MHB Universal Property for Coproducts in Ab

[Math Amateur]

I am reading Paolo Aluffi's book, Algebra: Chapter 0.

I am currently focused on Chapter II, Section 3: The Category Grp.

I need some help in getting started on Problem 3.3 in this section.

Problem 3.3 at the end of Chapter III, Section 3 reads as follows:https://www.physicsforums.com/attachments/4503I am slightly overwhelmed by this problem ... if someone can help me to understand what is required as well giving me significant help with the problem then I would be extremely grateful ...

To give MHB members an idea of the context and Aluffi's notation, I am npow providing Section 5.5 Coproducts ...View attachment 4504
https://www.physicsforums.com/attachments/4505

Hope someone can help ...

Peter

 

[caffeinemachine]

I am reading Paolo Aluffi's book, Algebra: Chapter 0.

I am currently focused on Chapter II, Section 3: The Category Grp.

I need some help in getting started on Problem 3.3 in this section.

Problem 3.3 at the end of Chapter III, Section 3 reads as follows:I am slightly overwhelmed by this problem ... if someone can help me to understand what is required as well giving me significant help with the problem then I would be extremely grateful ...

To give MHB members an idea of the context and Aluffi's notation, I am npow providing Section 5.5 Coproducts ...Hope someone can help ...

Peter

As such $G\times H$ cannot be the coproduct in the category of Abelain groups. It is rather meaningless to say that.

What one ought to say is that $G\times H$ coupled with the maps $i:G\to G\times H$ and $j\to G\times H$ is the coproduct of $G$ and $H$ in the category of abelian groups, where $i(g)=(g, e)$ for all $g\in G$, and similarly for $j$.

But nobody writes this for the sake of brevity.

Now. To verify this is very easy. What we need to do is that if we have any other group $M$ with homomorphisms $p:G\to M$ and $q\to M$, we need to show that there is a unique homomorphism $z:G\times H\to M$ such that $z\circ i=p$ and $z\circ j=q$ (draw the commutative diagram!).

If at all there is such a homomorphism in existence, then we must have $z(g, e)=p(g)$ and $z(e, h)=q(h)$ for all $g\in G$ and $h\in H$. And from here we are forced to have $z(g, h)=p(g)q(h)$ for all $(g, h)\in G\times H$.

So we have only one candidate for $z$. It is easy to verify that this candidate is indeed a homomorphism and we are done.

 

[Euge]

Hi Peter,

Since caffeinemachine has given a solution to the exercise, I'd like to check your understanding of the initial paragraph of section 5.5 where the author discusses products and coproducts in general. Given objects $A$ and $B$ in category $\mathcal{C}$, can you write or draw a typical object and morphism in $\mathcal{C}^{A,B}$? This is important, since you'll be able to use that knowledge to write out the definition of a coproduct of $A$ and $B$ being an initial object in $\mathcal{C}^{A,B}$ (with full diagrams).

 

[Euge]

Also, keep in mind that it's necessary for $G$ to be abelian in order for caffeinemachine's map $z$ to be a homomorphism. This is why the author emphasizes abelian. In the larger general category of groups, the coproducts are free products.

 

[Math Amateur]

Also, keep in mind that it's necessary for $G$ to be abelian in order for caffeinemachine's map $z$ to be a homomorphism. This is why the author emphasizes abelian. In the larger general category of groups, the coproducts are free products.

Euge, Caffeinemachine ... Thanks so much for your help ...

Wonderful to have such support in understanding abstract notions ...

Thanks again ...

Peter