Atiyah-Macdonald Proposition 4.7 We have the following proposition in Atiyah-Macdonald:

Let $\mathfrak{a}$ be a decomposable ideal in a ring $A$, let $\mathfrak{a} = \bigcap_{i=1}^n\mathfrak{q}_i$ be a minimal primary decomposition, and let $r(\mathfrak{q}_i) = \mathfrak{p}_i$. Then
$$
\bigcup_{i=1}^n \mathfrak{p}_i = \{x \in A : (\mathfrak{a}:x) \neq \mathfrak{a}\}.
$$
In particular, if the zero ideal is decomposable, the set $D$ of zero-divisors of $A$ is the union of the prime ideals belonging to $0$.

I am unsure about the proof. I understand that if $\mathfrak{a}$ is decomposible, then the image (which is 0) will be decomposible in $A/\mathfrak{a}$. But I don't see how the conclusion being true in the quotient allows us to lift the result to the whole ring.
If we know the result in the quotient, namely we know that: $\cup \overline{p_i}=\{\overline{x} \in A/\mathfrak{a}: (0:\overline{x})\neq 0\}$ how can we conclude that $\cup p_i=\{x \in A/: (\mathfrak{a}:x)\neq \mathfrak{a}\}$

There is a question that is similarly titled [here] (Proposition 4.7 in Atiyah-Macdonald) but we have different things we are confused about.

I guess an important point is that because $\mathfrak a\subseteq\mathfrak q_i\subseteq\mathfrak p_i$ for each $i$ we have that $\pi^{-1}(\overline{\mathfrak p_i})=\mathfrak p_i$ where $\pi$ is the projection $A\to A/\mathfrak a$ (in general you'll have that $\pi^{-1}(\overline I)=I+\mathfrak a$, so you get equality when $\mathfrak a\subseteq I$).
Now in general a pre-image of a union equals the union of the pre-images, so you get
$$\bigcup_i \mathfrak p_i=\bigcup_i\pi^{-1}(\overline{\mathfrak p_i})=\pi^{-1}(\bigcup_i\overline{\mathfrak p_i})=\pi^{-1}(\{\overline x\in A/\mathfrak a:(0:\overline x)\neq0\}),$$
and it's easy to check this last set is equal to $\{x\in A:(\mathfrak a:x)\neq\mathfrak a\}$.
