Lemma 99.12.1. Let $S$ be a scheme. Consider morphisms of algebraic spaces $Z \to B$ and $X \to B$ over $S$. If $X \to B$ is separated and $Z \to B$ is of finite presentation, flat, and proper, then there is a natural injective transformation of functors
which maps a morphism $f : Z_ T \to X_ T$ to its graph.
Proof. Given a scheme $T$ over $B$ and a morphism $f_ T : Z_ T \to X_ T$ over $T$, the graph of $f$ is the morphism $\Gamma _ f = (\text{id}, f) : Z_ T \to Z_ T \times _ T X_ T = (Z \times _ B X)_ T$. Recall that being separated, flat, proper, or finite presentation are properties of morphisms of algebraic spaces which are stable under base change (Morphisms of Spaces, Lemmas 67.4.4, 67.30.4, 67.40.3, and 67.28.3). Hence $\Gamma _ f$ is a closed immersion by Morphisms of Spaces, Lemma 67.4.6. Moreover, $\Gamma _ f(Z_ T)$ is flat, proper, and of finite presentation over $T$. Thus $\Gamma _ f(Z_ T)$ defines an element of $\mathrm{Hilb}_{Z \times _ B X/B}(T)$. To show the transformation is injective it suffices to show that two morphisms with the same graph are the same. This is true because if $Y \subset (Z \times _ B X)_ T$ is the graph of a morphism $f$, then we can recover $f$ by using the inverse of $\text{pr}_1|_ Y : Y \to Z_ T$ composed with $\text{pr}_2|_ Y$. $\square$
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