Lemma 101.3.3. Let $f : \mathcal{X} \to \mathcal{Y}$ be a morphism of algebraic stacks. Then
$\Delta _ f$ is representable by algebraic spaces, and
$\Delta _ f$ is locally of finite type.
Diagonals of morphisms of algebraic stacks are representable by algebraic spaces and locally of finite type.
Proof. Let $T$ be a scheme and let $a : T \to \mathcal{X} \times _\mathcal {Y} \mathcal{X}$ be a morphism. By definition of the fibre product and the $2$-Yoneda lemma the morphism $a$ is given by a triple $a = (x, x', \alpha )$ where $x, x'$ are objects of $\mathcal{X}$ over $T$, and $\alpha : f(x) \to f(x')$ is a morphism in the fibre category of $\mathcal{Y}$ over $T$. By definition of an algebraic stack the sheaves $\mathit{Isom}_\mathcal {X}(x, x')$ and $\mathit{Isom}_\mathcal {Y}(f(x), f(x'))$ are algebraic spaces over $T$. In this language $\alpha $ defines a section of the morphism $\mathit{Isom}_\mathcal {Y}(f(x), f(x')) \to T$. A $T'$-valued point of $\mathcal{X} \times _{\mathcal{X} \times _\mathcal {Y} \mathcal{X}, a} T$ for $T' \to T$ a scheme over $T$ is the same thing as an isomorphism $x|_{T'} \to x'|_{T'}$ whose image under $f$ is $\alpha |_{T'}$. Thus we see that
101.3.3.1
\begin{equation} \label{stacks-morphisms-equation-diagonal} \vcenter { \xymatrix{ \mathcal{X} \times _{\mathcal{X} \times _\mathcal {Y} \mathcal{X}, a} T \ar[d] \ar[r] & \mathit{Isom}_\mathcal {X}(x, x') \ar[d] \\ T\ar[r]^-\alpha & \mathit{Isom}_\mathcal {Y}(f(x), f(x')) } } \end{equation}
is a fibre square of sheaves over $T$. In particular we see that $\mathcal{X} \times _{\mathcal{X} \times _\mathcal {Y} \mathcal{X}, a} T$ is an algebraic space which proves part (1) of the lemma.
To prove the second statement we have to show that the left vertical arrow of Diagram (101.3.3.1) is locally of finite type. By Lemma 101.3.1 the algebraic space $\mathit{Isom}_\mathcal {X}(x, x')$ and is locally of finite type over $T$. Hence the right vertical arrow of Diagram (101.3.3.1) is locally of finite type, see Morphisms of Spaces, Lemma 67.23.6. We conclude by Morphisms of Spaces, Lemma 67.23.3. $\square$
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Comment #882 by Konrad Voelkel on
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