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
is representable (by schemes) and an open immersion and the “absolute” diagonal
is representable (by schemes).
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
is representable (by schemes) and an open immersion and the “absolute” diagonal
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
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
is closed in T. Now it is clear that V = T \setminus E works. \square
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