Lemma 101.6.6. Let $f : \mathcal{X} \to \mathcal{Y}$ be a separated (resp. quasi-separated, resp. quasi-DM, resp. DM) morphism of algebraic stacks. Then

1. given algebraic spaces $T_ i$, $i = 1, 2$ and morphisms $x_ i : T_ i \to \mathcal{X}$, with $y_ i = f \circ x_ i$ the morphism

$T_1 \times _{x_1, \mathcal{X}, x_2} T_2 \longrightarrow T_1 \times _{y_1, \mathcal{Y}, y_2} T_2$

is proper (resp. quasi-compact and quasi-separated, resp. locally quasi-finite, resp. unramified),

2. given an algebraic space $T$ and morphisms $x_ i : T \to \mathcal{X}$, $i = 1, 2$, with $y_ i = f \circ x_ i$ the morphism

$\mathit{Isom}_\mathcal {X}(x_1, x_2) \longrightarrow \mathit{Isom}_\mathcal {Y}(y_1, y_2)$

is proper (resp. quasi-compact and quasi-separated, resp. locally quasi-finite, resp. unramified).

Proof. Proof of (1). Observe that the diagram

$\xymatrix{ T_1 \times _{x_1, \mathcal{X}, x_2} T_2 \ar[d] \ar[r] & T_1 \times _{y_1, \mathcal{Y}, y_2} T_2 \ar[d] \\ \mathcal{X} \ar[r] & \mathcal{X} \times _\mathcal {Y} \mathcal{X} }$

is cartesian. Hence this follows from the fact that $f$ is separated (resp. quasi-separated, resp. quasi-DM, resp. DM) if and only if the diagonal is proper (resp. quasi-compact and quasi-separated, resp. locally quasi-finite, resp. unramified).

Proof of (2). This is true because

$\mathit{Isom}_\mathcal {X}(x_1, x_2) = (T \times _{x_1, \mathcal{X}, x_2} T) \times _{T \times T, \Delta _ T} T$

hence the morphism in (2) is a base change of the morphism in (1). $\square$

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