Module Structure Induced by Quotient Ring

Question

Proposition 7.7.15: Let $\mathfrak{a}$ be a two-sided ideal in the ring $A$. Then an $A$-module $M$ admits an $A/\mathfrak{a}$-module structure compatible with the given $A$-module structure if and only if $\mathfrak{a} \subseteq \mbox{Ann}(M)$. Such an $A/\mathfrak{a}$-module structure is unique if it exists. Finally, if $M$ and $N$ are $A/\mathfrak{a}$-modules, then every $A$-module homomorphism from $M$ and $N$ is an $A/\mathfrak{a}$ module homomorphism as well

Here is the proof my book offers of the bolded part:

Two $A/\mathfrak{a}$-module structures that induce the same $A$-module structure must be equal, because $\pi : A \to A/\mathfrak{a}$ is surjective.

I don't think I understand this reasoning. Does the proof say something like, if $\phi : A/\mathfrak{a} \times M \to M$ induces the same $A$-module structure as $\pi$, then they are the same? How are they the same? I don't understand...

Answer 1

Well, suppose you have two $A/\mathfrak{a}$-module structures $\phi,\phi':A/\mathfrak{a}\times M\to M$. The induced $A$-module structures are $\psi,\psi':A\times M\to M$ defined by $\psi(a,m)=\phi(\pi(a),m)$ and $\psi'(a,m)=\phi'(\pi(a),m)$. The claim is that if $\psi=\psi'$, then $\phi=\phi'$. This is immediate from the fact that $\pi$ is surjective: for any $(b,m)\in A/\mathfrak{a}\times M$ we can choose $a\in A$ such that $\pi(a)=b$ and then $$\phi(b,m)=\psi(a,m)=\psi'(a,m)=\phi'(b,m).$$

Answer 2

An $A/\mathfrak a$-module structure on $M$ is provided by a ring homomorphism $\bar\varrho:A/\mathfrak a\to\operatorname{End}(M)$. Let $\varrho:A\to\operatorname{End}(M)$ be the ring homomorphism which defines the $A$-module structure on $M$. The two module structures are compatible if and only if the following diagram commutes:

where $\pi$ is the projection onto the factor ring. Since $\pi$ is surjective (hence epic), there exists one and only one ring homomorphism $\bar\varrho:A/\mathfrak a\to\operatorname{End}(M)$ such that $\varrho=\bar\varrho\circ\pi$.