Consider a stack $\mathcal{X}$ over $\mathbb{C}$ as a category fibred in groupoids over the category of schemes. Let $\mathcal{X}^s$ be the $\pi_0$ of this category, i.e. objects of $\mathcal{X}^s$ are the objects in $\mathcal{X}$ and morphisms of $\mathcal{X}^s$ are the morphisms in $\mathcal{X}$ modulo automorphisms of objects. It "kills" the groupoid structure, so I think it is possible to consider $\mathcal{X}^s$ as a category fibred in sets over the category of schemes. Assume $\mathcal{X}^s$ is represented by a scheme. Should it be the coarse moduli space for $\mathcal{X}$?

3$\begingroup$ This should be relevant: mathoverflow.net/questions/70520/… $\endgroup$ – Mattia Talpo Feb 2 '13 at 14:21
Yes, this would imply that $\newcommand{\X}{\mathcal X}\X^s$ is the coarse moduli space, but I don't think this is the "right" question to ask  I believe that $\X^s$ will not even form a sheaf unless $\X$ happens to be a scheme/algebraic space to begin with.
Anyway, any morphism from a groupoid to a set factors through $\pi_0$ of the groupoid. This implies in particular that any morphism from $\X$ to an algebraic space factors through the presheaf $\X^s$. And the map $\X \to \X^s$ is a bijection on geometric points because it's in fact a bijection on $S$points for any scheme $S$. So if $\X^s$ is a scheme/algebraic space then it is the coarse moduli space.
Addendum. I think you are confused about some basic issues. Let us see why $BG^s$ is not the coarse moduli space of $BG$. Let $G$ be a nontrivial finite group, say.
Consider for simplicity the topological setting, so we have a topological space $X$ and an open cover $\{U_i\}$. If we have a $G$torsor on $X$ then we can restrict to a $G$torsor on each $U_i$, and on each overlap $U_i \cap U_j$ we have isomorphisms between the restrictions from $U_i$ and from $U_j$. These isomorphisms satisfy cocycle relation. Conversely, if we have $G$torsors on each $U_i$ and isomorphisms satisfying the cocycle relation, we can reconstruct a $G$torsor on the whole of $X$, unique up to canonical isomorphism. What this paragraph says is exactly that the functor $BG$ which sends a space to the groupoid of $G$torsors over it is a sheaf of groupoids, that is, a stack. (In the usual Grothendieck topology on the category of topological spaces, where open covers are, well, open covers. And when I call $BG$ a "functor" I should say "pseudofunctor" or "fibered category".)
On the other hand we can consider $BG^s$, which is now a priori just a presheaf of sets, mapping a space to the set of isomorphism classes of $G$torsors over it. If we have an isomorphism class of $G$torsor on $X$ then we get well defined isomorphism classes of $G$torsors on each $U_i$ with compatible restrictions to each $U_i \cap U_j$. But it is NOT true that if we have an isomorphism class of $G$torsor on each $U_i$ which agree on double overlaps, then we can reconstruct a unique isomorphism class on all of $X$: consider the case when $G$ is nontrivial on $X$ and $\{U_i\}$ is a trivializing cover! What this says is that $BG^s$ is in fact only a presheaf  it is NEVER a sheaf of sets. Put simply, one can not glue together isomorphism classes.
What this shows is in fact that if we sheafify $BG^s$, then we get a point. If we only remember isomorphism classes of torsors then every $G$torsor becomes equivalent to the trivial torsor on some open covering of your space, which means that these torsors are identified under sheafification.
The same arguments work verbatim in algebraic geometry, since every $G$torsor is locally trivial in the étale topology.
In any case, this is why I said above that this is not the "right" question to ask: it is not natural to expect $\X^s$ to be a sheaf in the first place. I would suggest reading Heinloth or Fantechi's notes on stacks (they are somewhere online) and thinking over just what question it is you want to ask.

$\begingroup$ And what if $\mathcal{X}$ is the classifying stack BG or, for example, the quotient stack $[\mathbb{A}/\mathbb{Z}_n]$? Isn't $\mathcal{X}^s$ the coarse moduli for $\mathcal{X}$ in these cases? $\endgroup$ – Nullstellensatz Feb 2 '13 at 17:10

$\begingroup$ The short answer is "no"  see the addendum above. $\endgroup$ – Dan Petersen Feb 3 '13 at 10:58

$\begingroup$ Thanks for the detailed answer, Dan. So, do I understand correctly that, if the sheafification of $\mathcal{X}^s$ is represented by a scheme, then it is a coarse moduli for $\mathcal{X}$? $\endgroup$ – Nullstellensatz Feb 3 '13 at 13:50

1$\begingroup$ @Scott: Thanks for the correction. @Nullstellensatz: You are correct that when the sheafification of $\X^s$ is representable, then that will be the coarse moduli space. For $\mathbb A^1$ divided by the $n$th roots of unity you have to be careful, the étale sheafification is not a scheme but the fppf sheafification is (and coincides with the scheme quotient). $\endgroup$ – Dan Petersen Feb 4 '13 at 8:08

2$\begingroup$ @Dan: No, the fppf sheafification of $[\mathbb{A}^1/\mathbb{Z}_n]$ is not a scheme. In general, $\mathcal{X}\to \mathcal{X}^s$ is always a gerbe: étale or fppf depending on in which topology the sheafification is done. If $\mathcal{X}^s$ happens to be a scheme/algebraic space (hence the cms) this implies that $\mathcal{X}\to \mathcal{X}^s$ and its diagonal are étale/flat and finitely presented. But the diagonal of $[\mathbb{A}^1/\mathbb{Z}_n]\to \mathbb{A}^1$ is not flat (b/c its inertia is not). Similarly, the fppf sheafification of $\mathcal{M}_{1,1}$ is not $\mathbb{A}^1$. $\endgroup$ – David Rydh Feb 6 '13 at 10:03