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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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