Abelianization of free product is the direct sum of abelianizations

I am interested in the result above, and would like to study a textbook that delves into it deeper.

However the standard algebraic topology textbooks (e.g. Hatcher/Munkres) and standard algebra books like Hungerford do not contain this result.

Is there any suitable book that I can consult? Thanks!

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    $\begingroup$ Do you really need a reference here? This is pretty straightforward to prove directly, since both are coproducts in their respective categories, and abelianization is a functor... $\endgroup$ – Steve D Jul 27 '18 at 17:02
  • $\begingroup$ @SteveD : In general, functors do not commute with coproducts though ... $\endgroup$ – Nicolas Hemelsoet Jul 27 '18 at 18:58
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    $\begingroup$ @Nicolas: this is a left adjoint functor though, so it does. I just didn't want to spell the whole thing out :) $\endgroup$ – Steve D Jul 27 '18 at 19:31

Let $\{G_i:i\in I\}$ be a collection of groups, $F$ be the free product of the $G_i$ and $A$ the direct sum of the abelianizations of the $G_i$. Free products, direct sums, and abelianizations all have universal properties. The universal property for the direct sum $A$ is that, for any abelian group $B$, we have

$$\mathrm{Hom}(A,B)=\prod_{i\in I}\mathrm{Hom}(G_i/[G_i,G_i],B)\text{.} $$

On the other hand, applying the universal property of abelianization, the free product, and finally abelianization again, gives

$$\mathrm{Hom}(F/[F,F],B)=\mathrm{Hom}(F,B)=\prod_{i\in I}\mathrm{Hom}(G_i,B)=\prod_{i\in I}\mathrm{Hom}(G_i/[G_i,G_i],B)$$.

These equalities (really canonical isomorphisms) constitute an isomorphism of functors $\mathrm{Hom}(A,-)\simeq\mathrm{Hom}(F/[F,F],-)$ from the category of abelian groups to the category of sets. Yoneda's lemma tells us that such an isomorphism is induced by a unique isomorphism $A\simeq F/[F,F]$.

  • $\begingroup$ Can the index set be infinite? (countable/uncountable) Is there any restriction? $\endgroup$ – yoyostein Aug 3 '18 at 15:19
  • $\begingroup$ Dear@yoyostein, Yes. $\endgroup$ – Keenan Kidwell Aug 3 '18 at 17:02

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