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Apr 13, 2017 at 12:57 history edited CommunityBot
replaced http://mathoverflow.net/ with https://mathoverflow.net/
Jul 11, 2010 at 18:57 vote accept Jan Weidner
Jul 10, 2010 at 18:05 comment added Jan Weidner Thanks, so, I am a bit confused. The functor $E(X)\rightarrow Desc(X,U)$ is monoidal, so no extra condition is needed, since it is automatically an monoidal equivalence whenever it is an equivalence?
Jul 10, 2010 at 14:29 comment added Chris Schommer-Pries @ Jan. Yes. If you have a monoidal functor, which is an equivalence of categories after forgetting the monoidal part, then it is in fact a monoidal equivalence. Hence fully-faithful and essentially surjective is enough.
Jul 9, 2010 at 19:14 comment added Jan Weidner Yes, I also thought about this, however I was confused and saw a problem which was no problem. Do you know whether there is a criterion when s monoidal functor is an monoidal equivalence? Is full, faithful and essential surjective enough?
Jul 9, 2010 at 16:52 history edited Harry Gindi
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Jul 9, 2010 at 15:46 answer added David Carchedi timeline score: 4
Jul 9, 2010 at 11:48 comment added Jeffrey Giansiracusa It looks like what you are asking for is that $E(X) \to Desc(X,U)$ is a monoidal equivalence.
Jul 9, 2010 at 11:44 history asked Jan Weidner CC BY-SA 2.5