3

In the comments to this answer, Martin Brandenburg told me that it is possible to derive the element structure of the coproduct $A\oplus B$ of two abelian groups $A$ and $B$, by knowing that such a coproduct exists and by using the universal property. In particular, it is not necessary to use an explicit description of it. Since I usually have to deal with situations in which I know the existence of a universal solution to a certain question, but I don't know how to explicitly describe it, I would be glad to learn how this could be done in this case.

For me, given two abelian groups $A,B$ their coproduct is an abelian group $Z$ together with two group homomorphisms $j_A:A\to Z$ and $j_B :B\to Z$ which is universal with respect to this property. That is to say, such that for any other abelian group $C$ together with two morphisms of abelian groups $f:A\to C$ and $g:B\to C$, there exists a unique morphism of abelian groups $\nabla(f,g):Z\to C$ such that $\nabla(f,g)\circ j_A = f$ and $\nabla(f,g)\circ j_B = g$.

I tried to solve the exercise Martin left to me. First of all, since we are in the category of abelian groups, we have the zero morphism and hence I can consider $\nabla(id_A,0):Z\to A$ and $\nabla(0,id_B):Z\to B$.

Now I need to assume that I know that products in the category of abelian groups exist and that they can be realized as the product of the underlying sets together with the component-wise composition law. Thanks to these assumptions, I know that there exists a unique group homomorphism $\Phi:Z \to A\times B$ such that $\pi_A\circ\Phi=\nabla(id_A,0)$ and $\pi_B \circ \Phi = \nabla(0,id_B)$. By composing, I can consider $i_A:=\Phi\circ j_A:A\to A\times B$ and $i_B:=\Phi\circ j_B :B \to A\times B$. Since $\pi_A\circ\Phi\circ j_A = id_A$ and $\pi_B\circ\Phi\circ j_B = id_B$, both $i_A$ and $i_B$ are injective (and $j_A$ and $j_B$ as well).

At this point I stopped because I realized that I am not "describing the element structure" of $A\oplus B$. I am showing that $A\oplus B$ is isomorphic to $A\times B$ with component-wise composition law (next step would have been to use the universal property of $A\times B$ to show that there exists a unique group homomorphism $\Psi:A\times B\to Z$ such that $\Psi \circ \Delta(id_A,0) = j_A$ and $\Psi \circ \Delta(0,id_B) = j_B$ and then proving that, by uniqueness and the explicit description of $A\times B$, $\Phi$ and $\Psi$ are each other inverses). Thus, I am providing an explicit realization of $A\oplus B$, which is not what has been asked me.

Can anybody suggest me a different approach?

Ender Wiggins
  • 2,982
  • 2
    I don't know if that's what Martin had in mind, but in this situation I would prove that $A\oplus B$ is also a (categorical) product of $A$ and $B$ (so in the sense of the universal property, not in the sense of product of sets with componentwise addition, although they agree), and then describe the elements of that guy with the universal property (knowing that an element of an abelian group $A$ is the same as a morphism $\mathbb Z\to A$) – Maxime Ramzi Dec 31 '19 at 10:36
  • @Max nice suggestion, but I am bit scared that you need the explicit description of $A\oplus B$ to show that it is, in fact, a biproduct. Anyway, I'll think on that, thanks. – Ender Wiggins Dec 31 '19 at 11:05
  • 4
    No, you don't. That's true in any additive category, so here since you're facing $\mathbf{Ab}$, the mother of all additive categories, it should be even easier. – Maxime Ramzi Dec 31 '19 at 11:10
  • You might profit from reading the first chapter of Borceux Handbook, volume II. Or Freyd's "Abelian categories", even better :) that book never gets old. – fosco Dec 31 '19 at 12:13
  • To flesh out Max's point, a biproduct in a category enriched in abelian groups is given by objects $a_1,a_2, b$ with maps $\iota_j:a_j\to b$ and $\pi_j:b\to a_j$ such that $\pi_j\circ \iota_j=\mathrm{id}{a_j}$, $\pi_j\circ \iota{j+1\pmod 2}=0$, and $\iota_1\pi_1+\iota_2\pi_2=\mathrm{id}_b$, all of which can be readily constructed from the assumption that $b$ is a coproduct or a product. – Kevin Carlson Dec 31 '19 at 21:04
  • If you know it is a categorical product, you can use the fact that the forgetful functor is a right adjoint to deduce that the underlying space is $A \times B$. – Connor Malin Jan 04 '20 at 22:33

0 Answers0