (\mathcal{S}'_ U)$ where $U = p'(z')$. Since $G_ U$ is essentially surjective we know that $z'$ is isomorphic, in $\mathcal{S}'_ U$, to an object of the form $G_ U(z)$ for some $z\in \mathop{\mathrm{Ob}}\nolimits (\mathcal{S}_ U)$. But morphisms in $\mathcal{S}'_ U$ are morphisms in $\mathcal{S}'$ and hence $z'$ is isomorphic to $G(z)$ in $\mathcal{S}'$. $\square$

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