# Space over Stack

It turns out that the result mentioned in this post is especially useful in theoretical considerations. For example consider the following statement

Given an 1-morphism \cX —> \cY of stacks in groupoids which is representable by algebraic spaces such that \cY is an algebraic stack, then \cX is an algebraic stack.

Proof: pick a scheme Y and a surjective smooth morphism Y —> \cY. By assumption the 2-fibre product Y x_{\cY} \cX is representable by an algebraic space X. The projection map X —> \cX is surjective and smooth as a base change and we win by the result mentioned above.

Of course the result can be proved in other ways as well, but it is quite pleasing how short the argument above is. This kind of thing is especially helpful because we intend to prove many results of this kind!

[Edit March 08: Here are some links to the result mentioned above and its improvement suggested by David Rydh in the comment below.]

## 2 thoughts on “Space over Stack”

1. More general, it is enough that the morphism is “algebraic”, i.e., “representable by algebraic stacks” or precisely “for any (affine) scheme/algebraic space/algebraic stack Y and Y->\cY the fiber product Y x_{\cY} \cX is an algebraic stack”.

• Ah, yes, of course, great! Please keep telling me these things. Thanks!