I was trying to prove the following lemma with two parts. Let $G$ be a locally profinite topological group. Denote the profinite completion of $G$ by $\hat{G} = \underset{N}{\varprojlim} G/N$, where $N$ runs over the normal open subgroups with finite index. Then,
(i) The canonical map $G \to \hat{G}$ is injective.
(ii) Let $H$ be a profinite group, and $f: G \to H$ injective continuous homomorphism with dense image. Then $f$ induces a topological isomorphism $\hat{G} \xrightarrow{\sim} H$.
Following is what I thought about (i) and my rough idea about (ii).
(i) $G \to \hat{G}$ is induced by the maps $G \to G/N$ where $N$ is normal, open, and of finite index. Hence, proving injectivity is equivalent to proving that the intersection of all normal, open subgroups of finite index is trivial. If $G$ is compact it is easy since being locally profinite, there is an open compact neighborhood basis of $0$, and by compactness each such basis element is of finite index and can be refined to be normal, which implies the result. But $G$ is only known to be locally compact. I am not sure how to proceed.
(ii) By the universal property of profinite completion we have the existence of such a map $\hat{G} \to H$. To prove surjectivity, let $h \in H$. Since $f$ is dense there exist a sequence $g_n \in G$ such that $g_n$ "converges to $h$" meaning $g_n$ is contained in a "decreasing" sequence of neighborhoods of $h$. Then since $\hat{G}$ is complete and hausdorff, there is a unique limit $g$ of $g_n$ in $\hat{G}$. Then $g$ should map to $h$, hence surjectivity. For injectivity, suppose there is a non-zero element $a$ in the kernel. $G \to \hat{G}$ is dense, so there is a sequence $a_n \in G$ such that its image "converge to $a$". Since $a \neq 0$ and $G \to \hat{G}$ is injective by part (i) above, $a_n$ can be chosen "bounded away from $0 \in G$". Now consider $f(a_n)$, by injectivity they are "bounded away from $0 \in H$". Since $H$ is complete and hausdorff $f(a_n)$ should have a unique limit in $H$ not equal to zero, which should be the image of $a$ under $\hat{G} \to H$, but it contradicts the fact that $a$ is in the kernel. So $a = 0$ hence injectivity.
In part (ii) I use terms such as convergence, boundedness without appealing to any metric. But I am sure that these arguments could be better expressed in the realm of topological groups – something that I am new to. It would be very useful if someone could help me refine my ideas.
Best Answer
Both of these statements are false. For (i), consider for instance $G=\mathbb{Q}$ with the discrete topology. This has no proper subgroups of finite index, so $\hat{G}$ is trivial. For (ii), consider $G=\mathbb{Z}$ (with the discrete topology). For any prime $p$, the inclusion map $\mathbb{Z}\to\mathbb{Z}_p$ is an injective continuous homomorphism with dense image, but the induced map $\hat{\mathbb{Z}}\to\mathbb{Z}_p$ is not injective (in fact $\hat{\mathbb{Z}}$ is the product $\prod_{q\text{ prime}}\mathbb{Z}_q$ and the induced map is the projection). Your proposed argument for injectivity is wrong because $(a_n)$ being bounded away from $0$ does not imply $(f(a_n))$ is bounded away from $0$. Indeed, for the inclusion $f:\mathbb{Z}\to\mathbb{Z}_p$, the sequence $a_n=p^n$ stays away from $0$ in $\mathbb{Z}$ but $f(a_n)$ converges to $0$ in $\mathbb{Z}_p$.