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Apr 12, 2013 at 6:45 vote accept Ulrich Pennig
Apr 11, 2013 at 21:22 comment added Ulrich Pennig Ah, but up to homotopy I can replace X by G.
Apr 11, 2013 at 21:03 comment added Ulrich Pennig Don't you need that the action of $G$ on $X$ is free to get the functor $F$?
Apr 11, 2013 at 20:57 history answered Oscar Randal-Williams CC BY-SA 3.0