Talk:Bimodule

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I believe this paragraph is false: "An R-S bimodule is actually the same thing as a left module over the ring R×Sop, where Sop is the opposite ring of S (with the multiplication turned around). Bimodule homomorphisms are the same as homomorphisms of left R×Sop modules. Using these facts, many definitions and statements about modules can be immediately translated into definitions and statements about bimodules. For example, the category of all R-S bimodules is abelian, and the standard isomorphism theorems are valid for bimodules."

Consider Z as the obvious Z-Z bimodule. Then the Z×Zop module structure on Z should be defined as (a, b)c = acb. But, for example, ((0, 1) + (1, 1))1 = (1, 2)1 = 2, while (0, 1)1 + (1, 1)1 = 0 + 1, so distributivity fails.

It is true, however, that an R-S bimodule can be regarded as a left module and vice versa.

mistakes: categories of bimodules and left modules[edit]

I believe the statement that "" for a commutative ring is wrong. These categories are not canonically equivalent: while any left -module can be made into a bimodule by taking the same action on the right, it is not true that any bimodule has the same action on left and right sides. So there is just a faithful monoidal functor .

Also, in this case, the monoidal structure on is indeed symmetric in an obvious way, but the monoidal structure on isn't! Here it might make sense to mention the fact that is not equivalent to as monoidal categories (different tensor product).

Can someone please fix these two mistakes in the article?

132.76.50.6 (talk) 08:59, 6 October 2015 (UTC)Inna Entova[reply]

Weak 2-category[edit]

Where it says:

> This is in fact a 2-category, in a canonical way...

I think it should say *weak* 2-category, since 1-cell composition is not strict as noted above, and noted here. — Preceding unsigned comment added by Jesuslop (talkcontribs) 08:54, 16 September 2018 (UTC)[reply]

Are there any plans to fix this mistake? 2003:DC:3F31:4121:4E6:7E16:B528:D086 (talk) 09:40, 13 February 2020 (UTC)[reply]