Understanding the $\mathfrak{a}$-adic completion of an $A$-module as a functor

Question

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I recently read the chapter 10 on Completions in Atiyah-MacDonald. They describe the $\mathfrak{a}$-adic completion $\hat{M}$ of an $A$-module $M$ as the inverse limit of an inverse system. They make a remark on page 109, that the function $M \mapsto \hat{M}$ is the object function of a funtor which is not neccessarily right exact in general.

I want to understand the arrow function of this functor and how composition works.

I'm thinking of an inverse system in the following manner, we start with a category $I$ of ordinals which can be viewed as $$ 0 \rightarrow 1 \rightarrow 2 \rightarrow \cdots $$ An inverse system associated to the $\mathfrak{a}$-adic completion of $M$ is the functor $M/{\mathfrak{a}^{i}}M: I^{op} \rightarrow A-Mod.$ Whose image can be viewed as $$ \cdots \rightarrow M/\mathfrak{a}^{2}M \rightarrow M/\mathfrak{a}M \rightarrow 0 $$ My guess on the arrow function is the following:

Given an arrow $p: M \rightarrow M''$, we can associate the fundamental system of neighborhoods of $M$ $$ M \supseteq \mathfrak{a}M \supseteq \mathfrak{a}^{2}M \supseteq \cdots $$ we can associate a similar fundamental system of neighborhoods on $M''$, namely $$ pM \supseteq p(\mathfrak{a}M) \supseteq p(\mathfrak{a}^{2}M) \supseteq \cdots $$ This gives an inverse system $M''/p(\mathfrak{a}^{i}M)$ associated to $M''$ which can be viewed as $$ \cdots \rightarrow M''/p(\mathfrak{a}^{2}M) \rightarrow M''/p(\mathfrak{a}M) \rightarrow 0 $$

This inverse system has an inverse limit $\hat{M''}$.

Skipping some details, this implies a natural transformation $\tau: M/\mathfrak{a}^{i}M \rightarrow M''/p(\mathfrak{a}^{i}M'')$, which is a map of inverse systems. The natural transformation, gives us a map $\hat{p}: \hat{M} \rightarrow \hat{M}''$.

The issue I have with this guess is that it does not seem to be independent of $M$, for example in the event of a sequence of arrows $M' \rightarrow M \rightarrow M''$. There is the Artin-Rees lemma to consider, but these maps may not be injective. I'm confused.

Answer 1

You're starting from the wrong side. Let $f\colon M\to N$ be a homomorphism of $A$-modules.

You can compose it with the canonical mapping $j\colon N\to \hat{N}$ and show it's continuous when $M$ is endowed with the $\mathfrak{a}$-adic topology. Indeed, if $U$ is an open neighborhood of $0$ in $\hat{N}$, we have that, for some $n$, $j^{-1}(U)\supseteq\mathfrak{a}^nN$. Then $$ (j\circ f)^{-1}(U)=f^{-1}(j^{-1}(U))\supseteq f^{-1}(\mathfrak{a}^nN) \supseteq \mathfrak{a}^nM $$

The universal property of the inverse limit tells you now that you can build a unique (continuous) homomorphism $\hat{f}\colon\hat{M}\to\hat{N}$ such that the diagram $$\require{AMScd} \begin{CD} M @>f>> N \\ @ViVV @VVjV \\ \hat{M} @>\smash{\hat{f}}>> \hat{N} \end{CD} $$ is commutative.

This uniqueness is what also shows that $\widehat{f\circ g}=\hat{f}\circ\hat{g}$, if $g\colon L\to M$ is another homomorphism.