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I need a proof of this well known fact:

If a group $G$ is given as a quotient of a free group $F/R$ then its pro-finite completion is given by the quotient $\hat{F}/ \bar{R}$ where $\bar{R}$ is the closure of $R$ in $\hat{F}$. (since $F$ embeds into $\hat{F}$ densely we can think of $R$ inside $\hat{F}$)

Thanks in advance.

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    $\begingroup$ The reason for which you can think of $R$ inside of $\hat F$ is not that $F$ embeds in $\hat F$ densely but simply that it injects there. $\endgroup$ – Mariano Suárez-Álvarez Oct 31 '10 at 23:12
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The profinite completion is the inverse limit of the $G/H$ for all normal subgroups $H$ of $G$ of finite index, that is the inverse limit of the $F/I$ for all normal subgroups $I$ of $F$ which are of finite index and contain $R$. So you have a continuous morphism $\hat{F} \rightarrow \hat{G}$, it is surjective and its kernel is the intersection of the above $I$, and it is easily seen to be $\bar{R}$. Thus you have a continuous bijective morphism $\hat{F}/\bar{R} \rightarrow \hat{G}$, and since $\hat{F}$ is compact (it is profinite) it is a homeomorphism, so it's an isomorphism of topological groups.

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