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As to why algebraic stacks are always assumed to be stacks in groupoids, there are several things I could say, but the honest answer is that I don't know the deep reason for this. … not stacks in groupoids. …

They only work with stacks that are locally finitely presented. …

From the definition it follows easily that tame stacks with affine moduli spaces have the property you require. … There are several different characterizations of tame stacks; see the paper "Tame stacks in positive characteristic" by Dan Abramovich, Martin Olsson and myself. …

It is certainly not true that $\mathcal X \to \mathcal X^H$ is a good moduli morphism, unless $H$ is linearly reductive, because when you push forward the cohomology of $H$ will come into play. On th …

Yes. The criterion for an Artin stack to be Deligne-Mumford is that it should have unramified diagonal (this is somewhere in Laumon Moret Bailly, I don't have it here). If the stack is fibered in sets …

When you post a question, it would be good if you include enough explanations not to force the interested reader to go search for a paper online. Anyway, the general question is: suppose that we have …

Then the induced morphism $\mathcal X\to \sqrt[k]{D/S}$ is proper, because both stacks are proper over $S$. It is also birational. … For example, when $D$ is the union of two smooth curves intersecting transversally, you take the fiber product of the root stacks of the two curves. …

I think that what they might have in mind is that for non-quasi-separated Deligne-Mumford algebraic stacks one should not assume that the diagonal is represented by schemes, but by algebraic spaces. … For quasi-separated Deligne-Mumford stacks this implies representability by schemes, but this is not true in general. …

I don't think this is true. Let $X$ be the quotient of the action of $\mathbb Q$ on $\mathbb R$ by translation. This is a sheaf, and its automorphism groups are trivial. Suppose that there exist a loc …

If $X \to Y$ is a proper morphism of DM stacks, where $X$ has finite inertia (the hypotheses in my paper are more stringent, but the theory has been refined since then), there exists a finite map $V \to …

In the smooth case, I think that the answer is positive over an arbitrary regular excellent base. The argument was in my PhD thesis; it was done over a field, but I think it adapts to this case. Cove …

Take the complement of a non-empty open subscheme, and use noetherian induction.

Tame Artin stacks (in the sense of Abramovich, Olsson and Vistoli, https://math.berkeley.edu/~molsson/tame.pdf) with quasi-projective moduli spaces will have property 2: the line bundle is the pullback … As to property 1, a line bundle will not be enough, but you can get a version for quotient tame Artin stacks using a generating sheaf (in the sense of Olsson and Starr, https://math.berkeley.edu/~molsson …

By a family of gerbes you mean, I suppose, a gerbe over $X \times \mathbb A^{1}$. In any case, it has a class in $\mathrm H^0(\mathbb A^1, \mathrm R^2 \mathrm{pr}_{2*}\mathbb G_{\rm m})$. Since $\math …

Consider the fiber product $S \times_{\frak M} M$; its projection onto $S$ is smooth and surjective, hence open. Since $S$ is quasi-compact, there exists a quasi-compact open subset $U$ of $S \times_{ …