The Stacks project

Proof. By the uniqueness property of a $2$-fibre product may assume that $\mathcal{X}' = \mathcal{Y}' \times _\mathcal {Y} \mathcal{X}$ as in Categories, Lemma 4.32.3. Let us prove properties (2)(a) and (2)(b) of Lemma 8.11.3 for $\mathcal{Y}' \times _\mathcal {Y} \mathcal{X} \to \mathcal{Y}'$.

Let $y'$ be an object of $\mathcal{Y}'$ lying over the object $U$ of $\mathcal{C}$. By assumption there exists a covering $\{ U_ i \to U\} $ of $U$ and objects $x_ i \in \mathcal{X}_{U_ i}$ with isomorphisms $\alpha _ i : G(y')|_{U_ i} \to F(x_ i)$. Then $(U_ i, y'|_{U_ i}, x_ i, \alpha _ i)$ is an object of $\mathcal{Y}' \times _\mathcal {Y} \mathcal{X}$ over $U_ i$ whose image in $\mathcal{Y}'$ is $y'|_{U_ i}$. Thus (2)(a) holds.

Let $U \in \mathop{\mathrm{Ob}}\nolimits (\mathcal{C})$, let $x'_1, x'_2$ be objects of $\mathcal{Y}' \times _\mathcal {Y} \mathcal{X}$ over $U$, and let $b' : F'(x'_1) \to F'(x'_2)$ be a morphism in $\mathcal{Y}'_ U$. Write $x'_ i = (U, y'_ i, x_ i, \alpha _ i)$. Note that $F'(x'_ i) = x_ i$ and $G'(x'_ i) = y'_ i$. By assumption there exists a covering $\{ U_ i \to U\} $ in $\mathcal{C}$ and morphisms $a_ i : x_1|_{U_ i} \to x_2|_{U_ i}$ in $\mathcal{X}_{U_ i}$ with $F(a_ i) = G(b')|_{U_ i}$. Then $(b'|_{U_ i}, a_ i)$ is a morphism $x'_1|_{U_ i} \to x'_2|_{U_ i}$ as required in (2)(b). $\square$