# Can two smooth categories be equivalent if their object manifolds aren't diffeomorphic or homotopy equivalent?

There is a category of "smooth categories", where the objects and the morphisms don't form sets, but manifolds (and there are some other conditions that I won't repeat here). Important examples are Lie groups and Lie groupoids such as the action groupoid of a homogeneous space.

The action groupoid of a finite $G$-set is equivalent to the stabiliser group of a point, seen as a 1-object category. I want to know whether this is the case in the smooth setting as well.

Given two equivalent smooth categories, what can we say about their object spaces? Doesn't the equivalence imply a homotopy equivalence? Or even a diffeomorphism?

• I suppose it depends on the definition of equivalence. – Zhen Lin Mar 16 '15 at 14:33
• @ZhenLin, what possible definitions are there, then? – Turion Mar 16 '15 at 15:21
• There is the obvious definition involving a pair of functors. As I understand it, the correct definition should be one that generalises "fully faithful and essentially surjective on objects", sometimes called "weak equivalence". But even "essentially surjective on objects" involves a subtlety. You should ask David Roberts. – Zhen Lin Mar 16 '15 at 15:25
• It is already not true that equivalent categories have isomorphic object sets (so this gives a counterexample in smooth categories where all of the manifolds are $0$-dimensional). I also don't see what the second paragraph has to do with the third. – Qiaochu Yuan Mar 16 '15 at 17:09
• @QiaochuYuan, my motivation is to understand whether smooth action groupoids are actually equivalent to their stabiliser Lie groups (viewed as a smooth category). I was hoping that the correct notion of equivalence forbids e.g. your example, so that what you say would hopefully not be a smooth equivalence. – Turion Mar 16 '15 at 18:00