Let $M$ be a left $R$-module.
(i) Prove that the map $\varphi_{M}: \operatorname{Hom}_{R}(R, M) \rightarrow M$, given by $\varphi_{M}: f \mapsto f(1)$, is an $R$-isomorphism.
Hint. Make the abelian group $\operatorname{Hom}_{R}(R, M)$ into a left $R$ module by defining $r f$ (for $f: R \rightarrow M$ and $r \in R$ ) by $r f: s \mapsto f(s r)$ for all $s \in R .$
(ii) If $g: M \rightarrow N$, prove that the following diagram commutes:
Conclude that $\varphi=\left(\varphi_{M}\right)_{M \in o b j(R M o d)}$ is a natural isomorphism from $\operatorname{Hom}_{R}(R, \square)$ to the identity functor on ${ }_{R} \mathbf{M o d}$.
[Compare with Example 1.16(ii).]