Pontryagin Duality and Injective Direct Limits Let $(G_n)_{n \in \Bbb{N}}$ be an injective direct limit of discrete groups. How does pontryagin duality play with injective direct limits? More precisely, what is $\widehat{\lim G_n}$ isomorphic to? Is it $\lim \widehat{G_n}$? If it helps, we may assume the $G_n$ are abelian groups.
EDIT: I suspect maybe they aren't isomorphic...I don't think that the direct limit of compact spaces is necessarily compact. Isn't $\Bbb{R}$ isomorphic to the injective direct limit of the compact intervals $[-n,n]$? If so, $\lim \widehat{G_n}$ probably won't always be compact, but $\widehat{\lim G_n}$ is always compact.
Since Pontryagin duality is a contravariant equivalence of categories, it turns colimits into limits.  So, if $G$ is the direct limit of a system $(G_n)$, the Pontryagin dual $\widehat{G}$ is the inverse limit of the dual system $(\widehat{G_n})$ (which is an inverse system, not a direct system, since the arrows got reversed) in the category of compact abelian groups.  Inverse limits in the category of compact abelian groups are computed by just taking the inverse limit as sets and giving it the inverse limit topology (compact Hausdorff spaces are closed under inverse limits of topological spaces) and group structure.
Here is another example. Consider a collection of cyclic
groups $G_n=\langle a\rangle$, where order of $a$ is $p^n$ and
a collection homomorphisms $a_n\rightarrow a_{n+1}^p$.
Then $\lim G_n=\mathbb{Z}(p^\infty)$ is the Pr$\ddot{\rm u}$fer $p$-group.
Since $\widehat{G_n}=G_n$,
we have also $\lim \widehat{G_n}=\mathbb{Z}(p^\infty)$.
However, it is easy to check that $\widehat{\mathbb{Z}(p^\infty)}=\widehat{\mathbb{Z}}_p$
is the group of $p$-adic integers.