prove that $Hom_R (_RM_S,_RN_T )$ is an $(S, T)$-bimodule. Let $R, S$ and $T$ be rings, $M$ an $(R, S)$-bimodule  and  $N$  an $(R, T)$-bimodule. 
If we define $sf$  and  $ft$ by  setting
$(sf) (m) = f (ms)$
and
$(ft) (m) = f (m) t$
for all  $f$  in  $\mathrm{Hom}_R (_RM_S,_RN_T )$ and $t\in T$, and $s\in S$, then the abelian group
Then prove that
$\mathrm{Hom}_R (_RM_S,_R N_T )$ is an $(S, T)$-bimodule.
 A: *

*$\mathrm{Hom}_R (_RM_S,_RN_T )$ is a left $S$-module.


The map $\cdot_S:S\times \mathrm{Hom}_R (_RM_S,_RN_T )\rightarrow \mathrm{Hom}_R (_RM_S,_RN_T )$ $s\times f\mapsto s\cdot_S f$, with 
$$(s\cdot_S f)(m):= f(ms)$$
endows $\mathrm{Hom}_R (_RM_S,_RN_T )$ with a left $S$-module structure as
$$(s_1s_2)\cdot_S f=s_1\cdot_S(s_2\cdot_S f); $$
this relation can be proved by applying both sides on any $m\in M$ and using the right $S$-module structure on $M$. However, we still need to prove that $s\cdot_S f$ is well defined, i.e.  $s\cdot_S f\in \mathrm{Hom}_R (_RM_S,_RN_T )$. We show that $s\cdot_S f$ is a left $R$-linear map. As
$$(s\cdot_S f)(rm):= f((rm)s)=(M~\text{is a bimodule})=f(r(ms))=(f~ \text{  is}~ R~ \text{linear})=rf(ms)=r(s\cdot_S f)(m),$$
we are done.


*

*$\mathrm{Hom}_R (_RM_S,_RN_T )$ is a right $T$-module.


The map $\cdot_T \mathrm{Hom}_R (_RM_S,_RN_T )\times T \rightarrow \mathrm{Hom}_R (_RM_S,_RN_T )$ $f\times t\mapsto  f\cdot_T$, with 
$$( f\cdot_T t)(m):= f(m)t$$
endows $\mathrm{Hom}_R (_RM_S,_RN_T )$ with a right $T$-module structure as
$$f\cdot_T (t_1t_2) =(f\cdot_T t_1)\cdot_T t_2; $$
this can be proved by applying both sides on any $m\in M$ and using the right $T$-module structure on $N$. Moreover $ f\cdot_T t\in \mathrm{Hom}_R (_RM_S,_RN_T )$: $R$-linearity can be prven by using the bimodule structure on $N$, as we did in the above subsection with due changes.


*

*$\mathrm{Hom}_R (_RM_S,_RN_T )$ is an $S-T$-bimodule.


We need to prove that the left and right module structures on $\mathrm{Hom}_R (_RM_S,_RN_T )$ are compatible, i.e.
$$(s\cdot_S f)\cdot_T t=s\cdot_S(f\cdot_T t), $$
for all $s\in S$, $f\in \mathrm{Hom}_R (_RM_S,_RN_T )$ and $t\in T$. Let us show it; for all $m\in M$ we have
$$((s\cdot_S f)\cdot_T t)(m)=(\text{def. of right $T$-module structure})(s\cdot_S f)(m)t
=(\text{def. of left $S$-module structure})=f(ms)t;$$
on the other hand
$$(s\cdot_S (f\cdot_T t))(m)=(\text{def. of left $S$-module structure}=(f\cdot_T t)(ms)=(\text{def. of right $T$-module structure})=f(ms)t.$$
We are done.
