Lemma 85.30.1. Let f : X \to S be a morphism of schemes. Let \pi : Y \to (X/S)_\bullet be a cartesian morphism of simplicial schemes, see Definitions 85.27.1 and 85.27.3. Then the morphism
j = (d^1_1, d^1_0) : Y_1 \to Y_0 \times _ S Y_0
defines an equivalence relation on Y_0 over S, see Groupoids, Definition 39.3.1.
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
(d^2_2, d^2_0) : Y_2 \to Y_1 \times _{d^1_0, Y_0, d^1_1} Y_1.
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
c : Y_1 \times _{s, Y_0, t} Y_1 \longrightarrow Y_1
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
(d^2_2, d^2_1) : Y_2 \to Y_1 \times _{d^1_1, Y_0, d^1_1} Y_1.
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
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