The Stacks project

Proof. We may assume that $\mathcal{T}_2$ is the category $\mathcal{S}_2 \times _{\mathcal{S}_1} \mathcal{T}_1$ described in Categories, Lemma 4.32.3. By Categories, Lemma 4.35.9 the faithfulness of $G, G'$ can be checked on fibre categories. Suppose that $y, y'$ are objects of $\mathcal{T}_1$ over the object $U$ of $\mathcal{C}$. Let $\alpha , \beta : y \to y'$ be morphisms of $(\mathcal{T}_1)_ U$ such that $G(\alpha ) = G(\beta )$. Our object is to show that $\alpha = \beta $. Considering instead $\gamma = \alpha ^{-1} \circ \beta $ we see that $G(\gamma ) = \text{id}_{G(y)}$ and we have to show that $\gamma = \text{id}_ y$. By assumption we can find a covering $\{ U_ i \to U\} $ such that $G(y)|_{U_ i}$ is in the essential image of $F :(\mathcal{S}_2)_{U_ i} \to (\mathcal{S}_1)_{U_ i}$. Since it suffices to show that $\gamma |_{U_ i} = \text{id}$ for each $i$, we may therefore assume that we have $f : F(x) \to G(y)$ for some object $x$ of $\mathcal{S}_2$ over $U$ and morphisms $f$ of $(\mathcal{S}_1)_ U$. In this case we get a morphism

in the fibre category of $\mathcal{S}_2 \times _{\mathcal{S}_1} \mathcal{T}_1$ over $U$ whose image under $G'$ in $\mathcal{S}_1$ is $\text{id}_ x$. As $G'$ is faithful we conclude that $\gamma = \text{id}_ y$ and we win. $\square$