Is every $S^{-1}A$-module homomorphism $f:N \to N'$ induced by an $A$-module homomorphism?

Question

Let $A$ be a commutative ring, $S \subseteq A$ a multiplicatively closed set and $S^{-1}A$ its localisation. Let $N$, $N'$ be $S^{-1}A$-modules and let $f:N \to N'$ be an $S^{-1}A$-module homomorphism.

So I previously asked if for every $S^{-1}A$-module $M$, there existed an $A$-module $\tilde{M}$ such that $S^{-1}\tilde{M} \cong M$. I was answered that in fact it did, in a very natural way, where the $A$-module $\tilde{M}$ is basically $M$ with the $A$-multiplication induced by the $S^{-1}A$-multiplication.

My question is: given $f:N \to N'$, the $S^{-1}A$-module homomorphism, there exist $\tilde{f}:\tilde{N} \to \tilde{N'}$ such that $S^{-1}\tilde{f}=f$?

When going in the other direction, I have that if $\tilde{M}$, $\tilde{M'}$ are $A$-modules and $\tilde{g}:\tilde{M} \to \tilde{M'}$ is $A$-linear, then \tilde{g} induces an $S^{-1}A$-module homomorphism, $S^{-1}\tilde{g}:S^{-1}\tilde{M} \to S^{-1}\tilde{M'}$ defined by $S^{-1}\tilde{g}(\frac{m}{s})=\frac{\tilde{g}(m)}{s} \in S^{-1}M'$.

Now, for my preceeding question, I have my $S^{-1}A$-module homomorphism $f$ such that $f(\frac{n}{s})=\frac{1}{s}f(\frac{n}{1})$ where I used $S^{-1}A$ linearity. However Im not sure that $f$ in this case, is necessairily induced by the $\tilde{f}:\tilde{N}\to \tilde{N'}$ which $\textit{would}$ be $S^{-1}\tilde{f}:S^{-1}\tilde{N} \to S^{-1}\tilde{N'}$ defined by $S^{-1}\tilde{f}(\frac{n}{s})=\frac{f(n)}{s}$. Putting it in other words, can I be sure that

$$f(\frac{n}{s}) =\frac{1}{s}f(\frac{n}{1})=\frac{f(n)}{s}=S^{-1}\tilde{f}(\frac{n}{s})$$

Answer 1

If $f\colon N\to N'$ is a homomorphism of $S^{-1}A$-modules, it is also a homomorphism of $A$-modules and it gets unchanged when you apply $S^{-1}$, because it is $S^{-1}A$-linear to begin with.

So this is essentially trivial.

I'm not saying that whenever $N=S^{-1}M$ and $N'=S^{-1}M'$, any homomorphism of $S^{-1}A$-modules $g\colon N\to N'$ is of the form $S^{-1}f$ for some $f\colon M\to M'$, though.

For instance, take $A=\mathbb{Z}$ and $S=\mathbb{Z}\setminus\{0\}$, so $S^{-1}\mathbb{Z}=\mathbb{Q}$. It is true that $S^{-1}\mathbb{Z}=\mathbb{Q}$, but it is not true that any $\mathbb{Q}$-module homomorphism $\mathbb{Q}\to\mathbb{Q}$ is of the form $S^{-1}f$, for some $f\colon\mathbb{Z}\to\mathbb{Z}$.

The argument in the first paragraph works because we are considering $M=N$ and $M'=N'$.