Showing that the product group of $G$ and $H$ satisfies the universal property for coproducts in the category of abelian groups $\mathbf{Ab}$

Question

I'm working on another problem of Aluffi's Algebra. Given the category $\mathbf{Ab}$ of abelian groups, the problem is to show that for any two groups $G$ and $H$ the product group $G\times H$ satisfies the universal property for coproducts in $\mathbf{Ab}$.

This amounts to showing that $G\times H$ is initial in the category $\mathbf{Ab}^{G,H}$. In other words we need to find a homomorphism $\sigma\in\text{Hom}(G\times H,K)$ and two suitable homomorphisms $\psi_G\in\text{Hom}(G,G\times H)$ and $\psi_H\in\text{Hom}(H,G\times H)$ such that for given $\varphi_G\in\text{Hom}(G,K)$ and $\varphi_H\in\text{Hom}(H,K)$ we have \begin{equation} \sigma\circ\psi_G=\varphi_G\\ \sigma\circ\psi_H=\varphi_H \end{equation} I think that defining \begin{equation} \psi_G(g):=(g,e_H)\\ \psi_H(h):=(e_G,h)\\ \sigma(g,h):=\varphi_G(g)\varphi_H(h) \end{equation} (where $e_G,e_H$ are the identities of $G$ and $H$ and the last product is in $K$) gives the homomorphism we want (checking that $\sigma$ is actually a group homomorphism requires commutativity, so this won't work in $\mathbf{Grp}$), but my problem is proving uniqueness of $\sigma$. Any hints?

Answer 1

The conditions $\sigma\circ\psi_{G}=\varphi_{G}$ and $\sigma\circ\psi_{H}=\varphi_{H}$ are determining for $\sigma$. This because:

$\sigma\left(g,h\right)=\sigma\left(\left(g,e_{H}\right)\left(e_{G},h\right)\right)=\sigma\left(g,e_{H}\right)\sigma\left(e_{G},h\right)=\sigma\circ\psi_{G}\left(g\right)\sigma\circ\psi_{H}\left(h\right)=\varphi_{G}\left(g\right)\varphi_{H}\left(h\right)$

Answer 2

Consider the two morphisms \begin{align} i&\colon G\to G\times H, &&i(x)=(x,1)\\ j&\colon H\to G\times H, &&j(y)=(1,y) \end{align} (what you called $\psi_G$ and $\psi_H$).

If $\alpha\colon G\to K$ and $\beta\colon H\to K$ are morphisms of abelian groups, then you can define $$ \sigma\colon G\times H\to K $$ by $$ \sigma(x,y)=\alpha(x)\beta(y) $$ and, of course, $\sigma\circ i=\alpha$, $\sigma\circ j=\beta$.

Since $G\times H$ is generated by the images of $i$ and $j$, because $(x,y)=i(x)j(y)$, the morphism $\sigma$ is unique, for we know what its action must be on a set of generators.