Let's suppose $R$ is a ring and $D \subset R$ is a multiplicative subset.
If we look at these two $D^{-1}R$-modules:
$D^{-1}(\operatorname{Hom}_R(M,N))$ and $\operatorname{Hom}_{D^{-1}R}(D^{-1}M,D^{-1}N)$ and consider the naturally defined homomorphism from $D^{-1}\operatorname{Hom}_R(M,N)) \rightarrow$ $\operatorname{Hom}_{D^{-1}R}(D^{-1}M,D^{-1}N)$
is there an example of $R, D, M, N$, when the homomorphism is not an isomorphism?
Set $R = \mathbb{Z}$, let $D = \mathbb{Z} \setminus \{0\}$, let $M = \bigoplus_{i \in \mathbb{Z}} R$ and $N = R$. Then \begin{align} D^{-1}\operatorname{Hom}_R(M,N)) &\simeq \textstyle \mathbb{Q} \otimes_{\mathbb{Z}} (\prod_{i \in \mathbb{Z}} \mathbb{Z}) \\ \operatorname{Hom}_{D^{-1}R}(D^{-1}M,D^{-1}N) &\simeq \textstyle \prod_{i \in \mathbb{Z}} \mathbb{Q} \end{align} where we use that localization commutes with direct sums and that \begin{align} \textstyle \operatorname{Hom}_{R}(\bigoplus_{i \in I} M_{i},N) \to \prod_{i \in I} \operatorname{Hom}_{R}(M_{i},N) \end{align} is an isomorphism for any collection of $R$-modules $\{M_{i}\}_{i \in I},N$. The natural map is not an isomorphism because every element of $\mathbb{Q} \otimes_{\mathbb{Z}} (\prod_{i \in \mathbb{Z}} \mathbb{Z})$ has bounded denominator, whereas elements of $\prod_{i \in \mathbb{Z}} \mathbb{Q}$ can have unbounded denominator.
