Canonical isomorphism between Cauchy sequence completion and inverse limit 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
 A: 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.
A: 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.
