I am wondering if this is valid reasoning:
Let $N$ and $N'$ be $R$-modules. I know that $0\rightarrow N\rightarrow N'$ is exact and want to show for a given multiplicative subset $S$ and that $0\rightarrow (S^{-1}R)\otimes_R N\rightarrow (S^{-1}R)\otimes_R N'$ is exact. Is valid to say that I if I show that that $ (S^{-1}R)\otimes_R N$ is isomorphic to $S^{-1}(R\otimes_R N)$, that because I know that $0\rightarrow R\otimes N\rightarrow R\otimes N'$ is exact, and therefore $0\rightarrow S^{-1}(R\otimes_R N)\rightarrow S^{-1}(R\otimes_R N')$ is exact, so by the isomorphism $0\rightarrow (S^{-1}R)\otimes_R N\rightarrow (S^{-1}R)\otimes_R N'$ is exact.
(I guess my question is when I say by isomorphism is this OK. Can I mess up the functions associated with the arrows?)