# Topology induced by the completion of a topological group

Let $G$ be an abelian topological group and let $\hat{G}$ be its completion, i.e. the group containing the equivalence classes of all Cauchy sequences of $G$. What exactly is the topology of $\hat{G}$?

• For each neighborhood $N$ of zero in $G$, define a neighborhood $\hat{N}$ in $\hat{G}$ consisting of those equivalence classes for which all sequences in the class are eventually in $N$. This is a base (of neighborhoods of zero) for the new topology. Commented Sep 8, 2012 at 16:06
• Another remark. Unless $G$ is metrizable, you cannot expect the "completion" by sequences to be complete in the sense of uniform space. So normally we would do completion by nets or by filters or similar. Commented Sep 8, 2012 at 16:07
• @GEdgar: How do we know that this $\hat{N}$ will be nonempty? Commented Sep 8, 2012 at 16:11
• It contains many constant sequences. For general $G$ it could happen that every cauchy sequence is eventually constant, so that the sequential completion is nothing new. Commented Sep 8, 2012 at 16:12
• @MSina: Do we really need a [completeness] tag? Commented Mar 3, 2013 at 18:16

formerly a remark

For each neighborhood $N$ of zero in G, define a neighborhood $\hat{N}$ in $\hat{G}$ consisting of those equivalence classes for which all sequences in the class are eventually in $N$. This is a base (of neighborhoods of zero) for the new topology.

• This construction of $\hat N$ is not well-defined. Consider $G = \mathbb{R}$ with its usual topology, $N = (-1,1)$. Then the Cauchy sequences $(1 - 1/2^n)_n$ and $(1 + (-1/2)^n)_n$ are equivalent, but the first one is an element of $\hat N$ while the second one is not. (Thanks to Tim Baumann for bringing this point to my attention and supplying this counterexample.) Commented Jan 15, 2016 at 16:12
• @IngoBlechschmidt... Formulation says "all sequences in the class". So in this example, the class of $(1-1/2^n)$ is not in $\widehat{N}$. Commented Jan 15, 2016 at 16:27
• Ah, okay! Thanks for the quick reply and sorry that I didn't read carefully enough. Commented Jan 15, 2016 at 16:34
• @GEdgar I just want to know what do you mean by "all sequence in the class eventually in $N$",is it if $[x_n] \in \hat{N}$ there is an $k$ such that whenever ${y_n} \in [x_n]$, $y_n \in N$ for all $n \geq k$, or for each element in $[x_n]$ $k$ may be different ? Commented May 14, 2020 at 1:29
• Are you talking about open neighborhoods here ? (If yes, then OK. If no, is it clear that just defining a system of neighborhoods defines a topology?)
– mnr
Commented Dec 15, 2020 at 11:44

$$\def\scrB{\mathscr{B}} \def\scrP{\mathscr{P}} \def\scrN{\mathscr{N}} \def\c{\mathrm{c}}$$I will give a more detailed explanation after GEdgar's answer. Most importantly, it is not immediately obvious that the translation invariant family of subsets obtained from GEdgar's answer induces a topology; some work must be done (see Lemma 3 below). (Many thanks to M W and his explanation on his comments here.) Additionally, I will provide other insights into $$\hat{G}$$: Lemma 7 gives an alternative definition for the topology on the completion, and Lemma 9 provides its universal property.

Given a set $$X$$, denote $$\scrP(X)$$ to the power set of $$X$$.

Definition 1. Let $$X$$ be a set and let $$\scrB:X\to\scrP(\scrP(X))$$ be a function. We say that $$\scrB$$ is a local basis system if for all $$x\in X$$ the following holds:

1. $$\scrB(x)\neq\varnothing$$,
2. For all $$B\in\scrB(x)$$, $$x\in B$$,
3. For all $$B\in\scrB(x)$$ and $$y\in B$$, there is $$C\in\scrB(y)$$ with $$C\subset B$$,
4. For all $$B,C\in\scrB(x)$$ there is $$D\in\scrB(x)$$ such that $$D\subset B\cap C$$.

It not difficult to prove that 3+4 is equivalent to “for all $$B,C\in\scrB(x)$$ and $$y\in B\cap C$$ there is $$D\in\scrB(y)$$ such that $$D\subset B\cap C$$.” Every local basis system is a neighborhood basis system too (a slightly different notion given by properties $$(i)$$-$$(iv)$$ here).

Here's an example: if $$G$$ is some abelian group and $$G\supset G_1\supset G_2\supset\cdots$$ is a filtration by subgroups, then defining $$\scrB:G\to\scrP(\scrP(G))$$ by $$\scrB(x)=\{x+G_n\mid n\geq 1\},\quad x\in G,$$ always gives a local basis system on $$G$$.

Recall that given a topological space $$X$$, a local basis at $$x\in X$$ is a family $$\mathcal{B}$$ of open neighborhoods of $$x$$ such that for all open neighborhoods $$U\subset X$$ of $$x$$ there is $$B\in\mathcal{B}$$ with $$B\subset U$$.

Local basis systems are an alternative way to characterize topologies:

Lemma 2. Let $$X$$ be a set and $$\mathscr{B}:X\to\mathscr{P}(\mathscr{P}(X))$$ be a function. The following are equivalent:

1. The set $$$$\label{eq:top gen SBL} τ_\mathscr{B}=\{U\subset X\mid \forall x\in U,\ \exists B\in\mathscr{B}(x)\text{ with } B\subset U\}$$$$ is a topology on $$X$$ and for all $$x\in X$$, it holds that $$\mathscr{B}(x)$$ is a local basis at $$x\in X$$ for $$(X,τ_\mathscr{B})$$.
2. There is a topology on $$X$$ for which $$\mathscr{B}(x)$$ is a local basis $$x$$ for all $$x\in X$$.
3. $$\mathscr{B}$$ is a local basis system on $$X$$.

Moreover, if these equivalent conditions hold, then $$\tau_\scrB$$ is the unique topology on $$X$$ for which $$\scrB(x)$$ is a local basis at $$x$$, for all $$x\in X$$.

I'll leave the proof to the interested reader (part of it it's done in p.8 here).

Now, we come back to our case of interest: Let $$G$$ be an topological abelian group, and denote $$\hat{G}$$ to its completion. For simplicity reasons (and because I don't know anything about filters and nets), assume $$G$$ is locally countably generated (also known as first-countable space). Then $$\hat{G}$$ is the set of Cauchy sequences in $$G$$ modulo the equivalence relation $$(x_n)\sim(y_n)$$ if and only if $$x_n-y_n\xrightarrow{n\to+\infty}0$$ (one checks this is an equivalence relation). Note that given a Cauchy sequence $$(x_n)$$ in $$G$$, the equivalent Cauchy sequences to $$(x_n)$$ are exactly those of the form $$(x_n+s_n)$$, where $$s_n\to 0$$. Given an open set $$U$$ of $$G$$, we define $$\tag{1}\label{subset} \hat{U}=\{[x_n]\in\hat{G}\mid \text{for all } (y_n)\in[x_n],\text{ we have }y_n\in U\text{ for almost all }n\}$$ (“Almost all” means “all but a finite set.”) The sum on $$\hat{G}$$ is defined by $$[x_n]+[y_n]=[x_n+y_n]$$ (one checks this is well-defined). Now, define $$\scrB:\hat{G}\to\scrP(\scrP(\hat{G}))$$ to be $$\label{neig_sys}\tag{2} \scrB(x)=\{x+\hat{U}\mid U\subset G\text{ is an open neighborhood of zero}\},$$ for $$x\in\hat{G}$$. It is left to prove:

Lemma 3. The function $$\scrB$$ is a local basis system on $$\hat{G}$$.

The topology on $$\hat{G}$$ is then the one obtained from Lemma 2. It readily follows that $$\hat{G}$$ is first-countable.

Proof. Conditions 1 and 2 from Definition 1 are trivial. Let $$(U_n)$$ be a countable local basis of $$G$$ at $$0$$. We can suppose that $$(U_n)$$ is decreasing.

Given any set $$S\subset G$$, define $$\hat{S}$$ as usual ($$\hat{S}$$ is the set of equivalence classes of Cauchy sequences such for each member on the class almost all its terms lie on $$S$$).

$$\underline{1^{\mathrm{st}}\text{ step}}\colon$$ Let $$S\subset G$$ be a subset set and let $$[x_n]\in \hat{S}$$. We claim there is an integer $$N$$ such that $$x_n+U_N\subset S$$ for all $$n\geq N$$. To look for a contradiction, suppose there's no such $$N$$. Then, for each $$N$$, there is $$s_N\in U_N$$ and $$n_N\geq N$$ such that $$x_{n_N}+s_N\not\in S$$. But $$[(x_{n_N})_N]\in \hat{S}$$ (for $$[(x_n)_n]\in\hat{S}$$ and $$n_N\to +\infty$$) and $$s_N\to 0$$, so contradiction. Now we claim that $$[x_n]+\hat{U}_N\subset\hat{S}$$. Let $$[y_n]\in\hat{U}_N$$ and $$s_n\to 0$$. Then $$x_n+y_n+s_n\in x_n+U_N\subset S$$ for all $$n\gg 0$$. That is, $$[x_n]+[y_n]=[x_n+y_n]\in\hat{S}$$.

$$\underline{2^{\mathrm{nd}}\text{ step}}\colon$$ We show 3+4 of Definition 1 by showing the equivalent condition stated after it. Let $$y,z\in \hat{G}$$ and let $$V,W\subset U$$ be open neighborhoods of zero. Let $$x\in (y+\hat{V})\cap(z+\hat{W})$$ (in particular, $$x-y\in\hat{V}$$ and $$x-z\in\hat{W}$$). We claim there is $$k$$ such that $$x+\hat{U}_k\subset (y+\hat{V})\cap(z+\hat{W})$$. To look for a contradiction, suppose that for all $$k$$, we have $$x+\hat{U}_k\not\subset (y+\hat{V})\cap(z+\hat{W})$$. Then at least one of the sets \begin{align*} \{k&\mid x+\hat{U}_k\not\subset y+\hat{V}\}\\ \{k&\mid x+\hat{U}_k\not\subset z+\hat{W}\} \end{align*} is infinite. Suppose it's the first one. Then $$x-y+\hat{U}_k\not\subset\hat{V}$$ for infinitely many $$k$$. But this is a contradicion since $$x-y\in\hat{V}$$ and by first step and since $$x-y+\hat{U}_{k+1}\subset x-y+\hat{U}_k$$. $$\square$$

This finishes the proof that $$\hat{G}$$ has a well-defined topology (hence, answering OP's question). We will answer a couple extra questions.

We remark that $$\hat{G}$$, along with this topology, is a topological group (proven here).

After, we note that the map $$\phi:G\to\hat{G}$$ sending $$x\in G$$ to the class of the sequence constantly $$x$$ is continuous: it suffices to verify that $$\phi$$ is continuous at the identity (see this), so we will show that $$\phi^{-1}(\hat{U})=U$$, where $$U\subset G$$ is an open neighborhood of zero. The containment $$(\subset)$$ is easy. For the converse, let $$x\in U$$ and pick $$s_n\to 0$$. Then $$x+s_n\to x$$ (sum is continuous in $$G$$), whence $$x+s_n\in U$$ for all $$n\gg 0$$; thus, $$\phi(x)\in\hat{U}$$.

Remark 4.

1. $$\hat{G}$$ is always Hausdorff: By Lemma 10.1 of Atiyah, MacDonald, Introduction to Commutative Algebra, we have to show that $$\bigcap_{\substack{U\subset\hat{G}\text{ open}\\ \text{nbhd of }0}}U =\bigcap_{\substack{U\subset G\text{ open}\\ \text{nbhd of }0}}\hat{U}$$ equals $$\{0\}$$. Let $$[x_n]\in\bigcap_{\substack{U\subset G\text{ open}\\ \text{nbhd of }0}}\hat{U}$$. Then $$x_n\in U$$ for all $$n\gg 0$$ and for all $$U\subset G$$ open neighborhoods of zero. That is, $$x_n\to 0$$, whence $$[x_n]=0$$.

2. Denote $$\phi:G\to\hat{G}$$ to the map that sends $$x\in G$$ to the class of the sequence constantly $$x$$. Then $$\phi(G)$$ is dense in $$\hat{G}$$: Since $$\hat{G}$$ is first-countable, it suffices to see that $$\phi(G)$$ is sequentially dense. Let $$x=[x_n]\in\hat{G}$$. Then $$\phi(x_n)\to x$$ in $$\hat{G}$$ iff $$\phi(x_n)-x\to 0$$. To prove the latter, let $$U\subset G$$ be an open neighborhood of zero. To show $$\phi(x_n)-x\in \hat{U}$$ for all $$n\gg 0$$, let $$s_n\to 0$$ and pick $$V\subset G$$ open neighborhood of zero such that $$V+V\subset U$$. Then $$x_m-x_n+s_n\in V+V\subset U$$ for all $$n,m\gg 0$$.

The topology on $$\hat{G}$$ here explained coincides with the one from GEdgar's answer:

Lemma 5. For any subset $$S\subset G$$, we have $$\hat{S}=\widehat{S^\circ}$$, where $$S^\circ$$ is the interior of $$S$$.

Proof. On the one hand, it is clear that $$\hat{S}\supset\widehat{S^\circ}$$, for $$\hat{S}\supset\hat{T}$$ for all subsets $$T\subset S$$. To prove the converse, let $$U_n\subset G$$ be a decreasing countable local basis of $$0$$ and let $$[x_n]\in\hat{S}$$. Then there is $$N$$ such that $$x_n+U_N\subset S$$ for all $$n\geq N$$ (this is the first step of last proof). In other words, $$x_n+U_N\subset S^\circ$$ for all $$n\geq N$$. Let $$s_n\to 0$$. Then $$s_n\in U_N$$ for all $$n\gg 0$$, whence $$x_n+s_n\in x_n+U_N\subset S^\circ$$ for all $$n\gg 0$$. That is, $$[x_n]\in\widehat{S^\circ}$$. $$\square$$

Exercise 6. Let $$x_n$$ be a Cauchy sequence in $$G$$. Let $$k_n$$ be natural numbers such that $$\lim_n k_n=+\infty$$. Then $$(x_{k_n})_n$$ is Cauchy and $$[x_n]=[x_{k_n}]$$ in $$\hat{G}$$.

The following gives an alternative way to define the topology on $$\hat{G}$$.

Lemma 7. For the topology on $$\hat{G}$$ (induced by Lemmas 2 and 3) the set $$\mathcal{B}=\{\hat{V}\mid V\subset G\text{ open}\}$$ is a basis.

Proof. Note that $$\mathcal{B}$$ is an abstract basis, for $$\bigcap_{i=1}^n\hat{V}_i=\widehat{\bigcap_{i=1}^nV_i}$$.

Let $$V\subset G$$ be open. Let $$x\in\hat{V}$$. As explained in the proof of Lemma 3, first step, there is an open neighborhood of zero $$U\subset G$$ with $$x+\hat{U}\subset\hat{V}$$.

Conversely, let $$[x_n]\in\hat{G}$$, suppose that $$W\subset G$$ is an open neighborhood of zero and pick $$[y_n]\in [x_n]+\hat{W}$$. We have to find an open $$V\subset G$$ with $$[y_n]\in\hat{V}\subset [x_n]+\hat{W}$$. Equivalently, such that $$\tag{3}\label{eq} [y_n]-[x_n]\in\hat{V}-[x_n]\subset\hat{W}.$$

As usual, suppose $$(U_n)$$ is a countable and decreasing local basis at $$0\in G$$.

$$\underline{\text{Claim i}}\colon$$ For each $$K$$ there is $$M(K)$$ such that $$[y_n]\in\widehat{y_m+U_K}$$ for all $$m\geq M(K)$$: Pick $$K'$$ such that $$U_{K'}+U_{K'}\subset U_K$$ (existence of such $$K'$$ can be deduced from the continuity of the sum in $$G$$). Pick an $$M$$ such that $$y_n-y_m\in U_{K'}$$ for all $$n,m\geq M$$. Pick $$s_n\to 0$$. Then, for $$m\geq M$$, $$y_n-y_m+s_n\in U_{K'}+U_{K'}\subset U_K,\quad \forall n\gg 0,$$ i.e., $$y_n+s_n\in y_m+U_K$$ for all $$n$$ big enough and $$m\geq M$$. Hence, $$[y_n]\in\widehat{y_m+U_K}$$.

$$\underline{\text{Claim ii}}\colon$$ There is $$N$$ such that $$y_m+U_K-x_n\subset W$$ for all $$n,m,K\geq N$$. Suppose not. Then, for each $$N$$, there are $$n_N,m_N,K_N$$ and $$s_N\in U_{K_N}$$ such that $$\lim_N\min(n_N,m_N,K_N)=+\infty$$ and $$y_{m_N}-x_{n_N}+s_N\not\in W$$. Since $$s_{N}\xrightarrow{} 0$$ and by exercise 6 we have $$[(y_{m_N}-x_{n_N})_N]=[(y_{m_N})_N]-[(x_{n_N})_N]=[(y_m)_n]-[(x_n)_n]\in\hat{W}$$, we arrive a contradiction.

$$\underline{\text{Claim iii}}\colon$$ There is $$N$$ such that $$\widehat{y_m+U_K}-[x_n]\subset\hat{W}$$ for all $$m,K\geq N$$. Pick the $$N$$ of claim ii. Let $$m\geq N$$ and pick $$[z_n]\in\widehat{y_m+U_K}$$. Let $$s_n\to 0$$. Then $$z_n+s_n-x_n\in y_m+U_K-x_n\subset W$$, for all $$n\gg 0$$, by claim ii. In other words, $$[z_n]-[x_n]\in\hat{W}$$, and we win.

We now can prove \eqref{eq}. Let $$N$$ be as in claim iii and pick $$K\geq N$$ and $$m\geq\max(K,M(K))$$, where $$M(K)$$ comes from claim i. Then $$[y_n]-[x_n]\in\widehat{y_m+U_K}-[x_n]\subset\hat{W}$$. $$\square$$

Next, we will investigate the universal property of $$\hat{G}$$.

Definition 8. A topological abelian group $$G$$ is sequentially complete if it is Hausdorff and every Cauchy sequence converges. In the case $$G$$ is first-countable we will simply say that it is complete if it is sequentially complete.

Note that the Hausdorff condition provides uniqueness of limit for (Cauchy) sequences. The reason for the terminology is because the term ‘complete’ is reserved for a different notion when the group is not first-countable (see Wikipedia or Bourbaki, General Topology, II.3).

The first-countable group $$\hat{G}$$ is complete: in Remark 4.1 we argue the Hausdorff property and convergence of Cauchy sequences is proven here.

Lemma 9 (Universal property of the completion of a first-countable abelian topological group). Let $$G$$ be a first-countable abelian topological group. The map $$\phi:G\to\hat{G}$$ is universal among morphisms $$G\to H$$ of topological abelian groups, with $$H$$ sequentially complete.

To prove the universal property, we'll use a nice topological property that topological groups have.

Remark 10. On a topological group $$G$$, every element $$g\in G$$ has a neighborhood basis of closed sets: it suffices to show it for $$g=e$$, the identity element. In turn, it suffices to show that for every neighborhood $$U\subset G$$ of $$e$$ there is a neighborhood $$V\subset G$$ of zero with $$\overline{V}\subset U$$. But we know that there is a neighborhood $$V\subset G$$ of $$e$$ with $$VV\subset U$$, and by the closure formula for topological groups, it follows that $$\overline{V}\subset VV$$.

Proof. (Thanks to Anne Bauval for pointing out the missing ingredient in the proof.) Let $$\psi:G\to H$$ be such a map. Uniqueness of a factorization $$\tilde{\psi}:\hat{G}\to H$$ follows from the facts that $$\hat{H}$$ is Hausdorff, $$\phi(G)$$ is dense in $$\hat{G}$$ (Remark 4) and this. It remains to show existence.

Note that continuity of $$\psi$$ implies that it sends Cauchy sequences in $$G$$ to Cauchy sequences in $$H$$. Hence, we can define a map \begin{align*} \tilde{\psi}:\hat{G}&\to H\\ [x_n]&\mapsto\lim_n\psi(x_n). \end{align*} Note that $$(\psi(x_n))$$ has a unique limit on $$H$$, for $$H$$ is Hausdorff. We leave as an exercise verifying that this definition for $$\tilde{\psi}$$ is independent of the choice of representative in $$[x_n]$$. Note that by continuity of $$+_H$$, the function $$\tilde{\psi}$$ is additive. It is left to verify continuity. It suffices to show continuity at zero (see this). In turn, since $$\hat{G}$$ is first-countable, it suffices to show sequential continuity at zero. Suppose $$x_n\in\hat{G}$$ converges to zero, where $$x_n=[x_{nm}]$$. This means that for every neighborhoood $$U\subset G$$ of the identity, $$x_n\in\hat{U}$$ for all $$n\gg 0$$. In particular, $$x_{nm}\in U$$ for all $$n,m\gg 0$$. By Remark 10, it suffices to show that for an closed neighborhood $$V\subset H$$ of zero we have that for all $$n\gg 0$$, $$V\ni\tilde{\psi}(x_n)=\lim_m\psi(x_{nm}).$$ We have $$x_{nm}\in\psi^{-1}(V)$$ for all $$n,m\gg 0$$, so $$\psi(x_{nm})\in V$$ for all $$n,m\gg 0$$. Hence, for all $$n\gg 0$$, we get $$\lim_m\psi(x_{nm})\in\overline{V}=V.$$ $$\square$$

Thus, completion is an endofunctor of the category of first-countable topological abelian groups: to each morphism $$f:G\to H$$ of this category it assigns the unique morphism $$\hat{f}:\hat{G}\to\hat{H}$$ that makes the diagram $$\require{AMScd} \begin{CD} G@>f>> H\\ @VVV@VVV\\ \hat{G}@>\smash{\hat{f}}>>\hat{H} \end{CD}$$ commute.

If $$G$$ is a group topologized by a filtration of subgroups $$(G_n)$$, there is a slight notational ambiguity when writing $$\widehat{G_n}$$. Namely, this can be interpreted either as the completion of the topological group $$G_n$$ or as the subset of $$\hat{G}$$ given by taking $$U=G_n$$ in \eqref{subset}. Each of these bears a topological group structure; the former by this and the latter because any subgroup of a topological group is a topological group with the subspace topology.

Taking $$H=G_n$$ in the following lemma implies that these two topological groups are the same.

Given a first-countable topological group $$G$$, for the next lemma, we'll denote $$G^\c$$ to the completion of $$G$$ (we reserve $$\widehat{(-)}$$ for the subset given in \eqref{subset}). Recall that a quotient group of a topological group is a topological group with the quotient topology (see here).

Lemma 11. Let $$G$$ be a first-countable topological group, and suppose $$H\subset G$$ is an open subgroup. Then:

1. the set $$\hat{H}\subset G^\c$$ is a subgroup,
2. we have a short exact sequence of topological groups $$0\to H^\c\to G^\c\to G/H\to 0,$$
3. the image of $$H^\c\to G^\c$$ equals $$\hat{H}$$ (defined in \eqref{subset}).
4. the morphism $$H^\c\to\hat{H}$$ is an isomorphism of topological abelian groups.

Proof. We leave proof of part 1 as an exercise.

We tweak a little our notation: for each subset $$S\subset G$$, we still denote $$\hat{S}$$ to the subset of $$\hat{G}$$ given by \eqref{subset}; but for a subset $$S\subset H$$, we will now denote $$\widetilde{S}$$ to the open subset of $$H^\c$$ induced by $$S$$. With these notations, we claim that for $$U\subset H$$ open,

$$\begin{gather}\tag{4}\label{open} \widetilde{U}=(i^\c)^{-1}(\hat{U})\\ i^\c(\widetilde{U})=\hat{U}, \end{gather}$$

where $$i^\c:H^\c\to G^\c$$ comes from $$i:G\to H$$. These equalities are not difficult to prove and we leave them as an exercise. Part 3 follows from the second equality applied to $$U=H$$.

Part 2: On the one hand, we note that $$G/H$$ is discrete (so is the quotient of a topological group by an open subgroup). In particular, $$G/H$$ is complete. We apply the completion functor to the short exact sequence $$0\to G\to H\to G/H\to 0$$ to obtain a sequence $$0\to H^\c\to G^\c\to G/H\to 0$$ of topological abelian groups. Here, the map $$G^\c \to G/H$$ sends $$[x_n]\in G^\c$$ to $$x_N+H$$, for some sufficiently big $$N$$ (note that $$x_n-x_m\in H$$ for all $$n,m\gg 0$$). Suppose $$[h_n]\in H^\c$$ maps to zero in $$G^\c$$. By the first equality in \eqref{open}, we have that $$[h_n]$$ lies in all open neighborhoods $$\subset H^\c$$ of zero. Since $$\hat{H}$$ is Hausdorff, this implies $$[h_n]=0$$ in $$H^\c$$. On the other hand, $$G^\c\to G/H$$ is onto for $$G\to G^\c\to G/H$$ is onto and the composite $$H^\c\to G^\c\to G/H$$ vanishes. Conversely, suppose $$[x_n]\in G^\c$$ maps to zero in $$G/H$$. This means that there is $$n_0$$ with $$x_n\in H$$ for all $$n\geq n_0$$. Hence $$[(x_n)_{n\geq n_0}]\in H^\c$$ is a preimage.

Part 4: $$H^\c\to \hat{H}$$ is an isomorphism for it is a continuous bijection and the inverse is also continuous (use the second equality of \eqref{open} plus either Lemma 7 or the fact that an additive map between topological groups is continuous iff it is continuous at zero). $$\square$$

• For your first step, it’s a little more subtle and I think you need a diagonalization argument. If you suppose there’s no such $N$, then the contradiction for each $n$ may not occur in the $n$th index.
– M W
Commented Sep 6, 2023 at 14:26
• @MW Thanks! I just edited it and corrected it. Commented Sep 6, 2023 at 15:07
• For those interested on the well-definedness of the completion of a ring and of a module—and the universal property they satisfy—, see this. Commented Sep 13, 2023 at 15:43