An emailed question asked I’m currently revising Commutative Algebra and I am going through the proof of proposition 5.4 in the lecture notes, which proves that if $0 \rightarrow M_1 \rightarrow M_2 \rightarrow M_3 \rightarrow 0$ is a short exact sequence, then $M_2$ is Artinian if and only if $M_1$ and $M_3$ are both Artinian. We have a short exact sequence so we know that $\alpha$ is injective and $\beta$ is surjective and $Im(\alpha)=Ker(\beta)$. During the proof it refers to $\alpha^{-1}$ and $\beta^{-1}$, but surely it only makes sense to talk about inverses if the maps are bijective? For example if $M_1$ was a submodule of $M_2$ and $\alpha$ was the inclusion map, then it wouldn’t make sense to talk about $\alpha^{-1}$ because elements in $M_2\setminus M_1$ don’t have a preimage in $M_1$ (obviously in this case a submodule would clearly be Artinian but still, the inverse doesn’t exist). Also in the proof of Theorem 4.12, I presume the maps $\alpha$ and $\beta$ are the localization maps i.e $\alpha(t)=t/1 \in A_p$. During the proof it refers to $\alpha^{-1}$ and $\beta^{-1}$, but again I’m not sure these have inverses because what is the image under $\alpha{^-1}$ of $r/s$? Perhaps you could tell me where I am going wrong. to which I replied If $f$ is any function from a set $A$ to a set $B$ and $C\subseteq B$, then we can define $f^{-1}(C)$ to be the subset of $A$ consisting of those $a \in A$ such that $f(a) \in C$ (and not just $\in B$). This is standard mathematical notation for what is known as the "inverse image of $C$ under $f$", and using it does not imply that $f$ is invertible. A follow-up question follwed: So in the case of Proposition 5.4, say $N_k$ is a submodule of $M_3$, then $\beta^{-1}(N_k)$ is the elements of $M_2$ which map into $N_k$. So assuming you’ve proved that these elements are a submodule, then the chain $\beta^{-1}(N_1)$, $\beta^{-1}(N_2)\dots$ terminates in $M_2$, so $\beta^{-1}(N_k)= \beta^{-1}(N_{k+1})\dots$ for some $k$. And because $\beta$ is surjective it follows that $\beta(\beta^{-1}(N_k))=N_k= \beta(\beta^{-1}(N_{k+1}))=N_{k+1}\dots$, so the chain in $M_3$ terminates. Is that correct? to which my reply was: Yes, you are completely correct. And it's not hard to see that the inverse image of a submodule (of the codomain) is a submodule of the domain.