Isomorphism between localizations (2) [duplicate]

Let $f:A\to B$ be a homomorphism of rings, $S$ be a multiplicatively closed subset of $A$ and $T=f(S)$. Then $S^{-1}B$ and $T^{-1}B$ are isomorphic as $S^{-1}A$-modules.

First we define the obvious homomorphism $\phi:S^{-1}B\to T^{-1}B$ by $\frac{b}{s}\mapsto \frac{b}{f(s)}$. I can derive the facts that $\phi$ is a homomorphism as well as surjectivity, but I am unsure about injectivity.

Suppose $\frac{b}{f(s)}=\frac{0}{1}$, then $\exists\ t=f(s')\in T$ such that $tb=f(s')b=0$. And now I am stuck, because what I need is $s''b=0$ for some $s''\in S$, but $f(s')\in T$, so it doesn't really help me.

Solution 1:

Injectivity. Let $\frac{r_1}{s_1},\frac{r_2}{s_2}\in S^{-1}B$, and suppose that \begin{align*} \phi\left(\frac{r_1}{s_1}\right)&=\phi\left(\frac{r_2}{s_2}\right) \\ \frac{r_1}{f(s_1)}&=\frac{r_2}{f(s_2)} \end{align*} So $\exists$ $t\in T$ such that \begin{equation*} t\left(f(s_2)r_1-f(s_1)r_2\right)=0 \end{equation*} Since $t\in T=f(S)$, $\exists$ $s\in S$ such that $f(s)=t$, and \begin{align*} f(s)\left(f(s_2)r_1-f(s_1)r_2\right)&=0 \end{align*} Since $B$ is an $A$ module via $f$ \begin{equation*} s\cdot (s_2\cdot r_1-s_1\cdot r_2)=0 \end{equation*} and thus $\frac{r_1}{s_1}=\frac{r_2}{s_2}$. So $\operatorname{ker}(\phi)=0$.