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Oct 27, 2009 at 0:49 comment added Qiaochu Yuan My mistake, then. I'm going more off of intuition than anything else.
Oct 27, 2009 at 0:49 history edited Qiaochu Yuan CC BY-SA 2.5
Fixed a mistake.
Oct 27, 2009 at 0:26 comment added Reid Barton The Gpd-enriched Hom(A, B) in Grp is obtained by just regarding A and B as categories (groupoids) with one element and taking functors as objects and natural transformations as morphisms. If you work this out then the objects are group homomorphisms A -> B and the automorphisms of an object f are the elements of B which commute with the image of f (more generally a morphism f -> g is an element b such that g = b f b^{-1}).
Oct 26, 2009 at 19:24 comment added Greg Muller Thats a bit of an odd fellow. So, if I am correct, then if there are no non-trivial morphisms from B to A, then Hom(A,B) is the action groupoid of tautological action of the symmetric group of the set Hom(A,B)?
Oct 26, 2009 at 19:15 comment added Qiaochu Yuan My understanding is this: let C = {A, B} be the subcategory of Grp consisting of two fixed groups A and B. Then Hom(A, B) can be thought of as the groupoid whose objects are morphisms from A to B and whose morphisms are invertible functors from C to C such that F(A) = A, F(B) = B.
Oct 26, 2009 at 19:06 comment added Greg Muller I'm somewhat confused by your definition of Gpd. Are the objects all morphisms from FIXED groups A and B, or are they morphisms between any two groups? In either case, what are the morphisms?
Oct 26, 2009 at 19:02 history answered Qiaochu Yuan CC BY-SA 2.5