Bimodule morphisms

Definition

If you had to guess the definition of a bimodule morphism, you'd guess correctly:

Definition of bimodule morphism

If M and N are (R,S)-bimodules, then a set map f:MN is a bimodule morphism if it is both a left R-module morphism and a right S-module morphism.

This is cheating slightly, of course. We should really say that a bimodule morphism f:MN consists of the data of a set map f:U(M)U(N) such that the following conditions hold:

In any case, we can now talk about the category of (R,S)-bimodules. What should we denote this category? It's not completely agreed upon. Some people denote it (R,S)-Bimod. Others denote is R-Bimod-S, or even R-Mod-S. Choose your favorite, make sure it's clear, and stick with it.

More than just hom-sets

As with R-modules, for any pair of (R,S)-bimodules M and N, the set of bimodule morphisms between them has the structure of an abelian group (using the addition in N).

There's a lot more to the story about bimodule morphisms, though. First suppose M is an (R,S)-bimodule and N is an (R,S)-bimodule. If we forget the right-actions and consider the left R-modules M and N, we can consider the set of left R-module morphisms, HomR(M,N). This set actually has the structure of an (S,S)-bimodule, as follows.

The addition in HomR(M,N) is defined through the addition in N. In other words, given R-module morphisms f,g:MN we define f+g:MN by (f+g)(m)=f(m)+g(m). Observe that f+g is indeed an R-module morphism. First, it is additive since f and g are additive; second, it is compatible with the R-actions since f and g are compatible with the R-actions.

The left S-action on HomR(M,N) is defined through the right S-action on M. In detail, for each R-module morphism f:MN and sS we define sf:MN by (sf)(m)=f(ms). Again, it is straightforward to verify that sf is an R-module morphism. Moreover, this really does define a left S-action on HomR(M,N), since

(s1s2f)(m)=f(m(s1s2))=f((ms1)s2)=(s2f)(ms1)=(s1(s2f))(m)

In other words, s1s2f=s1(s2f).

The right S-action on HomR(M,N) is defined through the right S-action on N. In detail, for each R-module morphism f:MN and sS we define fs:MN by (fs)(m)=f(m)s. It is once more simple to verify that fs is an R-module morphism, and that this really does define a right S-action on HomR(M,N).

In summary:

Hom-bimodules

For each (R,S)-bimodule M and (R,S)-bimodule N, the set of HomR(M,N) of left R-module morphisms between M and N (viewed as left R-modules) has the structure of an (S,S)-bimodule.

Similarly, for each (R,S)-bimodule M and (R,S)-bimodule N, the set HomS(M,N) of right S-module morphisms between M and N (viewed as right S-modules) has the structure of an (R,R)-bimodule.

A more careful approach?

We should really be careful here and use forgetful functors to move M and N into the category of left R-modules. Can you fill in the details?

Triples of bimodules and hom-sets

Suppose M is an (R,S)-bimodule, N is an (S,T)-bimodule, and P is an (R,T)-bimodule. By the above construction, the set HomT(N,P) has the structure of an (R,S)-bimodule. We can then consider the set of (R,S)-bimodule morphisms between M and HomT(N,P). This is the set

Hom(R,S)(M,HomT(N,P)).

This set will play a critical role with the tensor product construction on bimodules.


Suggested next notes

The 2-category of bimodules
Tensor Products II - Tensor products of bimodules