Let $R$ denote a (commutative and unital) ring and $S\subset R$ a multiplicative subset. Let $S^{-1}R$ denote the ring of fractions. I now want to consider the canonical projection $\pi: R\rightarrow S^{-1}R, r\mapsto \frac{r}{1}$. We are now supposed to show in our homework that $\pi$ is well-defined and for $s\in S: \pi(s)\in (S^{-1}R)^{*}$.
For well-definedness: Let $r=r'$, then $\pi(r) = \frac{r}{1}$ and $\pi(r') = \frac{r'}{1}$. Since $r=r'\Rightarrow \pi(r)=\pi(r')$. Is this short proof correct for well-definedness?
For $\pi(s)\in (S^{-1}R)^{*}$: I found in Grillet's Book (GTM), "Abstract Algebra", that the inverse is supposed to be $\frac{1}{s}$. Alright, but why doesn't this construction hold for all $r\in R$? I mean that $R$ itself is a multiplicative subset, i.e., $\forall a, b\in R: ab\in R$.
EDIT: @Don Thousand, alright, I see your point. $\frac{r}{1}$ is defined as the set $\{(\tilde r, \tilde s)\in R\times S \mid \exists u\in S: r\tilde su = \tilde r \cdot 1\cdot u = \tilde r u\}$. We also know that $\frac{r'}{1} = \{(\tilde r, \tilde s) \in R\times S\mid \exists u\in S: r' \tilde s u = \tilde r \cdot 1 \cdot u = \tilde r u \} \Rightarrow \frac{r}{1} = \frac{r'}{1}$. This is how I would have proved the well-definedness.
Kind regards, MathIsFun