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I'm having a bit of a hard time understanding a proof from Lang on why the completion and the inverse limit $\displaystyle \lim_{\longleftarrow}\ G/H_r$ are isomorphic. Here $\{H_r\}$ is a sequence of normal subgroups in $G$ with $H_r\supset H_{r+1}$ for all $r$.

Theorem 10.1 The completion and the inverse limit $\displaystyle \lim_{\longleftarrow}\ G/H_r$ are isomorphic under natural mappings.

Proof. We give the maps. Let $x=\{x_n\}$ be a Cauchy sequence. Given $r$, for all $n$ sufficiently large, by the definition of Cauchy sequence, the class of $x_n\mod H_r$ is independent of $n$. Let this class be $x(r)$. Then the sequence $(x(1),x(2),\dots)$ defines an element of the inverse limit. Conversely, given an element $(\bar{x_1},\bar{x_2},\dots)$ of the inverse limit, with $\bar{x_n}\in G/H_n$, let $x_n$ be a representative in $G$. Then the sequence $\{x_n\}$ is Cauchy. We leave to the reader to verify that the maps we have defined are inverse isomorphisms between the completion and the inverse limit.

I want to see if I'm understanding this correctly (unlikely). So $x(r)$ is the class of all elements $y\in G$ such that $x_ny^{-1}\in H_r$? Then $(x(1),x(2),\dots)$ is an element if the inverse limit as for $n\geq m$, we can take $f^n_m\colon G/H_n\to G/H_m$ to be the canonical homomorphism such that $\bar{x_n}$ in $G/H_n$ maps to $\bar{x_n}=\bar{x_m}$ in $G/H_m$ since $H_m\supset H_n$?

Also, to see that $\{x_n\}$ is Cauchy from $(\bar{x_1},\bar{x_2},\dots)$, I know that $f^n_m(\bar{x_n})=\bar{x_m}$ by definition of the inverse limit, so $\bar{x_n}=\bar{x_m}$ in $G/H_m$, so $\overline{x_nx_m^{-1}}=\bar{e}\in G/H_m$, so $x_nx_m^{-1}\in H_m$. Then given any $H_r$, for $n,m\geq r$, $x_nx_m^{-1}\in H_m\subset H_r$, and $\{x_n\}$ is Cauchy?

But then how exactly are the two maps inverses? I didn't quite understand the maps since Lang seems to map a Cauchy sequences $\{x_n\}$ to an element of the inverse limit, but the completion consists of equivalence classes mod the null sequences, so actual Cauchy sequences aren't even elements of the completion $C/C_0$?

Thank you for any clarifying details. I've been struggling to flesh this out for a while.

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    @yunone So what exactly does $x_n$mod$H_r$ mean? Sorry for bumping an old q btw2017-09-10

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It seems like you want to check both compositions. Let's try to do one of them.

Let $\{x_n\}$ be a Cauchy sequence in $G$. We send this to the sequence $(x(n))$ in the prescribed manner. To get back to the completion, for each $n$ we pick a representative $y_n$ of $x(n)$ in $G$. We need to show that $\{x_n\}$ and $\{y_n\}$ are equal in the completion, i.e. that $\{x_ny_n^{-1}\}$ is a null sequence. To verify this, fix an $r$. There is an $N \geqq r$ such that $n, m \geqq N$ implies $x_nx_m^{-1} \in H_r$.

Fixing an arbitrary $n \geqq N$, there is an $m \geqq N$ such that $x_mH_n = x(n)$. So $ f^n_r(x_mH_n) = x_mH_r = f^n_r(x(n)). $ From the Cauchy condition, we have $x_mH_r = x_nH_r$. And $y_nH_n = x(n)$ implies $ f^n_r(y_nH_n) = y_nH_r = f^n_r(x(n)). $ Thus $x_nH_r = y_nH_r$ for $n \geqq N$, so $\{x_ny_n^{-1}\}$ is indeed a null sequence.

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    Thanks for adding that to your answer. Talking about the other direction sounds quite helpful. I'm pretty busy this weekend, but I'm sure we'll both be online sometime next week so I'll try to set up a chat room for questions I have. Thanks very much for the time you've put into this.2011-08-20