# Representability of morphism of stacks

A morphism of Artin stacks $f:X\to Y$ over $\mathbb Q$ is representable by algebraic spaces if and only if its geometric fibres are algebraic spaces. I would like to know if one can use this to prove the following statement.

Let $f:X\to Y$ be a morphism of finite type separated DM stacks over $\mathbb Q$. Suppose that, for any geometric point $x$ of $X$ with $y= f(x)$, the induced morphism on stabilizers $Stab(x)\to Stab(y)$ is injective. Then $f:X\to Y$ is representable by algebraic spaces.

• Using pullback to an algebraic space over $Y$, your question is whether a DM stack $X$ (or more generally Artin stack) separated and finite type over an algebraic space $Y$ is itself an algebraic space when its geometric points have trivial Aut-schemes. This is true without char-0 hypotheses (so no need for Cartier!). See Theorem 2.2.5 in journals.cambridge.org/action/… (where Artin stacks are assumed to have diagonal separated and finite type); is this in the Stacks Project (maybe with weaker diagonal hypotheses)? – user74230 Mar 30 '15 at 1:40

This is http://stacks.math.columbia.edu/tag/04Y5 . I quote :

"

lemma

Let $S$ be a scheme contained in $Sch_{fppf}$. Let $f : \mathcal{X} \to \mathcal{Y}$ be a $1$-morphism of algebraic stacks over $S$. The following are equivalent:

1 for $U \in Ob((Sch/S)_{fppf})$ the functor $f : \mathcal{X}_U \to\mathcal{Y}_U$ is faithful,

2 the functor $f$ is faithful, and

3 $f$ is representable by algebraic spaces.

"

• Thank you for your answer. Just to be sure I'd like to ask one question. In my situation, $S = Spec \mathbb Q$ and I only know that condition 1 holds for $U= Spec \mathbb C$ (or $U=$ spectrum of an alg closed field of char zero). Do I understand correctly that this is enough? I guess to prove this we just replace the target by a geometric point $y$ of $Y$, right? – user234 Mar 29 '15 at 14:31
• Yes, it is enough to verify the property on geometric points : you need to check that the kernel say $K/U$ a is trivial, that is that the unit section $U\to K$ is an isomorphism. You can see these spaces as sheaves, and then use the fact that the geometric points form "a conservative family of points" see stacks.math.columbia.edu/tag/00YJ . – Niels Mar 29 '15 at 16:25
• @Niels: The difference is that base change on geometric objects along a map not "in the site" isn't generally computed by sheaf pullback for the representing functor. For example, if $X = Y \times \mathbf{A}^1$ over a scheme $Y$ then the scheme-theoretic pullback along geometric point $y:{\rm{Spec}}(k) \rightarrow Y$ is the scheme $\mathbf{A}^1_{k}$ whereas the fiber at $y$ of the functor represented by $X$ on the etale site is the constant sheaf over ${\rm{Spec}}(k)$ associated to the group of units in the strict henselization of $Y$ at $y$. So the SP reference seems not quite enough to me. – user74230 Mar 30 '15 at 14:59