There is a proof I don't get in Lang's Algebra:
(page 423) Theorem 3.3. Let $M$ be a Noetherian $A$-module. Let $N$ be a submodule. Then N admits primary decomposition.
Proof. We consider the set of submodules that do not admit a primary decomposition. If this set is not empty, then it has a maximal element because M is Noetherian. Let N be this maximal element. Then N is not primary and there exists $a \in A$ such that $a_{M / N}$ (This is $x \mapsto ax$ for $x \in M/N$) is neither injective nor nilpotent. The increasing sequence of modules $\ker a_{M / N} \subset \ker a^2_{M / N} \subset \ker a^3_{M / N} \subset \cdots$ stops, say at $a^r_{M / N}$. Let $\phi : M/N \to M/N$ be the endomorphism $\phi = a^r_{M / N}$. Then $\ker \phi = \ker \phi^2$. Hence $0 = \ker \phi \cap \text{im} \phi$ in $M/N$ and neither the kernel nor the image of $\phi$ is $0$. Taking the inverse image in M we see that N is the intersection of two submodules of M, unequal to N. We conclude, from the maximality of N, that each one of these submodules admits a primary decomposition, and therefore N admits one also, contradiction.
The sentence in bold is what I don't understand. The inverse image of what? Is there the canonical homomorphism involved somehow? And how would an inverse image show such a thing at all?