The Stacks project

Proof. We think of objects in $\mathcal{X} \times _\mathcal {Y} \mathcal{Z}$ as quadruples $(U, x, z, \alpha )$ where $\alpha : f(x) \to g(z)$ is an isomorphism over $U$, see Categories, Lemma 4.32.3. A morphism $(U, x, z, \alpha ) \to (U', x', z', \alpha ')$ is a pair of morphisms $a : x \to x'$ and $b : z \to z'$ compatible with $\alpha $ and $\alpha '$. Thus it is clear that if $f$ and $g$ are faithful, so is the functor $\mathcal{X} \times _\mathcal {Y} \mathcal{Z} \to \mathcal{Y}$. Now, suppose that $(a, b), (a', b') : (U, x, z, \alpha ) \to (U', x', z', \alpha ')$ are two morphisms of the $2$-fibre product. Then consider the equalizer $x'' \to x$ of $a$ and $a'$ and the equalizer $z'' \to z$ of $b$ and $b'$. Since $f$ commutes with equalizers (by Lemma 96.17.3) we see that $f(x'') \to f(x)$ is the equalizer of $f(a)$ and $f(a')$. Similarly, $g(z'') \to g(z)$ is the equalizer of $g(b)$ and $g(b')$. Picture

It is clear that the dotted arrow exists and is an isomorphism. However, it is not a priori the case that the image of $\alpha ''$ in the category of schemes is the identity of its source. On the other hand, the existence of $\alpha ''$ means that we can assume that $x''$ and $z''$ are defined over the same scheme and that the morphisms $x'' \to x$ and $z'' \to z$ have the same image in the category of schemes. Redoing the diagram above we see that the dotted arrow now does project to an identity morphism and we win. Some details omitted. $\square$