Canonical isomorphism between Cauchy sequence completion and inverse limit

Question

I'm studying chapter 10 of Atiyah Macdonald. The book introduces two ways to construct the completion of an abelian topological group: Equivalence classes of Cauchy sequences and inverse limit. I can see how these two are isomorphic as groups. However, the book doesn't explain how they're topologically equivalent (homeomorphic) and I'm unable to fill in the details. I'm looking for a reference that does so.

I've checked a few references. Most details are skipped. (Is it really that easy? I can't see it.) There are also many variations: Some take equivalence classes of Cauchy sequences modulo null sequences. The topology defined on the completion can also be expressed differently.

Here is how the book defines a Cauchy sequence in an abelian topological group:

$ (x_n) $ is a Cauchy sequence if for each neighbourhood $ U $ of $ 0 $, there is $ N $ such that $ n, m > N $ means $ x_n - x_m \in U$.

Given the definition above, I prefer a reference that doesn't use nets or metrics.

Given a chain of subgroups $G = G_0 \supset G_1 \supset G_2 \supset ...$, the book gives $ G $ the topology in which $\langle G_{n} \rangle $ is a neighbourhood basis of $ 0 $. The inverse limit considered is then $ \varprojlim_n G / G_n $.

Thanks

Answer 1

Outline of the proof.

Let $C$ be the group of Cauchy sequences (without the equivalence classes.)

If $\mathbf{a}=\{a_i\}$ is Cauchy, define $N_k(\mathbf {a})$ the be the least value so that for all $i,j\geq N_k(\mathbf a),\ a_i-a_j\in G_k$. This can be written as $a_i+G_k=a_j+G_k$ in $G/G_k$. Then define $\phi_k:C\to G/G_k$ by $\{a_i\}\to a_{N_k(\mathbf{a})}+G_k$. Show that this is well-defined, and map satisfies the inverse limit criterion. If $p_k:G/G_{k+1}\to G/G_k$ we need to show:

$$p_k\circ\phi_{k+1}=\phi_k$$

So the universal property shows that there is a homomorphism $\phi: C\to\varprojlim_n G / G_n$.

The next step is the to prove that the kernel is exactly the Cauchy sequences that are considered zero in your definition. I think that's easy.

So this shows an isomorphism of groups between these two constructs, but then you need to prove that they have the same open sets.

Answer 2

The topological group $\widetilde{G} := \varprojlim_n G/G_n$ is uniquely determined by the following universal property:

Let $B$ be an arbitrary topological group with continuous homomorphisms $p_n : B \to G/G_n$ such that all obvious triangles commute, i.e. $c_n \circ p_{n+1} = p_n$ for all $n$, where $c_n$ denotes the canonical (continuous!) homomorphism $G/G_{n+1} \to G/G_n$. Then there exists exactly one continuous homomorphism $\varphi : B \to \widetilde{G}$ such that again all obvious triangles commute, i.e. $d_n \circ \varphi = p_n$ for all $n$, where $d_n$ denotes the canonical (continuous!) homomorphism $\widetilde{G} \to G/G_n$.

You should probably draw these diagrams since the statement becomes very natural then. I would have done this, if xymatrix or tikz-cd was available here.

Now it is easy to show that $\widehat{G}$ (which denotes the group of Cauchy sequences of $G$ modulo zero sequences) satisfies this property and is therefore isomorphic to $\widetilde{G}$ as a topological group (in particular the isomorphism is a homeomorphism).

First of all we have canonical homomorphisms $d_n : \widehat{G} \to G/G_n$ which are easily seen to be continuous as $G/G_n$ is a discrete space and $d_n^{-1}(0) = \widehat{G_n}$ is open in $\widehat{G}$ for all $n$.

Let $B$ be an arbitrary topological group with homomorphisms $p_n$ as above. Consider an arbitrary element $x \in B$ and let $x_n \in G$ be a representative of $p_n(x)$. Then $(x_n)_{n \geq 1}$ is a Cauchy sequence in $G$ and any other choice of representatives leads to an equivalent sequence up to zero sequences. So we get a well defined map $\varphi : B \to \widehat{G}$ which is easily seen to be a homomorphism which satisfies the above properties, and is uniquely determined by them.

So it suffices to show that $\varphi$ is continuous. Since the topology of $\widehat{G}$ is given by the filtration $\widehat{G_1} \geq \widehat{G_2} \geq \widehat{G_3} \geq \dots$ we only have to show that $\varphi^{-1}(\widehat{G_n})$ is open in $B$ for all $n$. By looking at the definition we see that $\varphi^{-1}(\widehat{G_n}) = \bigcup_{m \geq n} p_m^{-1}(G_n/G_m)$. This set is open since all $p_m$ are continuous.