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Aug 28, 2013 at 12:32 answer added Buschi Sergio timeline score: 0
Jul 9, 2011 at 16:56 comment added Nacho Lopez This seems to be the old result of Kock on commutative monads (the monad here being $(A\otimes−)$). What is true is that TFAE: 1. Mod(A) is closed (resp monoidal) and the free A-module adjunction $F\dashv U:\mathbf{Mod}(A)\to\mathcal{V}$ is a closed adjunction (resp. monoidal adjunction). 2. A is commutative.
Feb 11, 2010 at 19:45 comment added Peter Arndt That is how I understood it - Tyler's newly edited answer gives now an easy counterexample to my original statement (the former one used things I didn't know). And Clark's proof works and tells me what I wanted to know.
Feb 11, 2010 at 1:23 comment added Tom Leinster OK, let's see if I have this right. Tyler's "no" and Clark's "yes" are both correct. Tyler's "no" answers the question posed. Clark's "yes" answers the same question but under the assumption of a (sensible) extra hypothesis. Right?
Feb 10, 2010 at 18:42 vote accept Peter Arndt
Feb 10, 2010 at 17:17 answer added Victor Ostrik timeline score: 6
Feb 10, 2010 at 11:43 comment added Clark Barwick Tom Leinster - one does have to require that the natural right action of the ambient category is compatible with the monoidal structure on the category of modules, but that's it.
Feb 10, 2010 at 11:01 comment added Tom Leinster I find this a really strange statement. Are the monoidal structure and the symmetry on A-Mod really not required to bear any kind of relationship to the monoidal structure and symmetry on your original category?
Feb 10, 2010 at 6:26 answer added Tyler Lawson timeline score: 4
Feb 10, 2010 at 1:43 answer added Clark Barwick timeline score: 12
Feb 9, 2010 at 21:13 history asked Peter Arndt CC BY-SA 2.5