Constructing a solenoid to satisfy its universal property as a projective limit over circles $\newcommand{\N}{\mathbb{N}}$
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The Pontryagin dual of $\Q$, $\hat{\Q}$ satisfies the universal property that it is the projective limit over a diagram whose objects are circles indexed by $\N$:  $\{S^1_{(n)} \}_{n \in \N}$, and whose morphisms are $\frac{m}{n}$th power maps $S^1_{(n)}  \to S^1_{(m)}$ whenever $n | m$. This projective limit is called a solenoid, denoted by $\S$.
Equivalently, $\S$ can be explicitly realized by considering that $\Z$ sits inside $\R \times \hat{\Z}$ (the product of the reals and the profinite integers) via the diagonal map. We then define $\S = (\R \times \hat{\Z})/\Z$. 
I wish to show that the latter definition satisfies the former universal property.
My thoughts are: Since $\hat{\Z}$ is the projective limit over the groups $\Z/n\Z$ under the projection maps $\Z/mZ \to \Z/nZ$ whenever $n|m$, we have maps from $\R \times \hat{\Z} \to \R \times \Z/n\Z \to \R/n\Z \cong S^1$, where the former morphism is given by the universal property of $\hat{\Z}$, and the latter morphism is given by multiplication. It is, however, unclear to me why these should descend to morphisms from $(\R \times \hat{\Z})/\Z \to S^1$ (It is possible I am missing something obvious). And it is not clear why the cone we obtain is minimal (i.e why $\S$ is indeed the projective limit).
Thank you in advance,
Maithreya
Remark: The topological group $\S$ can equivalently be characterized as the adele class group $\A_\Q/\Q$
 A: I find this easier to think about by using the fact that Pontryagin duality is a contravariant equivalence from the category of locally compact abelian groups to itself.  So rather than directly showing that $\mathbb{S}=\mathbb{R}\times\hat{\mathbb{Z}}/\mathbb{Z}$ is the projective limit in question, let us show that $\hat{\mathbb{S}}$ is the dual colimit, or equivalently that $\hat{\mathbb{S}}\cong\mathbb{Q}$ (since the colimit in question is just the colimit of the subgroups $\frac{1}{n}\mathbb{Z}\subset\mathbb{Q}$ under their inclusion maps).
To compute $\hat{\mathbb{S}}$, we use the fact that $\mathbb{S}$ is the cokernel of the map $\mathbb{Z}\to\mathbb{R}\times\hat{\mathbb{Z}}$ sending $1$ to $(1,1)$.  So, dually, $\hat{\mathbb{S}}$ is the kernel of the map $\mathbb{R}\oplus\mathbb{Q}/\mathbb{Z}\to S^1$ which is the quotient map $\mathbb{R}\to\mathbb{R}/\mathbb{Z}$ on the first coordinate and the inclusion map $\mathbb{Q}/\mathbb{Z}\to\mathbb{R}/\mathbb{Z}$ on the second coordinate.  This kernel is easily computed to be isomorphic to $\mathbb{Q}$, by sending $q\in\mathbb{Q}$ to $(q,-q+\mathbb{Z})\in\mathbb{R}\oplus\mathbb{Q}/\mathbb{Z}$.  (To be clear, I am talking about an isomorphism of topological groups, not just groups, where $\mathbb{Q}$ and $\mathbb{Q}/\mathbb{Z}$ both have the discrete topology and $\mathbb{R}$ has the usual topology.)
You can translate this argument back into a limit diagram the original setup as follows.  Since our isomorphism $\mathbb{Q}\to\hat{\mathbb{S}}$ sends $q$ to $(q,-q+\mathbb{Z})$, the inclusion $\frac{1}{n}\mathbb{Z}\to\mathbb{Q}\to\hat{\mathbb{S}}$ sends $\frac{a}{n}$ to $(\frac{a}{n},-\frac{a}{n}+\mathbb{Z})$.  The Pontryagin dual is then the map $\mathbb{S}\to S^1_{(n)}$ which sends $(x,y)\in\mathbb{R}\times\hat{\mathbb{Z}}$ to $\frac{x}{n}-\frac{y}{n}+\mathbb{Z}\in\mathbb{R}/\mathbb{Z}$ (here by $\frac{y}{n}$ I mean the fraction whose numerator is the image of $y$ in $\mathbb{Z}/n\mathbb{Z}$).  This map clearly vanishes on the diagonal $\mathbb{Z}\subset\mathbb{R}\times\hat{\mathbb{Z}}$, and so defines a map $\mathbb{S}\to S^1_{(n)}$.
