**Proof.**
We are going to use the fact that $\Delta _{X/S}$ is representable (by definition of an algebraic space) and that it satisfies properties (2) – (5), see Spaces, Lemma 64.13.1. Note that we have a factorization

\[ X \longrightarrow X \times _ Y X \longrightarrow X \times _ S X \]

of the diagonal $\Delta _{X/S} : X \to X \times _ S X$. Since $X \times _ Y X \to X \times _ S X$ is a monomorphism, and since $\Delta _{X/S}$ is representable, it follows formally that $\Delta _{X/Y}$ is representable. In particular, the rest of the statements now make sense, see Section 66.3.

Choose a surjective étale morphism $U \to X$, with $U$ a scheme. Consider the diagram

\[ \xymatrix{ R = U \times _ X U \ar[r] \ar[d] & U \times _ Y U \ar[d] \ar[r] & U \times _ S U \ar[d] \\ X \ar[r] & X \times _ Y X \ar[r] & X \times _ S X } \]

Both squares are cartesian, hence so is the outer rectangle. The top row consists of schemes, and the vertical arrows are surjective étale morphisms. By Spaces, Lemma 64.11.4 the properties (2) – (5) for $\Delta _{X/Y}$ are equivalent to those of $R \to U \times _ Y U$. In the proof of Spaces, Lemma 64.13.1 we have seen that $R \to U \times _ S U$ has properties (2) – (5). The morphism $U \times _ Y U \to U \times _ S U$ is a monomorphism of schemes. These facts imply that $R \to U \times _ Y U$ have properties (2) – (5).

Namely: For (3), note that $R \to U \times _ Y U$ is a monomorphism as the composition $R \to U \times _ S U$ is a monomorphism. For (2), note that $R \to U \times _ Y U$ is locally of finite type, as the composition $R \to U \times _ S U$ is locally of finite type (Morphisms, Lemma 29.15.8). A monomorphism which is locally of finite type is locally quasi-finite because it has finite fibres (Morphisms, Lemma 29.20.7), hence (5). A monomorphism is separated (Schemes, Lemma 26.23.3), hence (4).
$\square$

## Comments (2)

Comment #875 by Matthew Emerton on

Comment #901 by Johan on