Notes:Products and sums of groups

Recall

Given a pair of groups, $(A,*_A)$ and $(B,*_B)$ in the category of groups we define the following:

Product

• A wedge: $$\xymatrix{ A \\ S \ar[u]_{p_A} \ar[d]^{p_B} \\ B }$$ such that for each wedge $$\xymatrix{ & A \\ X \ar[ur]^{f_A} \ar[dr]_{f_B} & \\ & B}$$ there exists a unique arrow $\xymatrix{&\\X \ar[r]^m & S\\&}$ such that $$\xymatrix{ & A \\ X \ar[r]^m \ar[ur]^{f_A} \ar[dr]_{f_B} & S \ar[u]_{p_A} \ar[d]^{p_B} \\ & B}$$ commutes.

It is obvious that

Coproduct (sum)

• A wedge: $$\xymatrix{ A \\ S \ar@{<-}[u]^{i_A} \ar@{<-}[d]_{i_B} \\ B }$$ such that for each wedge $$\xymatrix{ A & \\ & X \ar@{<-}[ul]_{f_A} \ar@{<-}[dl]^{f_B} & \\ B&}$$ there exists a unique arrow $\xymatrix{&\\S \ar[r]^m & X\\&}$ such that $$\xymatrix{A \ar[dr]^{f_A} \ar[d]_{i_A}& \\ S \ar[r]^m & X \\ B \ar[u]^{i_B} \ar[ur]_{f_B} &}$$ commutes.

"Proof" that $(A\times B,*_{A\times B})$ is the product and coproduct of $A$ and $B$ (FALSE)

Here we define the operation: $*_{A\times B}:(A\times B)\times(A\times B)\rightarrow A\times B$ as follows:

• $*_{A\times B}:((a,b),(a',b'))\mapsto(aa',bb')$

(This means that I am claiming $S=A\times B$ with this operation)

Product (TRUE)

Let $(X,\cdot)$ be a group and let $f_A:X\rightarrow A$ and $f_B:X\rightarrow B$ be group homomorphisms be given.

• We must find a unique $m:X\rightarrow A\times B$ (also a group homomorphism) such that:
  1. $p_A\circ m=f_A$ and
  2. $p_B\circ m=f_B$

I claim that $A\times B$ is a group with this operation, and, $p_A$ and $p_B$ are defined as follows (and are group homomorphisms):

• $p_A:A\times B\rightarrow A$ by $p_A:(a,b)\mapsto a$ and
• $p_B:A\times B\rightarrow B$ by $p_B:(a,b)\mapsto b$

In this event, $A\times B$ is the categorical product of $A$ and $B$.

• Clearly we can define $m:X\rightarrow A\times B$ by $m:x\mapsto(f_A(x),f_B(x))$

Then $p_A(m(x))=p_A((f_A(x),f_B(x))=f_A(x)$ as required.

Coproduct (FALSE)

Let $(X,\cdot)$ be a group and let $f_A:A\rightarrow X$ and $f_B:B\rightarrow X$ be group homomorphisms be given.

• We must find a unique $m:A\times B\rightarrow X$ (also a group homomorphism) such that:
  1. $m\circ i_A=f_A$ and
  2. $m\circ i_B=f_B$

I claim that $A\times B$ is a group with this operation, and, $i_A$ and $i_B$ are defined as follows (and are group homomorphisms):

• $i_A:A\rightarrow A\times B$ by $i_A:a\mapsto(a,0)$ and
• $i_B:B\rightarrow A\times B$ by $i_B:b\mapsto(0,b)$

In this event $A\times B$ is the categorical coproduct of $A$ and $B$

• After a little thought I've come up with:
  • $m:A\times B\rightarrow X$ given by $m:(a,b)\mapsto f_A(a)\cdot f_B(b)$ HOWEVER is is clear that
    • $m:(a,b)\mapsto f_B(b)\cdot f_A(a)$ would also work just as well; and is distinct from the $m$ above unless $X$ is an Abelian group.
      • As $m(i_A(a))=m((a,0))=f_A(a)\cdot f_B(0)=f_A(a)\cdot e_B=f_A(a)$ as required, or:
        • $m(i_A(a))=m((a,0))= f_B(0)\cdot f_A(a)=e_B\cdot f_A(a)=f_A(a)$ depending on how you define $m$, similarly for $f_B$
• Thus we see that (unless the choice of $X$ is always an Abelian group) that there is no unique mediating arrow, and thus $A\times B$ cannot be the coproduct of $A$ and $B$.