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 $f:M\to N$ 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 $f:M\to N$ consists of the data of a set map $f:U(M)\to U(N)$ such that the following conditions hold:

$\varphi ({m}_{1}+{m}_{2})=\varphi ({m}_{1})+\varphi ({m}_{2})$, for all ${m}_{1},{m}_{2}\in M$

$\varphi (rm)=r\varphi (m)$, for all $r\in R$ and $m\in M$

$\varphi (ms)=\varphi (m)s$, for all $s\in S$ and $m\in M$

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 $(R,S)\mathbf{\text{-Bimod}}$. Others denote is $R\mathbf{\text{-Bimod-}}S$, or even $R\mathbf{\text{-Mod-}}S$. 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 $(R,{S}^{\prime})$-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, ${\mathrm{Hom}}_{R}(M,N)$. This set actually has the structure of an $(S,{S}^{\prime})$-bimodule, as follows.

The addition in ${\mathrm{Hom}}_{R}(M,N)$ is defined through the addition in $N$. In other words, given $R$-module morphisms $f,g:M\to N$ we define $f+g:M\to N$ 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 ${\mathrm{Hom}}_{R}(M,N)$ is defined through the right $S$-action on $M$. In detail, for each $R$-module morphism $f:M\to N$ and $s\in S$ we define $s\cdot f:M\to N$ by $(s\cdot f)(m)=f(ms)$. Again, it is straightforward to verify that $s\cdot f$ is an $R$-module morphism. Moreover, this really does define a left $S$-action on ${\mathrm{Hom}}_{R}(M,N)$, since

In other words, ${s}_{1}{s}_{2}\cdot f={s}_{1}\cdot ({s}_{2}\cdot f)$.

The right ${S}^{\prime}$-action on ${\mathrm{Hom}}_{R}(M,N)$ is defined through the right ${S}^{\prime}$-action on $N$. In detail, for each $R$-module morphism $f:M\to N$ and ${s}^{\prime}\in {S}^{\prime}$ we define $f\cdot {s}^{\prime}:M\to N$ by $(f\cdot {s}^{\prime})(m)=f(m){s}^{\prime}$. It is once more simple to verify that $f\cdot {s}^{\prime}$ is an $R$-module morphism, and that this really does define a right ${S}^{\prime}$-action on ${\mathrm{Hom}}_{R}(M,N)$.

In summary:

Hom-bimodules

For each $(R,S)$-bimodule $M$ and $(R,{S}^{\prime})$-bimodule $N$, the set of ${\mathrm{Hom}}_{R}(M,N)$ of left $R$-module morphisms between $M$ and $N$ (viewed as left $R$-modules) has the structure of an $(S,{S}^{\prime})$-bimodule.

Similarly, for each $(R,S)$-bimodule $M$ and $({R}^{\prime},S)$-bimodule $N$, the set ${\mathrm{Hom}}_{S}(M,N)$ of right $S$-module morphisms between $M$ and $N$ (viewed as right $S$-modules) has the structure of an $({R}^{\prime},R)$-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 ${\mathrm{Hom}}_{T}(N,P)$ has the structure of an $(R,S)$-bimodule. We can then consider the set of $(R,S)$-bimodule morphisms between $M$ and ${\mathrm{Hom}}_{T}(N,P)$. This is the set