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Suppose you have a group $F$, and a characteristic subgroup $K\le F$. Under what conditions on $F$ and $K$ is the closure $\overline{K}$ inside the profinite completion $\widehat{F}$ also characteristic?

I'm mostly happy to assume that $F$ is finitely generated, though I'd like to know as much about this as possible.

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Here's one sufficient condition: If every surjection $F\rightarrow F/K$ is equivalent up to an automorphism of $F/K$, then $\overline{K}$ is characteristic in $\widehat{F}$.

Proof: There's a bijection between $Hom(F,F/K)$ and $Hom(\widehat{F},F/K)$. Thus, by our hypothesis, every surjection $\widehat{F}\rightarrow F/K$ is also equivalent up to $Aut(F/K)$, and hence they all have the same kernel $\overline{K}$, but the intersection of all kernels of surjections to $F/K$ is characteristic, hence $\overline{K}$ is characteristic.

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    $\begingroup$ I'd suggest to unaccept your own answer to encourage new answers be posted here as well, rather than forming a duplicate with mathoverflow.net/questions/295277/… $\endgroup$ – YCor Mar 15 '18 at 11:13

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