Let $0\to G^{\prime}\to G\to G^{\prime\prime}\to 0$ be an exact sequence of abelian groups. Suppose $G$ is a topological group and then give topologies on $G^{\prime}$ and $G^{\prime\prime}$ which are induced by the topology on $G$. If $H$ is topological abelian group, $\widehat{H}$ denotes its completion using Cauchy sequences.
Is the sequence of completions $0\to \widehat{G^{\prime}}\to \widehat{G}\to \widehat{G^{\prime\prime}}\to 0$ still exact as abelian groups?
For a topological abelian group $G$ with filtration of subgroups $G=G_0\supseteq G_1\supseteq G_2\supseteq\cdots\supseteq G_n\supseteq \cdots$, if $\{G_n\}$ is a basic system of neighborhoods of $0$, then from Corollary 10.3 of Atiyah's Introduction to Commutative Algebra, we know that the above sequence of completions is exact. So in general if the topology on $G$ is not so good we don't have exactness of taking completion, and there will exist many counterexamples, right? Could you provide an easy counterexample or some references?
Thanks!
Here is an easy counterexample. The injective morphism $\mathbb{Z}\hookrightarrow \mathbb{Q}$ does not remain injective after, say, taking $p$-adic completions. In general, the completion functor for $R$-modules is neither left-exact nor right-exact, and need not be exact in the middle as well. It is only true that the completion functor preserves surjections. For more details see for example here.