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Let $F_2$ be the free group of rank 2, and $\widehat{F_2}$ its profinite completion. Abelianization gives an exact sequence $$1\rightarrow[\widehat{F_2},\widehat{F_2}]\rightarrow \widehat{F_2}\rightarrow\widehat{\mathbb{Z}}^2\rightarrow 1$$

Surely this can't be split right? Is there a reason why it can't be? (Maybe some kind of group cohomology thing?)

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  • $\begingroup$ I don't really understand the phrasing of the question. You seem both very confident that this is true and also admitting that you don't know a reason why it would be true. Which is it? $\endgroup$ – Qiaochu Yuan Nov 21 '15 at 0:21
  • $\begingroup$ @QiaochuYuan Well (imho) it would be pretty amazing if it were split, so by the principle of "it seems too good to be true", I believe it isn't true. I hope I answered your question. I'm not really sure what you mean by "true". Does your "true" mean "not split?" $\endgroup$ – oxeimon Nov 21 '15 at 0:47

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