Proving the universal propetry of the coproduct for groups in $\mathrm{Ab}$ This question is from Alufi's algebra text.

Show that if $G, H$ are abelian groups, then $G \times H$ satisfies
  the universal property for coproducts in $\mathrm{Ab}$.


My attempt:
Define $i_G:G \to G \times H$ and $i_H:H \to G \times H$ as follows:
$$i_G:g \mapsto (g,e_H)$$ 
$$i_H:h \mapsto (e_G,h)$$
We'll show that these injections together with $G \times H$ satisfy the universal property of the coproduct in Ab.
Let $\varphi_1 :G \to Z$ and $\varphi_2 :H \to Z$ be morphisms to an arbitrary abelian group Z (object in the category Ab).
We're looking for a morphism $\psi$ that satisfies the following equations:
$$\psi \circ i_G = \varphi_1$$
$$\psi \circ i_H = \varphi_2$$
In order for $\psi$ to be a morphism the following must hold for any $(g,h) \in G \times H$:
$$\psi(g,h) = \psi(ge_G,he_H) = \psi(g,e_H)\psi(e_G,h) = (\psi \circ i_G(g)) \cdot (\psi \circ i_H(h)) = \varphi_1(g) \cdot \varphi_2(h)$$
(a) We've showed that if the desired morphism exists it must be that $\psi = \varphi_1(g) \cdot \varphi_2(h)$ (in other words we’ve shown that if it exist it is unique).  
Now we show that it is indeed a morphism.
let $(g_1,h_1), (g_2,h_2) \in G \times H$. Using commutativity of $Z$:
$$\psi(g_1,h_1) \cdot \psi(g_2,h_2) = \varphi_1(g_1) \cdot \varphi_2(h_1) \cdot \varphi_1(g_2) \cdot \varphi_2(h_2) = \varphi_1(g_1) \cdot \varphi_1(g_2) \cdot \varphi_2(h_1) \cdot \varphi_2(h_2) = \varphi_1(g_1g_2) \cdot \varphi_2(h_1h_2) = \psi(g_1g_2,h_1h_2)$$
(b) Thus we proved it is a morphism in the category Ab.
By (a) and (b) there exist a unique morphism $\psi$ satisfying our requirements hence we conclude that $G \times H$ satisfies the universal property of the coproduct.  
Is that about right?
 A: It is indeed 'about right'. I should use additive notation in abelian groups and zeros for the identities, but that is not a necessity. My order would be:
0) Defining $\psi:G\times H\rightarrow Z$ by $\psi\left(g,h\right):=\varphi_{1}\left(g\right)+\varphi_{2}\left(h\right)$. Then to be shown:
1) $\psi$ is a morphism, i.e. $\psi\left(\left(g_{1},h_{1}\right)+\left(g_{2},h_{2}\right)\right)=\psi\left(g_{1},h_{1}\right)+\psi\left(g_{2},h_{2}\right)$
2) $\psi\circ i_{G}=\varphi_{1}$ and $\psi\circ i_{H}=\varphi_{2}$
, i.e. $\psi\circ i_{G}\left(g\right)=\psi\left(g,0_{H}\right)=\varphi_{1}\left(g\right)+\varphi_{2}\left(0_{H}\right)=\varphi_{1}\left(g\right)$
and $\psi\circ i_{H}\left(h\right)=\psi\left(0_{G},h\right)=\varphi_{1}\left(0_{G}\right)+\varphi_{2}\left(h\right)=\varphi_{2}\left(h\right)$
3) $\psi$ is unique, i.e. if $\psi':G\times H\rightarrow Z$ is a
morphism with $\psi'\circ i_{G}=\varphi_{1}$ and $\psi'\circ i_{H}=\varphi_{2}$
then $\psi=\psi'$. This by showing that $\psi'\left(g,h\right)=\varphi_{1}\left(g\right)+\varphi_{2}\left(h\right)$
