Category theoretical proof that Abelianization commutes with products of group Let $Grps$ and $Ab$ be the categories of groups and abelian groups respectively and lets denote $\times,\sqcup, \oplus$ the product and coproduct in Grps and the biproduct in $Ab$. If I want to prove for groups $G,H \in Grps$ that there is an isomorphism $(G\sqcup H)_{ab}\cong G_{ab}\oplus H_{ab}$, where $(-)_{ab}$ denotes the abelianization, there is a wonderful category theoretical proof using adjunction: We call $U:Ab\to Grps$ the forgetful functor, then we have the adjunction $Hom_{Ab}(G_{ab},A)=Hom_{Grps}(G,U(A))$ for $G \in Grps, A \in Ab$, using and the Yoneda Lemma we can easily prove this:
$$\begin{align*}
Hom_{Ab}((G\sqcup H)_{ab},A)&=Hom_{Grps}(G\sqcup H,U(A))\\&=Hom_{Grps}(G,U(A))\times Hom_{Grps}(H,U(A))\\
&=Hom_{Ab}(G_{ab},A)\times Hom_{Ab}(H_{ab},A)\\
&=Hom(G_{ab}\oplus H_{ab},A)\\
\end{align*} $$
which implies that $(G\sqcup H)_{ab}\cong G_{ab}\oplus H_{ab}$.
My question is if there is a similar proof for the  statement that  $(G\times H)_{ab}\cong G_{ab}\oplus H_{ab}$
 A: The Yoneda lemma argument given shows that a left adjoint takes a binary coproduct to a binary coproduct. Similarly, you can show that it takes  an initial object to an initial object, and an epimorphism to an epimorphism.
That is where the similarities end. The rest of the argument requires two facts: that the trivial group $\{e\}$ being a "zero" (i.e. both an initial and a terminal) object, and that the canonical morphisms $G_k\to G_1\times G_2$ for $k=1,2$ are jointly epimorphic (you can show this using the relation $(g_1,g_2)=(g_1,e_{G_2})(e_{G_1},g_2)$ where $e_{G_k}$ are units of $G_k$).
The argument goes as follows.
First, given indexed objects $G_k$ let $\delta_{ij}\colon G_i\to G_j$ be the zero morphism if $i\neq j$ and the identity morphism otherwise. Since a left adjoint preserves initial objects, it also preserves zero objects, hence zero morphisms, and hence the $\delta_{ij}$.
Second, the universal properties of product and coproduct imply there is a unique morphism $G_1\sqcup G_2\to G_1\times G_2$ such that all four morphisms $\delta_{ij}$ for $i,j\in\{1,2\}$ factor as $G_i\to G_1\sqcup G_2\to G_1\times G_2\to G_j$, where the first map is the canonical map from a summand to a coproduct, and the last map is the canonical projection to a factor of a product. The canonical morphisms $G_k\to G_1\times G_2$ being jointly epimorphic means exactly that the morphism $G_1\sqcup G_2\to G_1\times G_2$ is an epimorphism.
Third, under the abelianization functor, the $\delta_{ij}$ morphisms get sent to the morphisms
$\delta_{ij}\colon G_i^{ab}\to G_1^{ab}\otimes G_2^{ab}\to(G_1\sqcup G_2)^{ab}\to (G_1\times G_2)^{ab}\to G_1^{ab}\otimes G_2^{ab}\to G_j^{ab}$. From preservation of coproducts and epimorphisms, we have that $G_1^{ab}\otimes G_2^{ab}\to(G_1\sqcup G_2)^{ab}$ is an isomorphism and $(G_1\sqcup G_2)^{ab}\to(G_1\times G_2)^{ab}$ is an epimorphism, whence $G_1^{ab}\otimes G_2^{ab}\to(G_1\times G_2)^{ab}$ is an epimorphism.
But from the definition of the biproduct, we also have that $G_1^{ab}\otimes G_2^{ab}\to(G_1\sqcup G_2)^{ab}\to (G_1\times G_2)^{ab}\to G_1^{ab}\otimes G_2^{ab}$ is the identity, hence so is $G_1^{ab}\otimes G_2^{ab}\to(G_1\times G_2)^{ab}\to G_1^{ab}\otimes G_2^{ab}$. In particular, the epimorphism $e\colon G_1^{ab}\otimes G_2^{ab}\to (G_1\times G_2)^{ab}$ has a retraction $r\colon (G_1\times G_2)^{ab}\to G_1^{ab}\otimes G_2^{ab}$ and so is an isomorphism, i.e. $r\circ e=\mathrm{id}$ implies $e\circ r\circ e=e=\mathrm{id}\circ e$, and by $e$ being an epimorphism implies $e\circ r=\mathrm{id}$.
