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We know that the abelianization of a free product is the direct sum, for example $Ab(\mathbb{Z}*\mathbb{Z})=\mathbb{Z}\oplus\mathbb{Z}$.

Is there a “canonical” (or even non-canonical) operator that reverses the process of Abelianization? (For instance, it may goes from direct sum to free product?)

Thanks.

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    $\begingroup$ If you want a group with a given abelianization $A$, one choice is $A$. That is boring, but you're unlikely to get an interesting answer in a canonical way. $\endgroup$ – KCd Apr 22 at 6:08
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The abelianization functor, from the category of groups to the category of abelian groups, has a right adjoint, which is the forgetful functor, in other words the functor that sends an abelian group $A$ to the group $A$ itself. I guess this is as close to an "inverse" as you can reasonably hope.

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    $\begingroup$ Thanks for this answer. $\endgroup$ – yoyostein Apr 22 at 6:23

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