I have found this online notes from Columbia University. However, I have a question (maybe very silly) about the proof.
It says that by the universal property, the map $f'$ induces a unique $S^{-1}R$-module homomorphism. But should it be an $R$-module homomorphism?
I think maybe I messed up with the universal property. I always assume that if we have an $R$-bilinear map from $M\times N$ to $L$, then it induces an $R$-module homomorphism from $M\otimes_RN$ to $L$. How could we induce a $S^{-1}R$ homomorphism if we have $\otimes_R$ instead of $\otimes_{S^{-1}R}$?
PS It says that the map $f'$ is bilinear. In the sense of $S^{-1}R$-bilinear or $R$-bilinear? I suppose it should be the latter, is it?
Thank you very much in advance!
You are 100% right that the map induced by the universal property is, a priori, just an $R$-linear map.
However, the following lemma shows it doesn't matter:
Lemma. Let $f: M \to N$ be an $R$-module map, where $M$ and $N$ are $S^{-1}R$-modules. Then in fact $f$ is $S^{-1}R$-linear.
This is because $s f(\frac{r}{s} m) = rf(m) $ in $N$, so, multiplying both sides by $1/s$ gives that $f$ is $S^{-1}R$-linear.