This question already has an answer here:

I mean groups and not abelian groups. In Grp, categorically, he product is the cartesian product and the coproduct is the free product. So what place takes the direct sum? Can it be defined categorically in Grp or any similar category? I would be surprised if can't be defined using limits of co/cones or something similar. If not, are they really useful (the direct sum of groups) for something?


marked as duplicate by TomGrubb, Kevin Carlson, Namaste group-theory Oct 1 '18 at 9:34

This question has been asked before and already has an answer. If those answers do not fully address your question, please ask a new question.

  • $\begingroup$ Not very satisfying but you can say that it represents $G\mapsto \{f\in \displaystyle\prod_{i\in I}\hom (G,G_i) \mid \forall x: \mathbb{Z}\to G, $ for all but finitely many $i, f_i\circ x =0 \}$. This is of course not very satisfying because I mimicked elements with $\mathbb{Z}$ ($\mathbb{Z}$ can be defined either as the object that represents the forgetful functor; or as an initial $1$-generated group, $1$-generated being categorically definable in terms of coproducts) $\endgroup$ – Max Sep 30 '18 at 17:49
  • $\begingroup$ Yes, Qiaochu’s answer to the linked question answers this. $\endgroup$ – Kevin Carlson Sep 30 '18 at 19:59
  • $\begingroup$ @KevinCarlson : it doesn't seem to answer "can it be defined categorically ?" $\endgroup$ – Max Sep 30 '18 at 21:02

The direct sum $\bigoplus_i G_i$ is the "commutative coproduct": it's the universal group admitting a map from each of the $G_i$ whose images commute. The same construction for a finite collection of $k$-algebras produces the tensor product of $k$-algebras.

  • $\begingroup$ Thanks! That was exactly what I wanted to know. And the same construction for M and N (left and right) modules, gives the tensor product of them, right? $\endgroup$ – dami Oct 1 '18 at 18:44
  • $\begingroup$ No. The universal property there is about bilinear maps as usual. $\endgroup$ – Qiaochu Yuan Oct 1 '18 at 19:32

Not the answer you're looking for? Browse other questions tagged or ask your own question.