$\Bbb{R}/n\Bbb{Z}$ is isomorphic to $A_\Bbb{Q}/(\Bbb{Q}+C_n)$. Let $A_\Bbb{Q}$ be the adele group of $\Bbb{Q}$. Let $C_n=\{x \in A_\Bbb{Q}: x_\infty=0 \text{ and }x_p \in p^{\operatorname{ord}_p(n)}\Bbb{Z}_p \text{ for prime }p\}$. I want to show that $\Bbb{R}/n\Bbb{Z}$ is isomorphic to $A_\Bbb{Q}/(\Bbb{Q}+C_n)$. 
We can construct a map $q$ from $\Bbb{R}/n\Bbb{Z}$ to  $A_\Bbb{Q}/(\Bbb{Q}+C_n)$ sending $x$ to the class of the adele with $x$ as the Archimedean component and zero for the non archimedian component. I showed that this map is well defined and injective. But I cannot show that this map is surjective. Please help...
 A: So the overarching question is showing that the quotient $A_{\mathbb{Q}}$ can be identified with the pro-universal covering of $\mathbb{R}/\mathbb{Z}$ $$A_{\mathbb{Q}}/\mathbb{Q} = \varprojlim_n\, \mathbb{R}/n\mathbb{Z},$$the projective limit being taken over the set of natural integers ordered by the divisibility order.
There is an exact sequence $$0 \to \mathbb{R} \times \hat{\mathbb{Z}} \to A_\mathbb{Q} \to \bigoplus_p \mathbb{Q}_p/\mathbb{Z}_p \to 0$$where $\bigoplus_p \mathbb{Q}_p/\mathbb{Z}_p $ is the subgroup of $\prod_p \mathbb{Q}_p/\mathbb{Z}_p$ consisting of sequences $(x_p)$ whose members $x_p \in \mathbb{Q}_p/\mathbb{Z}_p$ vanish for almost all $p$. Consider now the homomorphism between the two exact sequences:
$$\require{AMScd}
\begin{CD}
0 @>>> 0 @>>> \mathbb{Q} @>>> \mathbb{Q} @>>> 0\\
@VVV     @VVV     @VVV     @VVV     @VVV\\
0 @>>> \hat{\mathbb{Z}} \times \mathbb{R} @>>> A_\mathbb{Q} @>>> \bigoplus_p \mathbb{Q}_p/\mathbb{Z}_p @>>> 0
\end{CD}
$$
Because the middle vertical arrow is injective and the right vertical arrow is surjective with kernel $\mathbb{Z}$, the snake lemma induces an exact sequence$$0 \to \mathbb{Z} \to \hat{\mathbb{Z}} \times \mathbb{R} \to A_{\mathbb{Q}}/\mathbb{Q} \to 0,$$ $\mathbb{Z}$ being diagonally embedded in $\hat{\mathbb{Z}} \times \mathbb{R}$. In other words, there is a canonical isomorphism$$A_\mathbb{Q}/\mathbb{Q} \to (\hat{\mathbb{Z}} \times \mathbb{R})/\mathbb{Z}.$$Dividing both sides by $\hat{\mathbb{Z}}$, we get an isomorphism $$A_{\mathbb{Q}}/(\mathbb{Q} + \hat{\mathbb{Z}}) \to \mathbb{R}/\mathbb{Z}.$$By the same argument, for every $n \in \mathbb{N}$ we can identify the covering $$A_\mathbb{Q}/(\mathbb{Q} + n\hat{\mathbb{Z}}) = A_\mathbb{Q}/(\mathbb{Q} + C_n)$$ of $A_\mathbb{Q}/(\mathbb{Q} + \hat{\mathbb{Z}})$ with the covering $\mathbb{R}/n\mathbb{Z}$ of $\mathbb{R}/\mathbb{Z}$. Hence $A_{\mathbb{Q}}/\mathbb{Q}$ is the pro-universal covering of $\mathbb{R}/\mathbb{Z}$.
