Let $A$ be a commutative ring and $S\subseteq A$ a multiplicatively closed subset. Let $M$ be an $A$-module. We can form the rings/modules of fractions $S^{-1}A$ and $S^{-1}M$ respectively. See Atiyah Commutative Algebra page 38-39.
In the book, proposition 3.3 reads
The operation $S^{-1}$ is exact, i.e., if $M^{\prime} \stackrel{f}{\rightarrow} M \stackrel{g}{\rightarrow} M^{\prime \prime}$ is exact at $M$, then $S^{-1} M^{\prime} \stackrel{S^{-1 f}}{\longrightarrow} S^{-1} M \stackrel{S^{-1} g}{\longrightarrow} S^{-1} M^{\prime \prime}$ is éxact at $S^{-1} M$.
The author proves it directly by investigating the kernels and images. I wonder whether this can be proved functorially, using the fact that left adjoints are right exact, and right adjoints are left exact.
Clearly $S^{-1}$ defines a functor from the category of $A$-modules to the category of $S^{-1}A$-modules. For the other direction, recall that we have a canonical map $\psi:A\to S^{-1}A$, which sends $a\in A$ to the class $a/1$. Hence, $\psi$ turns any $S^{-1}A$-module to an $A$-module, by restriction of scalars, i.e., define $a\cdot n=\psi(a)n$. This defines a functor $\psi^*$ going the opposite direction. Would this functor be the left/right/both adjoints to $S^{-1}$? If so, this would prove the claim; if not, how can I find the left/right adjoints to $S^{-1}$?
I suspect that $\psi^*$ would be right adjoint to $S^{-1}$, but maybe not left adjoint. I'm not super sure. Any help would be appreciated!