Let $\phi:R\rightarrow S$ be a homomorphism and $S$ an $R-$algebra. Moreover, let $T\subseteq R$ be a multiplicative group. Prove that $T^{-1}S$ is a $T^{-1}R$ algebra and that $T^{-1}S\simeq \phi(T)^{-1}S$.
I was able to prove the first part (straight from definition). The other part is where I'm having problems. I defined $\varphi:T^{-1}S\rightarrow \phi(T)^{-1}S$, $\varphi(\frac{s}{t})=\frac{s}{\phi(t)}$. This is obviously a surjective homomorphism but I wasn't able to prove that its $1-1$. I assumed $\frac{s}{\phi(t)}=\frac{s'}{\phi(t')}$, therefore there exists $\phi(m)$ s.t $\phi(m)(\phi(t')s-\phi(t)s')=0$ but wasn't able to take it from there.
Another approach I tried was showing that both $T^{-1}S$ and $\phi(T)^{-1}S$ are solutions to the universal property of the same localization problem, but that didn't work out either.
Any help would be appreciated.
What does $\frac{s}{t} = \frac{s'}{t'}$ mean? That there exists $m \in T$ such that $m \cdot (t' \cdot s - t \cdot s') = 0$. Here the $\cdot$ denotes the $R$-action on $S$. But by definition $r \cdot s = \phi(r)s$, so this equation is simply a restatement of $\phi(m)(\phi(t')s - \phi(t)s') = 0$.