The Stacks project

Lemma 79.12.7. Let $S$ be a scheme. Let $f : X \to Y$ be a morphism of algebraic spaces over $S$. The diagonal of $(X/Y)_{fin} \to Y$

\[ (X/Y)_{fin} \longrightarrow (X/Y)_{fin} \times _ Y (X/Y)_{fin} \]

is representable (by schemes) and an open immersion and the “absolute” diagonal

\[ (X/Y)_{fin} \longrightarrow (X/Y)_{fin} \times (X/Y)_{fin} \]

is representable (by schemes).

Proof. The second statement follows from the first as the absolute diagonal is the composition of the relative diagonal and a base change of the diagonal of $Y$ (which is representable by schemes), see Spaces, Section 65.3. To prove the first assertion we have to show the following: Given a scheme $T$ and two pairs $(a, Z_1)$ and $(a, Z_2)$ over $T$ with identical first component satisfying 79.12.0.1 there is an open subscheme $V \subset T$ with the following property: For any morphism of schemes $h : T' \to T$ we have

\[ h(T') \subset V \Leftrightarrow \Big(T' \times _ T Z_1 = T' \times _ T Z_2 \text{ as subspaces of }T' \times _ Y X\Big) \]

Let us construct $V$. Note that $Z_1 \cap Z_2$ is open in $Z_1$ and in $Z_2$. Since $\text{pr}_0|_{Z_ i} : Z_ i \to T$ is finite, hence proper (see Morphisms of Spaces, Lemma 67.45.9) we see that

\[ E = \text{pr}_0|_{Z_1}\left(Z_1 \setminus Z_1 \cap Z_2)\right) \cup \text{pr}_0|_{Z_2}\left(Z_2 \setminus Z_1 \cap Z_2)\right) \]

is closed in $T$. Now it is clear that $V = T \setminus E$ works. $\square$

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