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Apr 13, 2017 at 12:58 history edited CommunityBot
replaced http://mathoverflow.net/ with https://mathoverflow.net/
May 2, 2010 at 20:06 history edited Dan Ramras CC BY-SA 2.5
added link
Apr 27, 2010 at 5:40 history edited Dan Ramras CC BY-SA 2.5
added 37 characters in body
Apr 27, 2010 at 5:40 comment added Dan Ramras Ah, I see. I think this saved me from a fair amount of time-wasting. I'll leave the answer up, as a warning...
Apr 27, 2010 at 4:52 comment added Mariano Suárez-Álvarez If the action is not free on objects, you have choices when picking the $g$ which makes the morphisms "match up".
Apr 27, 2010 at 4:52 comment added Reid Barton What if I take C = BH, where H is a group with an action of G? I think the quotient category will be B(H with the relations x = gx for all x in H imposed), whose set of morphisms will be different from the set of orbits for the action of G on H as a set.
Apr 27, 2010 at 4:42 history answered Dan Ramras CC BY-SA 2.5