Sum of the completion of a topological abelian group is continuous

Question

Paraphrasing Atiyah, MacDonald, Introduction to Commutative Algebra, Chapter 10, p. 102: let $G$ be a topological abelian group. Then:

I understand everything what is on this excerpt. Also, the set $\hat{G}$ has a topology explained here. My question is: with this topology, is $\hat{G}$ a topological group? On the one hand, if we denote $\hat{U}\subset \hat{G}$ to the open subset induced by an open subset $U\subset G$, then one has $-\hat{U}=\widehat{-U}$, so I am actually asking:

Is the sum operation $\hat{G}\times\hat{G}\to\hat{G}$ continuous?

None of the previous questions I've found on MSE asking about $\hat{G}$ address this issue. The following are my thoughts on the problem:

We make some reductions.

1. Denote $G'$ to the set of Cauchy sequences with terms in $G$. Denote $G'_0\subset G'$ to the subgroup of Cauchy sequences converging to zero. Then $\hat{G}=G'/G'_0$. We endow $G'$ with the following topology: the open sets are of the form $U'\subset G'$, were $U\subset G$ is open and $U'$ is the set of Cauchy sequences $x$ such that for all $y\in G'$ equivalent to $x$, it holds $y_n\in U$ for almost all $n$. On the other hand, note that $\pi^{-1}(\hat{U})=U'$, where $\pi:G'\to\hat{G}$ is the natural projection.

2. A quotient group of an abelian topological group by an arbitrary subgroup is an abelian topological group with the quotient topology.

3. If $A$ is any abelian group and $\nu:A\times A\to A$ is the sum, then $\nu^{-1}(S)=\bigcup_{x\in A}S\times(S+x)$.

Combining 1, 2, 3 (plus the observation $-(U')=(-U)'$), it suffices to show that $U'+x$ is open in $G'$, for all $U\subset G$ open and all $x\in G'$. My question is: Why? Let $y\in U'+x$. We want to find an open set $V\subset G$ such that $y\in V'\subset U'+x$. We have $y=z+x$, where $\tilde{z}_n\in U$ for almost all $n$ and all $\tilde{z}\in G'$ equivalent to $z$. But now... what? The only tools at hand are Cauchyness of $x,z$ plus continuity of $+_G:G\times G\to G$, although I am unsure how one can exploit them.

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(Now with the corrected definition for $U'$, it is not clear to me that doing the proof in $G'$ is easier than doing the proof in $\hat{G}$.)

Answer 1

[*Update, corrected a problem with the original argument.]

Unfortunately point #3 in the question appears incorrect, since certainly for $s_1,s_2\in S$ and $x\in A$ we cannot conclude $s_1+(s_2+x)\in S$.

Regardless, we can prove continuity of the sum as follows.

For a Cauchy sequence $x_n$, denote $[x_n]$ its equivalence class in $\hat{G}$. Note a key fact we will use is that the representatives of $[x_n]$ are precisely those sequences of the form $x_n+s_n$ for some sequence $s_n\to 0$.

Now let $\nu\colon \hat{G}\times\hat{G}\to \hat{G}$ be the sum. We wish to show that for an arbitrary basis member $[z_n]+\hat{W}$ of $\hat{G}$, $\nu^{-1}([z_n]+\hat{W})$ is open. To this end, suppose $([x_n],[y_n])\in \nu^{-1}([z_n]+\hat{W})$, or equivalently, $[x_n+y_n-z_n]\in\hat{W}$.

Let $U_n$ be a basis of $0$ in $G$, and assume WLOG the sequence $U_n$ is decreasing. We first claim there is some $N\in\mathbb N$ for which $$x_n+U_N+y_n+U_N-z_n+U_N\subseteq W\tag{1}$$ for all $n\geq N$. Indeed, to see this, observe that otherwise we would have for each $N\in \mathbb N$ sequences $a_n^N,b_n^N,c_n^N\in U_N$ where frequently, $$x_n+a_n^N+y_n+b_n^N-z_n+c_n^N\notin W\text{.}$$ We could then take diagonal sequences $a_{k_n}^n$, $b_{k_n}^n$, and $c_{k_n}^n$, each converging to $0$, such that $$x_{k_n}+ a_{k_n}^n+ y_{k_n}+ b_{k_n}^n- z_{k_n}+ c_{k_n}^n\notin W\text{.}$$ but since $a_{k_n}^n+ b_{k_n}^n+ c_{k_n}^n\to 0$, this implies $$[x_n+y_n-z_n]=[x_{k_n}+ y_{k_n}- z_{k_n}]\notin \hat{W}\text{,}$$ a contradiction.

Now we claim $$([x_n]+\hat{U}_N)\times([y_n]+\hat{U}_N)\subseteq \nu^{-1}([z_n]+\hat{W})\text{.}\tag{2}$$ To see this, let $[a_n],[b_n]\in \hat{U}_N$. Then whenever $s_n\to 0$, we have that eventually $s_n\in U_N$, so by (1) we get that eventually $$x_n+a_n+y_n+b_n-z_n+s_n\in W\text{,}$$ from which it follows that $$[x_n]+[a_n]+[y_n]+[b_n]-[z_n] = [x_n+a_n+y_n+b_n-z_n]\in \hat{W}\text{.}$$

From (2) and the fact that $([x_n],[y_n])$ was an arbitrary member of $\nu^{-1}([z_n]+\hat{W})$ we conclude that $\nu^{-1}([z_n]+\hat{W})$ is open, and so $\nu$ is continuous.