Let $R,S,T$ be rings, $_RM_S$ be a $R$-$S$-bimodule and $_RN_T$ a $R$-$T$-bimodule. Besides we have $$H=\text{Hom}_R(M,N).$$ We can turn $H$ into a $S$-$T$-bimodule via $$(s.f)(m):=f(m.s)\quad\text{and}\quad (f.t)(m):=f(m).t \qquad \forall f\in H,m\in M,s\in S,t\in T.$$
This is a calculation I have already done. Now I want to show, that $$\text{Hom}_R(_RR_R,_RN_T)~\cong~_RN_T$$ as $R$-$T$-bimodules.
My idea is, to show that $$\phi:\text{Hom}_R(_RR_R,_RN_T)~\rightarrow~_RN_T,~f\mapsto f(1_R)$$ is a bimodule-isomorphism.
At first, we have to show, that this is a bimodule-homomorphism:
- $\phi(f+g)=f(1_R)+g(1_R)=\phi(f)+\phi(g)\quad\forall f,g\in H$
- $\phi(f.t)=(f.t)(1_R)=f(1_R).t=\phi(f).t\quad\forall f\in H,t\in T$
- $\phi(r.f)=(r.f)(1_R)=f(1_R.r)=...=r.f(1_R)=r.\phi(f)\quad\forall f\in H,r\in R$
My problem is, that I can't fill the $\dots$ in the third point. How would this work?
We have $$1_R\cdot r=r=r\cdot 1_R$$ Since $f$ is a left module homomorphism, $$f(r\cdot 1_R)=r\cdot f(1_R)$$ If I understand correctly this is what you are trying to prove.