Lemma 99.13.2. The diagonal

$\Delta : \mathcal{S}\! \mathit{paces}'_{fp, flat, proper} \longrightarrow \mathcal{S}\! \mathit{paces}'_{fp, flat, proper} \times \mathcal{S}\! \mathit{paces}'_{fp, flat, proper}$

is representable by algebraic spaces.

Proof. We will use criterion (2) of Algebraic Stacks, Lemma 94.10.11. Let $S$ be a scheme and let $X$ and $Y$ be algebraic spaces of finite presentation over $S$, flat over $S$, and proper over $S$. We have to show that the functor

$\mathit{Isom}_ S(X, Y) : (\mathit{Sch}/S)_{fppf} \longrightarrow \textit{Sets}, \quad T \longmapsto \{ f : X_ T \to Y_ T \text{ isomorphism}\}$

is an algebraic space. An elementary argument shows that $\mathit{Isom}_ S(X, Y)$ sits in a fibre product

$\xymatrix{ \mathit{Isom}_ S(X, Y) \ar[r] \ar[d] & S \ar[d]_{(\text{id}, \text{id})} \\ \mathit{Mor}_ S(X, Y) \times \mathit{Mor}_ S(Y, X) \ar[r] & \mathit{Mor}_ S(X, X) \times \mathit{Mor}_ S(Y, Y) }$

The bottom arrow sends $(\varphi , \psi )$ to $(\psi \circ \varphi , \varphi \circ \psi )$. By Proposition 99.12.3 the functors on the bottom row are algebraic spaces over $S$. Hence the result follows from the fact that the category of algebraic spaces over $S$ has fibre products. $\square$

Comment #2327 by Matthieu Romagny on

In the statement of the lemma, the word "is" is missing before the word "representable".

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