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This fact is Corollary 16.4.1 on page 40 of the course notes from (the first semester of) Brian Conrad's course on linear algebraic groups. In the notes there is a proof of this fact is modern language, over an arbitrary field. The notes can be found here: http://math.stanford.edu/~conrad/252Page/handouts/alggroups.pdf

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Let us take $V=Q_p[T]$ and $L_2 = \oplus_{i \geq 0} Z_p \cdot T^i$ and $L_1 = Z_p \cdot p \oplus (\oplus_{i \geq 0} Z_p (T^i + p T^{i+1}))$. The element $x = 1 \cdot p - p \cdot (1+pT) + p^2 \cdot (T+pT^2) - \cdots$ belongs to the $p$-adic completion of $L_1$ and is nonzero in it, but its image in the $p$-adic completion of $L_2$ is zero. This should give ...

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This is not a complete answer, but a conditional one. If you assume that $\bigcap_{N\in\mathbb N} \pi^N\left(L_2/\pi^n L_1\right)=0,$ or, what amounts to the same: $\bigcap_N(\pi^NL_2+ L_1)=L_1$, then the map is injective. It suffices to show that the map $\hat L_1\to \hat L_2$ is injective, where $L_j=\lim_{\leftarrow}L_j/\pi^n$. So let $l=(l_n)$ be in ...

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To summarize some of what is stated in the comments above, there is a related question (posed as Question 4.8.5b on p. 156 of Ribes-Zalesskii) of whether torsion profinite groups have bounded exponent. As noted in the comments, your question is equivalent to this question due to Zelmanov's solution of the restricted Burnside problem, which implies that ...

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Let's take the group $G = \mathbb Z^T$, where $T$ is an uncountable set. Is $G$ weakly separable? Yes: If $U$ is any neighborhood of $0$ in $G$, then there exists a finite set $T_0 \subset T$ so that $$U \supseteq V := \{\phi \in G \colon \phi(t)=0 \text{ for all } t \in T_0\}$$ Countably many translates of $V$ cover $G$: namely translates by all $\psi ... 19 No. Take$G=SO(5)$and$H=SO(3)$, both of which have fundamental group$\mathbb{Z}/2$. I claim that there is no continuous map$f: G\to H$which induces the identity homomorphism. If there were, then$f$would induce a nontrivial homomorphism$f_\ast: H_1(G;\mathbb{Z}/2)\to H_1(H;\mathbb{Z}/2)$, and a graded ring homomorphism$f^\ast: ...

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Not always. Take $G=H=\mathrm{PGL}(n,\mathbb{R})$, with $n\geq 7$. Every nontrivial continuous homomorphism $G\rightarrow G$ is a Lie automorphism; the group of such automorphisms is generated by inner automorphisms and $A\mapsto {}^tA^{-1}$, so it acts on $\pi _1(G)$ through {$\pm 1$}. On the other hand $\pi _1(G)=\mathbb{Z}/n$ has automorphisms $\neq ... 18 Here is a counter-example with$G$homeomorphic to$\mathbb R^2$. Let$f:\mathbb R\to\mathbb R$be a discontinous additive homomorphism (constructed using a Hamel basis of$\mathbb R$over$\mathbb Q$). Define a group operation$*$on$\mathbb R^2$by $$(x,y)*(x',y') = (x+x'e^{f(y)},y+y') .$$ This groups is a semidirect product of$\mathbb R$and$\mathbb ...

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No, a counterexample—courtesy of Klaus Schmidt—is the semidirect product $SO(2) \ltimes \mathbb{R}^2$ taken with respect to the defining action of $SO(2)$ on $\mathbb{R}^2$. The von Neumann kernel of this group is $\mathbb{R}^2$, which follows from the results in “The structure of homomorphisms of connected locally compact groups into compact groups” by A. ...

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