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.
Best Answer
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.