Pontryagin dual of an inverse limit

Question

For a locally compact Hausdorff abelian topological group $G$, let $G^\vee = \mathrm{Hom}_{cts}(G, \mathbb{R}/\mathbb{Z})$ denote its Pontryagin dual, endowed with the compact-open topology. It is more or less easy to show that, if $(M_i)_i$ is a directed system of such groups, then $$ (\varinjlim M_i)^\vee \cong \varprojlim M_i^\vee $$ with the obvious morphisms on the RHS. I am interested in the somehow dual question: if $\{M_i\}$ is an inverse system of such groups, is it true that $$ (\varprojlim M_i)^\vee \cong \varinjlim M_i^\vee $$ ? On one hand, Pontryagin duality (the double dual of something is isomorphic to that thing) would seem to indicate that the above is true: $$ (\varprojlim M_i)^\vee \cong (\varprojlim M_i^{\vee\vee})^\vee \color{blue}{\cong} (\varinjlim M_i^\vee)^{\vee\vee} \color{blue}{\cong} (\varinjlim M_i^\vee). $$

However this seems to contradict the following example: consider the inverse system $(\mathbb{Z})_i$ with $i \in \mathbb{N}$ and transition maps equal to multiplication by e.g. $p$ a prime. Here $\mathbb{Z}$ is given the discrete topology. Then on one hand we have $\mathbb{Z}^\vee \cong \mathbb{R}/\mathbb{Z}$ and on the other $\varprojlim \mathbb{Z} = \{0\}$. How can it be that $$ (\varprojlim \mathbb{Z})^\vee = \{0\}^\vee = 0 \stackrel{?}{\cong} \varinjlim \mathbb{R}/\mathbb{Z} $$ ? If my computations are correct, the transition maps on the RHS are multiplication by $p$, and that direct limit certainly does not seem trivial.

P.S.: I guess for the two duality properties of limits stated at the beginning, one needs to ask the corresponding direct/inverse limit to be locally compact (in order to take duals). I'm not sure if that is really an extra condition or always the case, but it holds in the example at any rate.

Edit: In the two isomorphisms highlighted in blue, one is tacitly assuming that the inverse/direct limit in question is indeed (locally compact, abelian) Hausdorff to apply Pontryagin duality. This is what fails in the example.

Answer 1

If $\mathbf{LCA}$ denotes the category of locally compact abelian groups, then the Pontryagin dual establishes an equivalence of categories $\mathbf{LCA}^{\mathrm{op}}\to\mathbf{LCA}$. In particular, this means that sending LCA's to their duals will send limits to colimits and colimits to limits (inverse / direct limits being another term for these). Thus, the answer to your original question is yes, and your argument for proving this is correct.

Edit: originally, I had the arrows on backwards in the computation.

Regarding your example, the limit of $\dots\xrightarrow{\times p}\mathbb Z\xrightarrow{\times p}\mathbb Z$ is indeed trivial (like you mentioned), and the dual diagram $\mathbb R/\mathbb Z\to\mathbb R/\mathbb Z\to\dots$ is given by multiplication by $p$ again, like you mentioned. However, the colimit of that diagram inside $\mathbf{LCA}$ is subtle, as it has to still be locally compact and Hausdorff. This means it can't be necessarily computed in the ambient category of general topological groups.

If we take the colimit in topological groups, the resulting group will not be Hausdorff: if we take $p=3$ for concreteness, then the element $\frac12$ (being fixed under multiplication by $p$) will be nonzero in the colimit in topological groups, but there is no open set separating it from the neutral element. For any open neighbourhood of $\frac12$ contains some element of the form $\frac n{3^m}$ for $n$, $m$ integers, which vanishes in the colimit. However, we can correct this colimit to be a colimit in $\mathbf{LCA}$ by applying the "Hausdorffification" done in this answer, and it will (has to!) turn out that the correction will make the colimit trivial.