I am reading algebra chapter 0 and have two questions about the proof:
$\mathbf{z}^{\oplus n}$ is a free abelian group on A={1,...,n} .
Proof: we frist define a function $j:A\rightarrow $$\mathbf{z}^{\oplus n}$ by $j(i):=(0,...,0, \underset{i-th \ place}{1}$ $,0,...,0)$ $\in \mathbf{z}^{\oplus n}$, so every element of $\mathbf{z}^{\oplus n}$ can be written uniquely in the form $\sum_{i=1}^{n}m_i j(i)$.
Now let $f:A\rightarrow G$ be any fuction from $A={1,...,n}$ to an abelian group G.We define $\phi:\mathbf{z}^{\oplus n} \rightarrow G$ by $$\phi(\sum_{i=1}^{n}m_i j(i)):=\sum_{i=1}^{n}m_i f(i).$$to make the digram 
commute. And it's obviously that $\phi$ is a homomorphism.
There are my quetions:
1.$j$ is injective, for all $i\in A$,I can get $\phi(j(i))=f(i)$. But what's the conditon of the element in $\mathbf{z}^{\oplus n}$ and G but not the image of $i\in A$? I think they also must make $\phi$ is a homomorphism but needn't make the digram commute,for their preimages are not involved in this process.Right?
2.However, the operation in $G$ of this proof is "addition".Is the proof still right if the oepration in $G$ is multiplication? Is there a way to proof the claim that satisfy all commutative binary operation ?
The commutativity of the diagram just means that $\phi$, restricted to the elements of $A$, coincides with $f$ or, equivalently, that every group homomorphism from $A$ to a group $G$ extends (uniquely) to a group homomorphism from $\mathbb{Z}^{\oplus n}$, or, as sometimes one says, that every group homomorphism from $A$ to $G$ factors uniquely through $\mathbb{Z}^{\oplus n}$. Your assertion that the diagram is commutative only on the elements on $A$ just does not make sense, because $A$ is a part of that diagram, and commutativity means that you only have to consider elements of $A$. To have commutativity, you have to start from $A$ and reach $G$ following the two different paths you have, (the one through $\mathbb{Z}^{\oplus n}$ and that which goes directly to $G$) then check that the result is the same. What you can do "outside" $A$ is checking that, with your definition of $\phi$, the function $\phi$ is a group homomorphism, but this comes almost automatically from the definition itself.
As for the second question, a group is a set with just a single binary operation defined on it. You can call it addition or multiplication, and then use a coherent notation. Since the groups involved are abelian, that is they satisfy the commutative property, it is customary to use the additive notation with $+$. If you want, you can just shift to product notation on $G$, writing $f(i)^{m_i}$ in place of $m_i f(i)$ and $\prod$ in place of $\sum$.