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Let $\mathcal{M}$ be an algebraic stack, and let $X$ be its coarse moduli space (assume it exists as a scheme).

We know that $h_X(Spec(k))=\mathcal{M}(Spec(k))$ if $k$ is algebraically closed. Is there anything intelligent we can say about $h_X(U)$ for a general scheme $U$?

For example, can you come up with an algorithm for knowing what $h_X(U)$ is that would be considerably easier than constructing $X$?

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  • $\begingroup$ I'm no expert, but maybe is it something like $h_X(U)=\pi_0(\mathcal{M}(U))$ i.e. the set of isomorphism classes of objects of the groupoid $\mathcal{M}(U)$? Or is my guess totaly mistaken? $\endgroup$
    – Qfwfq
    Jul 16, 2011 at 22:08
  • $\begingroup$ @unknowngoogle: that sounds interesting! Can anyone confirm? $\endgroup$ Jul 16, 2011 at 22:18

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This is more an answer to the comment of unknowngoogle.
• The object U ↦ π0(ℳ(U)) that you described is the initial presheaf to which the stack ℳ maps.
• One can also consider the sheafification of U ↦ π0(ℳ(U)), which is the initial sheaf to which ℳ maps.
• Finally, there is the (possibly non-existent) initial representable sheaf (=scheme) to which ℳ maps.

In general, all those are different.
They are already different in the case of the affine line modded out by the involution x ↦ -x.

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    $\begingroup$ Dear Andre -- The last statement you make here is sensitive to the topology. What you say is correct for the etale topology, but your second and third objects are the same for the fppf topology. This is because the quotient map (to the scheme quotient) is a flat cover (though not an etale cover) and the corresponding equivalence relation is the one induced by your involution. Therefore this scheme quotient is the sheaf quotient. $\endgroup$
    – JBorger
    Jul 17, 2011 at 12:28
  • $\begingroup$ Thank you James. Very relevant comment. I was indeed thinking of the étale topology. $\endgroup$ Jul 17, 2011 at 20:51
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    $\begingroup$ If you take the plane instead of the line, with the involution $(x,y) \mapsto (-x,-y)$, the quotient map is not flat anymore. $\endgroup$
    – Angelo
    Jul 17, 2011 at 21:03
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What André says is absolutely correct. The functor represented by the moduli space does not have any reasonable description, except in very particular cases. Given a map $U \to X$, it can be hard work to decide whether it comes from an object of $\mathcal M(U)$.

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