Question on Proof of ismorphism between product and coproduct under Abelian group

In Steve Awodey's book Category Theory, he has some proof that seems wrong to me on page $$60$$, he mentioned a proposition that

Proposition $$3.11.$$In the category $$Ab$$ of abelian groups, there is a canonical isomorphism between the binary coproduct and product,

$$A+B \cong A \times B.$$

And his proof goes as follows,

Proof. To define an arrow $$\theta:A+B \to A \times B,$$ we need one $$A \to A \times B$$ (and one $$B \to A \times B$$), so we need arrows $$A \to A$$ and $$A \to B$$ (and $$B \to A$$ and $$B \to B$$). For these, we take $$1_A : A \to A$$ and the zero homomorphism $$0_B : A \to B$$ (and $$0_A:B \to A$$ and $$1_B : B \to B$$). Thus, all together, we get $$\theta = [\langle1_A,0_B\rangle,\langle0_A,1_B\rangle]:A+B \to A \times B.$$ Then given any $$(a,b) \in A+B,$$ we have $$\theta(a,b) = [\langle1_A,0_B\rangle,\langle0_A,1_B\rangle](a,b) \\ =\langle1_A,0_B\rangle(a)+\langle0_A,1_B\rangle (b) \\ = \langle1_A(a),0_B(a)\rangle+\langle0_A(b),1_B(b)\rangle \\ = (a,0_B) + (0_A,b) \\= (a+0_A,0_B+b)\\ = (a,b)$$

My question is that: where did he exactly use the assumption that the group $$A$$ and $$B$$ are abelian. He did not explicitly stated the fact, so I am a little bit suspicious about the proof.

And another thing, he leave the identity $$[f,g](a,b) = f(a)+_X g(b)$$ (page $$60$$) as an exercise. I did not get the way to prove it. If anyone has the proof, please share in the post. Thank you.

• @RghtHndSd How so? – Zack Ni Feb 8 '19 at 2:53
• I really don't like this proof either. When he says $(a,b) \in A + B$ he comes awfully close to just assuming that the coproduct is the product. – Chessanator Feb 8 '19 at 2:53
• @Chessanator Yeah I have to agree on it. But is there any better proof, though? – Zack Ni Feb 8 '19 at 2:54
• @RghtHndSd There's a canonical morphism from coproduct to product in any category with a zero, but it's not necessarily an isomorphism. In particular, the coproduct of non-abelian groups is not the product. – Chessanator Feb 8 '19 at 3:06

If I was proving this, I'd construct the inverse $$\varphi: A \times B \rightarrow A + B$$ to the morphism $$\theta$$ that Awodey gave by using the projections $$\pi_A: A \times B \rightarrow A$$, $$\pi_B: A \times B \rightarrow B$$ and the inclusions $$\iota_A:A \rightarrow A+B$$, $$\iota_B:B \rightarrow A+B$$.
I would define $$\varphi = \iota_A \pi_A + \iota_B \pi_B$$. (This is where abelian-ness is used. You can't define the addition of group homomorphisms like this unless the groups are commutative.)
You can check for yourself that this is an inverse to $$\theta$$.
Edited: I've also just realised that this helps with your second question about the exercise, about $$[f,g](a,b)$$. $$(a,b)$$ is an element of $$A \times B$$, but $$[f,g]$$ is a morphism out of $$A + B$$. Awodey has implicitly used the fact that they are isomorphic again.
In other words, there's an implied instance of $$\varphi$$ in the expression: it should be $$[f,g] \varphi (a,b)$$.