We call a functor $F: \mathcal{A} \rightarrow \mathcal{B}$ a strong isomorphism if there exists a functor $G: \mathcal{B} \rightarrow \mathcal{A}$ with $G F=1_{\mathcal{A}}$ and $F G=1_{\mathcal{B}}$. If $R$ is a ring, show that $\operatorname{Hom}_{R}(R, \square):{ }_{R} \operatorname{Mod} \rightarrow{ }_{R}$ Mod (which is naturally isomorphic to $1_{R \text { Mod }}$, by Exercise $2.13$ on page 66 ) is not a strong isomorphism. Conclude that strong isomorphism is not an interesting idea.