Date: Wed, 17 Jan 1996 10:00:03 -0500 From: Michael Barr <barr@triples.math.mcgill.ca> - - Date: Wed, 17 Jan 1996 12:22:32 +0000 - From: Steve Vickers <sjv@doc.ic.ac.uk> - - >Has anyone proved that if you take an "algebra" (actually monoid) object - >in a monoidal biclosed category that has equalizers and coequalizers, - >then the category of two-sided modules for that algebra is again a - >monoidal biclosed category. Mac Lane did this in 1965 when everything - >is symmetric (the tensor, the algebra and the modules) under the (surely - >irrelevant) assumption that the original category is also abelian. The - >fact is certainly true, but writing down the proof would be rather - >painful. - - The question made me think - as perhaps it was meant to - of modules over - non-commutative rings, and I wondered whether its scope was unnecessarily - restricted in considering just 2-sided modules over a single algebra. We - also know that there can be a rich and well-behaved structure for bimodules - over two algebras: - - (A,B)-bimod tensor (B,C)-bimod is (A,C)-bimod - B - - (A,B)-bimod hom (A,C)-bimod is (B,C)-bimod - A - - (A,B)-bimod hom (C,B)-bimod is (C,A)-bimod - B - - Does anyone know how to state the question to cover this more general - context? But then what works for algebras ought also to work for - algebroids, i.e. enriched categories. - - Steve Vickers. - - - - Actually, if the ground tensor is not symmetric, you cannot even define the notion of (A,B) bimodule. You can define that of left A, right B bimodule and that seems to be that. Perhaps that is what Steve means when he says (A,B) bimodule. In that case, he is, of course, correct. In fact, in writing this up, I find it less confusing to do the more general situation exactly as described. But I would rather not have to write it up if someone has already done it as it is rather unpleasant. Michael