Localization commutes with Hom for finitely presented modules I am trying to solve an exercise given in Vakil's Algebraic Geometry notes. Suppose $M$ is a finitely presented $A$-module. The $M$ fits inside an exact sequence $A^q\rightarrow A^p\rightarrow M\rightarrow 0$. I'd like to understand why in this case we get an isomorphism $S^{-1}\text{Hom}_A(M,N)\cong \text{Hom}_{S^{-1}A}(S^{-1}M,S^{-1}M)$. This problem is towards the beginning of the book, so in particular there should be a way to solve it without heavy duty commutative algebra.
So far, I've only come up with the following: We can use the universal property of localization of modules so that for any map from $\text{Hom}_A(M,N)$ to $\text{Hom}_{S^{-1}A} (S^{-1}M,S^{-1}M)$ (in which the elements of $S$ are invertible), there exists a unique map from  $S^{-1}\text{Hom}_A(M,N)$ to $\text{Hom}_{S^{-1}A}(S^{-1}M,S^{-1}M)$.
However, what should this map explicitly be? Is this the way to go about showing these two are isomorphic?
EDIT: There is a question about the same problem, but I am specifically asking about how to construct a map between the two sets. The solution in the related question uses facts about flat modules which I am trying to eschew.

Regarding the problem, solve it for $M=A^p$ and then use the exact sequence. Regarding the map, the universal property of localization tells you what it is: if $f\in \text{Hom}_A(M,N)$, then the image of $f/s$ is the map taking $m/t$ to $f(m)/st$.
