Showing that the natural map into the completion is continuous

Question

Let $M$ be an $A$-module and $M=M_0 \supset M_1 \supset \cdots$ a sequence of submodules, which we define to be a fundamental system of neighborhoods of $0$. Thus we make $M$ into a topological group. Define $\hat{M}$ to be the completion of $M$, i.e. the inverse limit of the system $\left\{M/M_n\right\}$. We give to $M/M_n$ the discrete topology, $\prod M/M_n$ the product topology and $\hat{M}$ the subspace topology. Then we have a natural map $\psi: M \rightarrow \hat{M}$ such that $\psi(x)=(x + M_n)_n$.

Question: How can i show that $\psi$ is continuous? Effort: It is enough to show that for any neighborhood $\hat{N}_{\psi(x)}$ of $\psi(x)$, $\psi^{-1}(\hat{N}_{\psi(x)})$ is a neighborhood of $x$. But because $\hat{M}$ has the discrete topology, $\psi(x)$ itself is a neighborhood of $\psi(x)$. How do i compute $\psi^{-1} (\psi(x))$ and subsequently show that it is a neighborhood of $x$?

If you define $\hat{M}$ via the Cauchy sequences approach, then continuity of $M\to\hat{M}$ is proven in the paragraph before Remark 4 here. — Elías Guisado Villalgordo, Sep 11 '23 at 10:15

score 1 · Accepted Answer · answered Mar 19 '13 at 14:06

1

$\tilde{M}$ is not discrete.

Using the universal properties of subspaces and products (or directly of projective limits) in $\mathsf{Top}$, it is enough to prove that each projection $M \to M/M_n$ is continuous. This means that for every $x \in M$ the preimage of $[x]$, i.e. $x + M_n$, is open. But this holds by definition of the topology on $M$.

answered Mar 19 '13 at 14:06

Martin Brandenburg

163,620

Why is $\hat{M}$ not discrete? It has the subspace topology of a discrete space...What am i missing here? – Manos Mar 19 '13 at 14:12
Infinite products of discrete spaces are not discrete. For example, ${0,1}^{\mathbb{N}}$ is the Cantor set. – Martin Brandenburg Mar 19 '13 at 14:20
I see, thanks +1 – Manos Mar 19 '13 at 14:23

Showing that the natural map into the completion is continuous

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