Isomorphism from completion of topological abelian group to the inverse limit $\varprojlim G/G_n$ induced by a filtration of subgroups

Question

Let $G$ be a topological abelian group and let $\widehat{G}$ denote its completion (i.e. equivalence classes of Cauchy sequences). Let $G_n$ be a descending sequence of subgroups, i.e. $G = G_0 \supset G_1 \supset \dots$ such that $G_n$ form a countable neighborhood basis of $0$.

Let $\bar{x} \in \widehat{G}$ and let $x$ be its representative so that $x$ is Cauchy in $G$. We know that for a given $k$, $x_n$ is constant in $G/G_k$ for $n$ large enough, say $x_n = \xi_k$ in $G/G_k$. So from the sequence $x$ we get a new sequence $\xi$.

Define $\varphi_n : \widehat{G} \to G/G_n$ as $x \mapsto \xi_n$. Then define $\varphi: \widehat{G} \to \prod_n G/ G_n$ as $x \mapsto \xi$. Now we want to show that $\mathrm{Im}\varphi \subset \displaystyle \lim_{\longleftarrow} G/G_n$, i.e. $\theta_i \circ \varphi_i = \varphi_{i-1} $.

For this note that if $x_n = \xi_k$ in $G/G_k$ and $x_n = \xi_{k-1}$ in $G/G_{k-1}$ then for the projections $\theta_{i}: G/G_i \to G/G_{i-i}$ we get $\theta_{i} \varphi_{i}(x) =\theta_{i} \xi_{i} = \xi_{i-1} = \varphi_{i-1}(x)$.

Now to show that $\varphi$ is an isomorphism we want to find a two-sided inverse. I was thinking that if $(\xi_n) \in \displaystyle \lim_{\longleftarrow} G/G_n$ then I could just define $\psi: (\xi_n) \mapsto \overline{(\xi_n)}$ where $\overline{(\xi_n)}$ denotes the equivalence class of $(\xi_n)$ in $\widehat{G}$ but this all feels a bit shaky.

What is the correct definition of inverse of $\varphi$? Thanks for your help.

Answer 1

You could show that your map $\phi$ (which is easily seen to be a group morphism) is injective and surjective.

Injective: if $\phi(x)=0$ in $\varprojlim G/G_n$, then for all $n$, $\phi_n(x)=0$ in $G/G_n$, so for all $n$, eventually $x_i\in G_n$, meaning $(x_i)=0$ in $\hat{G}$.

Surjective: given $\xi=(\xi_n)\in \varprojlim G/G_n$, take lifts $x_n\in G$ of $\xi_n\in G/G_n$; then note that $(x_n)$ is Cauchy in $G$, so yields an element in $\hat{G}$ which by construction gets mapped to $\xi$ by $\phi$.

You probably meant to prove $\hat{G}=\varprojlim G/G_n$ as topological abelian groups. So to finish, you could then prove that $\phi$ is an open map. Alternatively you can just prove that the inverse (which is actually constructed in the proof of surjectivity) is continuous.

Answer 2

I think you're close to the right map, but let's be careful. If $(x_n) \in \varprojlim G/G_n$ then define a sequence $(y_n)$ of elements in $G$ by taking $y_n$ to be any lift of $x_n$, and then take the equivalence class of this inside $\hat G$. This class does not depend on the choices of lifts, for if $(z_n)$ is another such sequence then $(y_n) - (z_n)$ is a null sequence since $y_n - z_n \in G_N$ for all $n \geq N$.

Let me know how proving that these are inverse goes.

Answer 3

We don't need to construct an isomorphism between the two. We can instead show that $\hat{G}$ satisfies the universal property of $\varprojlim G/G_n$ as follows:

Let our inverse system be $(G/G_n, \theta_{nk})$ (where $n \geq k$), satisfying $\pi_k = \theta_{nk} \circ \pi_n$ where $\pi_n : \hat{G} \to G/G_n$ is the map taking a representative of a Cauchy sequence in $\hat{G}$ and "modding" it by $G_n$ so that all its terms above index $n$ are zero. Clearly, $\pi_n$ are surjective. (We will use this vital property later to construct our unique homomorphism that makes our diagram commute.)

Now let $H$ be any (topological?) Abelian group and let $\pi_n^\prime :H\to G/G_n$ be maps such that $\pi_n^\prime = \theta_{mn} \circ \pi_m^\prime$ for all $m \geq n$. Then we construct a homomorphism $\alpha : H \to \hat{G}$ as follows: Let $h \in H$. Then $\pi_k^\prime (h) \in G/G_k$. Since $\pi_k$ are surjective there exists $g \in \hat{G}$ such that $\pi_k(g) = \pi_k^\prime (h)$. Define $\alpha (h) = g$.

To conclude the proof we need to verify that $\alpha$

(i) is well-defined: let $g,g^\prime$ both be in $\hat{G}$ and such that $\pi_k(g) = \pi_k(g^\prime) = \pi^\prime_k (h)$. We claim that then $g-g^\prime \to 0 $ in $\hat{G}$. Since $g$ and $g^\prime$ are equal "mod $G_k$" they agree on the first $k$ terms. By induction they agree on all terms and hence $g-g^\prime$ is the zero sequence.

(ii) is a group homomorphism i.e. $\alpha (h + h^\prime) = \alpha (h) + \alpha (h^\prime)$:

$\alpha (h) = g$ where $g \in \hat{G}$ is such that $\pi_k (g) = \pi_k^\prime(h)$ and $\alpha (h^\prime) = g^\prime$ where $\pi_k (g^\prime) = \pi_k^\prime (h^\prime)$ hence $\alpha (h + h^\prime) = g + g^\prime$ where $g + g^\prime$ is such that $\pi_k(g + g^\prime) = \pi_k (g) + \pi_k (g^\prime) = \pi_k^\prime (h) + \pi_k^\prime (h^\prime) $ and hence $\alpha (h) + \alpha (h^\prime) = g + g^\prime = \alpha(h + h^\prime)$.

(iii) makes the diagram commute: by definition.

(iv) is unique: let $\alpha^\prime$ be such that $\pi_n \circ \alpha^\prime = \pi^\prime_n$ for all $n$. Then $\pi_n \circ \alpha^\prime = \pi^\prime_n = \pi_n \circ \alpha $ which (by same reasoning as in (i)) means that $\alpha$ and $\alpha^\prime$ agree on the first $n$ terms and then by induction they agree on all terms.