Prove that the product satisfies the universal property for coproducts in Ab. I refer to this question.
The question is

Is there an intuitive way to understand why finite products and coproducts in Ab coincide, while the same is not true in Grp?

The author of the question has added an attempt of his to solve the question, which is the same as my attempt. 


*

*Is the attempt correct?

*Why does this not work for non-abelian groups? It seems to me that the argument would work for any group!


The attempt is
I defined the inclusion functions $\iota_G:G⟶G×H$ and $\iota_H:H⟶G×H$ by $\iota_G(g)=(g,1_G)$ and $\iota_H(h)=(1_H,h)$. Suppose $φ:G⟶K$ and $ψ:H⟶K$ are homomorphisms. To show $G×H$ is a coproduct in Ab, I then need to construct a unique function $τ:G×H⟶K$ such that $φ=τ∘ιG$ and $ψ=τ∘ιH$. As such, I defined by $τ$ by $τ(g,1_H)=φ(g)$ and $τ(1_G,h)=ψ(h)$. I think this should work since for all $(g,h)∈G×H$ we have $(g,h)=(g,1_H)∗(h,1_H)$, so by defining $τ$ so that $τ(g,h)=τ(g,1_H)∗τ(h,1_H)$ everything should work.
 A: Notice that if $G=C_2=\langle\alpha\rangle, H=C_2=⟨\beta⟩$, then $C_2\times C_2$ is the Klein group $V$, the smallest non-cyclic group. If you take for $K$ the group of all words of the form $ab$, $baba$, ..., that means of (possibly empty) alternating sequences of $a$'s and $b$'s, with concatenation of words as product (where you may have to reduce the product, e.g. $ab*baba=a(bb)aba=(aa)ba=ba $ ), then sending $α$ to $a$ and $β$ to $b$ gives group homomorphisms (since $\varphi(α)*\varphi(α)=aa=()=\varphi(α*α)=\varphi(e)$ ). Now, $(α,β)$ has two representations as a product, namely $(α,e)(e,β)$ and $(e,β)(α,e)$, and these are mapped to different values $ab$ or $ba$, respectively.
A: Be careful. Your $\tau$ is a homomorphism if and only if $\phi(g)$ and $\psi(h)$ commute for all $g \in G$ and $h \in H$. In fact, $G \times H$ represents the functor of homomorphisms $G \to ?$ and $H \to ?$ that commute pointwise. In other words, $G \times H = (G \coprod H) / \langle \{gh=hg\}_{g \in G, h \in H} \rangle$.
