I'm stuck at the last step on this problem:
Let $S,T$ be multiplicatively closed subsets of a commutative ring $A$ such that $S\subset T$. Let $\phi:S^{-1}A\rightarrow T^{-1}A$ be the homomorphism which maps each $a/s\in S^{-1}A$ into $a/s$ considered as an element of $T^{-1}A$.
I want to show v) $\Rightarrow$ i) where
v) Every prime ideal which meets $T$ also meets $S$.
i) $\phi$ is bijective.
My approach: We know that the set of zero-divisors $Z$ in $A$ is a union of prime ideals. Then, $$x/s\in\ker(\phi)\Leftrightarrow 0_{T^{-1}A}=\phi(x/s)=x/s$$ so for every $t\in T$ exists a $t'\in T$ with, $$x/s=0/t\Leftrightarrow t'(xt-0s)=0\Leftrightarrow t'tx=0$$ so now I split it in two cases:
a) If $T\cap Z=\emptyset$ and since $tt'\in T$ must be $x=0$ so $x/s=0$.
b) If $T\cap Z\neq\emptyset\Rightarrow S\cap Z\neq\emptyset$, but I don't know how to use it to prove that $x/s=0$.
Injectivity: $a/s=0/1$ in $T^{-1}A$ iff $\exists\ t\in T$ such that $ta=0$. We want to show that $a/s=0/1$ in $S^{-1}A$, that is, there is $s'\in S$ such that $s'a=0$. If not, then $\operatorname{Ann}(a)\cap S=\emptyset$ and let $P$ be a prime ideal of $A$ such that $P\supseteq\operatorname{Ann}(a)$ and $P\cap S=\emptyset$. But $t\in P\cap T$, a contradiction.
Surjectivity: for $t\in T$ there is $a\in A$ such that $at\in S$. Otherwise, $Rt\cap S=\emptyset$ and therefore there is a prime ideal $P\supseteq RT$ such that $P\cap S=\emptyset$. But then $P\cap T=\emptyset$, a contradiction.