The Stacks project

Proof. By Categories, Lemma 4.35.15 we see that a fibre product in $\mathcal{X}$ is any commutative square lying over a fibre product diagram in $(\mathit{Sch}/S)_{fppf}$. Similarly for $\mathcal{Y}$. Hence (1) is clear.

Let $x \to x'$ be the equalizer of two morphisms $a, b : x' \to x''$ in $\mathcal{X}$. We will show that $f(x) \to f(x')$ is the equalizer of $f(a)$ and $f(b)$. Let $y \to f(x)$ be a morphism of $\mathcal{Y}$ equalizing $f(a)$ and $f(b)$. Say $x, x', x''$ lie over the schemes $U, U', U''$ and $y$ lies over $V$. Denote $h : V \to U'$ the image of $y \to f(x)$ in the category of schemes. The morphism $y \to f(x)$ is isomorphic to $f(h^*x') \to f(x')$ by the axioms of fibred categories. Hence, as $f$ is faithful, we see that $h^*x' \to x'$ equalizes $a$ and $b$. Thus we obtain a unique morphism $h^*x' \to x$ whose image $y = f(h^*x') \to f(x)$ is the desired morphism in $\mathcal{Y}$. $\square$