If $$f:\mathscr{X} \to \mathscr{Y}$$ is a map between (possibly ineffective) orbifolds (in the sense of differentiable stacks, or orbifold groupoids), does it follow that $f$ induces a map between their underlying effective orbifolds? At first, I thought it should always be true, but now that I think about it, you might need $f$ to be open (in this case, I can prove it). Is this known? (by this I mean, is it known to always hold, even without this open assumption?) Note, this really has nothing to do with the differentiable structure, so you may as well ask this for proper etale topological stacks, in fact, I doubt properness plays a role.
1

$\begingroup$ Remind me the definition of an effective groupoid.... $\endgroup$ – David Roberts Nov 13 '10 at 23:55

$\begingroup$ @David: In terms of orbifold groupoids, by using the fact that it is etale and proper, you can show that for every point $x \in G_0$ in the groupoid, there exists a neighborhood $U$ on which $G_x$ acts, where $G_x$ is the stabilizing group (i.e. Hom_G(x,x)). (This is done by taking a local inverse of the source map that goes through $g$ and applying the target map). $G$ is effective if each of these actions are faithful. $\endgroup$ – David Carchedi Nov 14 '10 at 1:10

$\begingroup$ Even better, look at this other question I asked: mathoverflow.net/questions/39531/… $\endgroup$ – David Carchedi Nov 14 '10 at 1:50
2
$\begingroup$
$\endgroup$
Not true:
Take X to be a nontrivial Z/2gerbe on S^2, take Y to be a faithful vector bundle over X, and take f to be the inclusion of the zero section.
The effective quotient of X is S^2, and it has no map back to X. In particular, it has no map to Y that's compatible with f.

$\begingroup$ Thanks Andre. In fact, I realized this is not true later, but forgot to answer my own question on here. I give some sufficient conditions for a map to descend however in the preprint you are reading. I call these conditions "etale invariance" $\endgroup$ – David Carchedi Nov 23 '10 at 23:09