Atiyah-Macdonald problem 3.8 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 $x/s=0$.
 A: 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.
A: I wonder if this can be done "element-less" by universal arrows. In fact, the arrow $\phi$ is just the arrow given by universal property, since $S\subseteq T$. Then, given any prime $\mathfrak{p}$ such that $\mathfrak{p}\cap S=\varnothing$, and hence $\mathfrak{p}\cap T=\varnothing$, consider the multiplicative part $Q=A-\mathfrak{p}$, and $\bar{Q}$ its image inside $S^{-1}A$: by exercise 3.3 we know that 
$$Q^{-1}A=(SQ)^{-1}A\cong\bar{Q}^{-1}S^{-1}A=(S^{-1}A)_\mathfrak{p}$$
The same holds for $Q$ and $T$:
$$Q^{-1}A=(TQ)^{-1}A\cong\bar{Q}^{-1}T^{-1}A=(T^{-1}A)_\mathfrak{p}$$
And the isomorphisms are the canonical arrows. Thus we get that
$$(S^{-1}A)_\mathfrak{p}\cong (T^{-1}A)_\mathfrak{p}$$
Via the canonical arrows. The arrow is the one induced by $\phi$ localizing, and being locally mono/epi at every prime it must be so at the global level, thus $\phi$ is an iso. 
