The Stacks project

Proof. Note that $j$ is a monomorphism. Namely the composition $Y_1 \to Y_0 \times _ S Y_0 \to Y_0 \times _ S X$ is an isomorphism as $\pi $ is cartesian.

Consider the morphism

This works because $d_0 \circ d_2 = d_1 \circ d_0$, see Simplicial, Remark 14.3.3. Also, it is a morphism over $(X/S)_2$. It is an isomorphism because $Y \to (X/S)_\bullet $ is cartesian. Note for example that the right hand side is isomorphic to $Y_0 \times _{\pi _0, X, \text{pr}_1} (X \times _ S X \times _ S X) = X \times _ S Y_0 \times _ S X$ because $\pi $ is cartesian. Details omitted.

As in Groupoids, Definition 39.3.1 we denote $t = \text{pr}_0 \circ j = d^1_1$ and $s = \text{pr}_1 \circ j = d^1_0$. The isomorphism above, combined with the morphism $d^2_1 : Y_2 \to Y_1$ give us a composition morphism

over $Y_0 \times _ S Y_0$. This immediately implies that for any scheme $T/S$ the relation $Y_1(T) \subset Y_0(T) \times Y_0(T)$ is transitive.

Reflexivity follows from the fact that the restriction of the morphism $j$ to the diagonal $\Delta : X \to X \times _ S X$ is an isomorphism (again use the cartesian property of $\pi $).

To see symmetry we consider the morphism

This works because $d_1 \circ d_2 = d_1 \circ d_1$, see Simplicial, Remark 14.3.3. It is an isomorphism because $Y \to (X/S)_\bullet $ is cartesian. Note for example that the right hand side is isomorphic to $Y_0 \times _{\pi _0, X, \text{pr}_0} (X \times _ S X \times _ S X) = Y_0 \times _ S X \times _ S X$ because $\pi $ is cartesian. Details omitted.

Let $T/S$ be a scheme. Let $a \sim b$ for $a, b \in Y_0(T)$ be synonymous with $(a, b) \in Y_1(T)$. The isomorphism $(d^2_2, d^2_1)$ above implies that if $a \sim b$ and $a \sim c$, then $b \sim c$. Combined with reflexivity this shows that $\sim $ is an equivalence relation. $\square$