$\tau: S^{-1} (\operatorname{Hom}_A(M, N)) → \operatorname{Hom}_{S^{-1}A}(S^{-1}M, S^{-1}N)$ is injective if $M$ is a finitely generated $A$-module 
Let $A$ be a ring.  Let $M$ and $N$ be two $A$-modules, $S$ is a multiplicatively closed subset of $A$. Show that there is a homomorphism of $S^{-1}A$-modules $$\tau: S^{-1} (\operatorname{Hom}_A(M, N)) → \operatorname{Hom}_{S^{-1}A}(S^{-1}M, S^{-1}N)$$ and this is injective if $M$ is a finitely generated $A$-module. Is it true if $M$ is not finitely generated?

To give an "counter example" to the claim in case $M$ is not finitely generated I guess one can take $A=\mathbb{Z}, S=\mathbb{Z}\backslash \{0\}, N = M = \bigoplus_{n\geq 2}\mathbb{Z}/n\mathbb{Z}$, but I am not sure how to show that the map is not injective. For proving the main claim I have no clue. Can anyone give me some hints? Thank you in advance.
 A: Hint. In your example $S^{-1}M=0$, but $S^{-1}\operatorname{Hom}_A(M,N)\ne0$.
A: This is partially an answer for your question: 
If we consider the additional condition that $A$ is noetherian, we'll be able to prove that $\tau$ is injective (in fact under this condition it is possible to show that $\tau$ is $S^{-1}A$-isomorphism):
Since under new hypothesis, $A$ is noetherian and $M$ is a finitely generated $A$-module we know that the following exact sequence exists where $K_{0}$ is finitely generated:
\begin{equation}
0\longrightarrow K_{0}\longrightarrow F_{0}\longrightarrow M\longrightarrow 0
\end{equation}
where $F_{0}$ is a finitely generated free $A$-module.
Now, by constructing the following free resolution for M:
\begin{equation}
 F_{1}\longrightarrow F_{0}\longrightarrow M\longrightarrow 0
\end{equation} 
where $F_{1}$ is a finitely generated free $A$-module, we'll have the following diagram:
\begin{array}
 00 & {\rightarrow} & S^{-1}Hom_{A}(M, N) & {\rightarrow} & S^{-1}Hom_{A}(F_{0}, N) & {\rightarrow} & S^{-1}Hom_{A}(F_{1}, N) \\
 & & \downarrow{\tau} & & \downarrow{\alpha_{2}} & & \downarrow{\alpha_{1}} \\
0 & {\rightarrow} & Hom_{S^{-1}A}(S^{-1}M, S^{-1}N) & {\rightarrow} & Hom_{S^{-1}A}(S^{-1}F_{0}, S^{-1}N) & {\rightarrow} & Hom_{S^{-1}A}(S^{-1}F_{1}, S^{-1}N)   
\end{array}
where $\alpha_{1}$ and $\alpha_{2}$ are isomorphisms, since both $F_{0}$ and $F_{1}$ are free $A$-modules.
Now by expanding the diagram in the following from:
\begin{array}
 00 & {\rightarrow} & 0 & {\rightarrow} & S^{-1}Hom_{A}(M, N) & {\rightarrow} & S^{-1}Hom_{A}(F_{0}, N) & {\rightarrow} & S^{-1}Hom_{A}(F_{1}, N) \\
\downarrow{\alpha_{4}} & & \downarrow{\alpha_{3}} & & \downarrow{\tau} & & \downarrow{\alpha_{2}} & & \downarrow{\alpha_{1}} \\
0 & {\rightarrow} & 0 & {\rightarrow} & Hom_{S^{-1}A}(S^{-1}M, S^{-1}N) & {\rightarrow} & Hom_{S^{-1}A}(S^{-1}F_{0}, S^{-1}N) & {\rightarrow} & Hom_{S^{-1}A}(S^{-1}F_{1}, S^{-1}N)   
\end{array}
where $\alpha_{4}$ and $\alpha_{3}$ are zero homomorphisms.
Now according to one of the theorm in theory of modules (if you don't know this theorem ask about it and I'll explain more!), since $\alpha_{2}$ and $\alpha_{3}$ are injective and $\alpha_{4}$ is surjective, $\tau$ is injective.
